Current Electricity Notes JEE Main and Advanced

CURRENT ELECTRICITY

C H A P T E R

1 8

LEARNING OBJECTIVES

(i) Current through a given area of a conductor is the net charge passing per unit time through the area. Current is a scalar although we represent current with an arrow. Currents do not obey the law of vector addition. That current is a scalar also follows from it’s definition. The current I through an area of cross-section is given by the scalar product of two vectors:

I J. A  , where J and A are vectors.

Current density j gives the amount of charge flowing per second per unit area normal to the flow, Jnqv_d

where n is the number density (number per unit volume) of charge carriers each of charge q, and v_d is the drift velocity of the charge carriers. For electrons q = – e. If J is normal to a cross-sectional area A and is constant over the area, the magnitude of the current I through the area is nev_d A.

(ii) Ohm’s law: The electric current I flowing through a substance is proportional to the voltage V across its ends, i.e., V  I or V = RI, where R is called the resistance of the substance. The unit of resistance is ohm: 1 = 1 VA–1.

Homogeneous conductors like silver or semiconductors like pure germanium or germanium containing impurities obey Ohm’s law within some range of electric field values. If the field becomes too strong, there are departures from Ohm’s law in all cases. Equation E Janother statement of Ohm’s law, i.e., a conducting material obeys Ohm’s law when the resistivity of the material does not depend on the magnitude and direction of applied electric field.

(iii) The resistance R of a conductor depends on its length  and constant cross-sectional area A through the relation, R

A 

 , where , called resistivity is a property of the material and depends on temperature and pressure. Electrical resistivity of substances varies over a very wide range. Metals have low resistivity, in the range of 10–8  m to 10–6  m. Insulators like glass and rubber have 1022 to 1024 times greater resistivity. Semiconductors like Si and Ge lie roughly in the middle range of resistivity on a logarithmic scale.

(iv) In the temperature range in which resistivity increases linearly with temperature, the temperature coefficient of resistivity á is defined as the fractional increase in resistivity per unit increase in temperature.

(v) (a) Total resistance R of n resistors connected in series is given by R = R₁ + R₂ +... + R_n (b) Total resistance R of n resistors connected in parallel is given by

1 2 n

1 1 1 1

... R R R  R .

(vi) Kirchhoff’s Rules – (a) Junction Rule: At any junction of circuit elements, the sum of currents entering the junction must equal the sum of currents leaving it. (b) Loop Rule: The algebraic sum of changes in potential around any closed loop must be zero. Kirchhoff’s junction rule is based on conservation of charge and the outgoing currents add up and are equal to incoming current at a junction. Bending or reorienting the wire does not change the validity of Kirchhoff’s junction rule.

(vii) The Wheatstone bridge is an arrangement of four resistances – R₁, R₂, R₃, R₄. The null-point condition is given by

3 1 2 4 R R R  R

using which the value of one resistance can be determined, knowing the other three resistances.

(viii) The potentiometer is a device to compare potential differences. Since the method involves a condition of no current flow, the device can be used to measure potential difference; internal resistance of a cell and compare emf’s of two sources.

INTRODUCTION

If there is a life’s blood in our technology, it’s most assurely electricity–electricity coursing along wire veins, delivering power and information, trickling through a metal nervous system just as it tricklets through our own. The ordered flow of charge is called electric current, whether we’re talking about electrons propelled down a wire by a battery or protons hurled through space by an exploding star. And currents carry energy. Much of the energy we “consume” is delivered by electricity conveniently on tap at wall outlets everywhere, refrigerators, Computers, heart-lung machines, all plug into the electric stream and draw energy from it.

Electric circuits, while very useful, can also be hazardous. To reduce the danger inherent in using circuits, proper electrical grounding is necessary.

Appliances are connected to a wall socket via a three-prong plug that provides safe electrical grounding. The third prong connects the metal casing directly to a copper rod driven into the ground or to a copper water pipe that is in the ground. This arrangement protects against electrical shock in the event that a broken wire touches the metal casing, for charge would flow through the casing, through the third prong, and into the ground, returning eventually to the generator. No charge would flow through the person’s body, since the copper rod provides much less electrical resistance than does the body.

Serious and sometimes fatal injuries can result from electric shock. The severity of the injury depends on the magnitude of the current and the parts of the body through which the moving charges pass. The amount of current that causes a mild tingling sensation is about 0.001A. Currents on the order of 0.01-0.02A can lead to muscle spasms, in which a person “can’t let go” of the object causing the shock. Currents of approximately 0.2A are potentially fatal because they can make the heart fibrillate, or beat in an uncontrolled manner. Substantially larger currents stop the heart completely. However, since the heart often begins beating normally again after the current ceases, the larger currents can be less dangerous than the smaller currents that cause fibrillation. ELECTRIC CURRENT

The electric current through a section is defined as the rate of charge flow past the section. Thus, if q charge passes across a certain section in time t, then the current i through that section (fig.) is given by i q

t 

q For nonuniform flow to charge, the average current < i > over a certain time interval t, will be

q i t     

where q is the total charge flown in time t, and instantaneous current, will be

t 0 q dq i lim t dt       _{ }  

The SI unit of current is ampere, which consists of a charge flow of 1C per second. Thus, 1 A 1C/1s

Conventional direction of flow of current is opposite to the flow of electrons or in a conductor the direction of flow of positive charge gives the direction of current. Electric current is a scalar quantity.

If in a conductor n electrons pass in t second from any point them total charge passing through that point in t second is q = ne Current through the conductorI = q

t = ne

t Thus for 1 Ampere current n = 1

1 6 10. x 19

= 6.25 × 1018 electronsecond

Number of electron flowing in conductor in time t n = I x t e

For a given conductor current does not change with change in cross-sectional area. In the following figure i₁ = i₂

i1

i2

i1 = i2

Current due to translatory motion of charge : If n particle each having a charge q,

pass through a given area in time t then i nq t  + + + + + +

If n particle each having a charge q pass per second per unit area, the current associated with cross-sectional area A is, i = nqA If there are n particle per unit volume each having a charge q and moving with velocity v, the current thorough, cross section A is, i = nqvA.

Current due to rotatory motion of charge : If a point charge q is moving in a circle of radius r with speed v (frequency , angular speed  and time period T) then corresponding current

q qv q i qv T 2 r 2       

Current density (J)

Current density at any point inside a conductor defined as a vector having magnitude equal to current per unit area surrounding that point. Remember area is normal to the direction of charge flow (or current passes) through that point.

1. Current density is given by J di nˆ dA  

2 If the cross-sectional area is not normal to the current, but makes an angle  with the direction of current then di

J di JdA cos JdA i JdA

dA cos

      





 

 

3. _{If current density J} _{is uniform for a normal cross-section A} then J i A 

4. Current density J is a vector quantity. It’s direction is same as that of E . It’s S.I. unit is amp/m² and dimension [L–2_A]

5. In case of uniform flow of charge through a cross-section normal to it as i = nqvA J i A  = nqv

6. Current density relates with electric field as J  E E    

, where  = conductivity and  = resistivity or specific resistance of substance.

ELECTRIC CURRENT IN CONDUCTOR :

(1) Conductors are substances through which electric charges can flow easily.

(2) They are characterised by presence of a large number of free electrons (1029 electrons per m³).

(3) The number density of free electrons in a conductor is same throughout the conductor. This is because free electrons experience repulsive force between them and conductor allows movement of free electrons. Thus, free electrons are evenly scattered throughout the volume of conductor.

(4) These free electrons transport electric charge so are called as conduction electrons. (a) Behaviour of conductor in absence of applied potential difference :

(i) The free electrons present in a conductor gain energy from temperature of surrounding and move randomly in a conductor.

(ii) The speed gained by virtue of temperature is called as thermal speed of an electron.

2 rms 1 3 mv kT 2 2 so thermal speed rms 3kT v m

 where m is mass of electron At room temperature T = 300 K, v_rms = 105 m/sec.

(iii) The average distance travelled by a free electron between two consecutive collisions is called as mean free path . ( ~ 10 A°) mean free path total distance travelled

number of collisions  

(iv) The time taken by an electron between two successive

collisions is called as relaxation time . ( ~ 10–14 s) e–

Random motion of electrons

relaxation time total time taken number of collisions  

(v) The thermal speed can be written as vT

 



(v) In absence of applied potential difference electrons have random motion. The average displacement and average velocity is zero. There is no flow of current due to thermal motion of free electrons in a conductor.

(b) Behaviour of conductor in presence of applied potential difference : (i) When two ends of a conductors are joined to a battery then one end is at higher potential and another at lower potential. This produces an electric field inside the conductor from point of higher to lower potential i.e. E = V/L.

- - - -- - - -L HP LP V E

(ii) The field exerts an electric force on free electrons causing acceleration of each electron. F m a e E       so acceleration a e E m   _ 

(iii) Drift velocity is the average uniform velocity acquired by free electrons inside a metal by the application of an electric field which is responsible for current through it. Drift velocity is very small it is of the order of 10–4 m/s as compared to thermal speed (105m/s) of electrons at room temperature.

Using _v_{u }_{a t} we have _vd e E m



 _

  (vd ~ 10–4 m/sec)

(iv) The direction of drift velocity for electrons in a metal is opposite to that of applied field E. (5) Relation between current and drift velocity

Let n be number density of free electrons and A be area of cross-section of conductor. Number of free electrons in conductor of length L = nAL

Total charge on these free electrons q = neAL

Time taken by drifting electrons to cross conductor t = L/v_d

 Current d d v q I neAL neAv t L      or I = neAvd.

The current flowing through a conductor is directly proportional to the drift velocity (I v_d)

(6) The current density d

eE ne² J nev ne E A m m         _{ }  _{ }     s o J  E or J =  E where 2 ne m 

  is specific conductivity of conductor which depends on temperature and nature of material.

J E

 

  is microscopic form of ohm’s law..

(7) The drift velocity depends on nature of metal through , applied potential difference, length of conductor.

d

eE eV v

m mL

   

(8) When a steady current flows through a conductor of non-uniform section drift velocity varies inversely with area of cross-section d 1 v A       

(9) If diameter (d) of a conductor is doubled, then drift velocity of electrons inside it will not change.

(10) The rise of temperature causes increase in v_rms and hence a decrease in  and relaxation time  causing a decrease in drift velocity. (11) Mobility of a charge carrier is defined as drift velocity acquired per unit electric field.

Mobility µ vd e e m₂

E m mne ne

  

   

The unit is m2 V–1 s–1 and dimensions are M–1 T2 A1

The mobility depends on applied potential difference, length of conductor, number density of charge carriers, current in conductor, area of cross-section of conductor.

Example 1 :

A typical wire for laboratory experiments is made of copper and has a radius 0.815 mm. Calculate the drift velocity of electrons in such a wire carrying a current of 1A, assuming one free electron per atom.

Sol. Equation I Q qnAv_d t



 

 relates the drift velocity to the number density of charge carriers, which equals the number density of copper atoms n_a. We can find n_a from the mass density of copper, its molecular mass, and Avogadro’s number.

The drift velocity is related to the current and number density of charge carries : d

I v

nqA 

The number density of atoms n_a is related to the mass density _m. Avogadro’s number N_A, and the molar mass M. For copper r =  = 8.93 g/cm3 and M = 63.5 g/mol 3 23 m A a N (8.93 g / cm ) (6.02 10 atoms / mol) n M 63.5 g / mol     22 3 28 3 a n 8.47 10 atoms / cm 8.47 10 atoms / m  The magnitude of the charge is e, and the area is q = e related to the radius r of the wire A = r2

Substituting numerical values yields v_d : d ₂ a 1 1 v nqA _{n e r}    5 d _{28 3 19 2} 1 C / s v 3.54 10 m / s (8.47 10 m ) (1.6 10 C) (0.000815m)          Example 2 :

In a certain particle accelerator, a current of 0.5mA is carried by a 5 MeV proton beam that has a radius of 1.5 mm. (a) Find the number density of protons in the heat. (b) In the beam hits a target, how many protons hit the target in 1s ?

Sol. (a) The number density is related to the current, charge, cross-sectional area, and speed : n I qAv  We find the speed of the protons from their kinetic energy :

19 2 6 13 1 1.6 10 J K mv 5 MeV 5 10 eV 8 10 J 2 1eV          

Use m = 1.67 × 10–27 _{kg for the mass of a proton, and solve for the speed :} 13 7 27 2K (2) (8 10 J) v 3.10 10 m / s m 1.67 10 kg         Substitute to calculate n : 3 19 3 2 7 I 0.5 10 A n qAv (1.6 10 C / proton) (1.5 10 m) (3.10 10 m / s)           n = 1.43 × 1013 proton/m³

(b) The number of protons that hit the target in 1s is related to the total charge Q that hits in 1s and the proton charge q : Q

N q  

The charge Q that strikes the target in some time t is the current times the time : Q t (0.5mA) (1s) 0.5 mC

    

The number of protons is then :

3 15 19 Q 0.5 10 C N 3.13 10 protons q 1.6 10 C / proton          Check the result :

The number of protons hitting the target in time t is also the number in the volume Av t . Then

NnAv t . Substituting n = (I/qAv) then gives NnAv t (1 / qAv) (Av) t  Q / q, which is what we used in part (b) Example 3 :

The current density at a point is j 2 104 A₂ j m

 

_  _

 

 

. Find the rate of charge flow through a cross-sectional area S such that

(i) 2

S(2cm ) j, (ii) 2

S(4cm ) i and _{ˆ ˆ} 2

S(2i3j cm ) Sol. The rate of charge flow = current

(i) Current i(2 10 A / m ) j.(2 10 4 2 ˆ  4m ) j2 ˆ2A [Using j.j 1]ˆ ˆ (ii) Current i(2 10 A / m ) j.(4 cm ) i 4 2 ˆ 2 ˆ0 [Using j.iˆ ˆ0] (iii) Current i(2 10 A / m ) j.(2i 4 2 ˆ ˆ3 j) 10ˆ  4m26A

Example 4 :

Two boys A and B are sitting at two points in a field. Both boys are sitting near assemblence of charged balls each carrying charge +3e. A throws 100 balls per second towards B while B throws 50 balls per second towards A. Find the current at the mid point of A and B.

Sol. Let mid point be C as shown

Charge moving to the right per unit time = 100 × 3e = 300e

Charge moving to the left per unit time = 50 × 3e = 150e A C B

100e

50e

Movement of charge per unit time is 300e –150e = 150e towards right I = 150e = 150  1.6  10–19 A = 2.4 × 10–17 A.

TRY IT YOURSELF

Q.1 Current of 4.8 amperes is flowing through a conductor. The number of electrons per second will be – (A) 3 × 1019 (b) 7.68 × 1021 (c) 7.68 × 1020 (d) 3 × 1020

Q.2 When the current i is flowing through a conductor, the drift velocity is v. If 2i current is flowed through the same metal but having double the area of cross-section, then the drift velocity will be –

(A) v/4 (b) v/2 (c) v (d) 4v

Q.3 Every atom makes one free electron in copper. If 1.1 ampere current is flowing in the wire of copper having 1mm diameter, then the drift velocity (approx.) will be (Density of copper = 9 × 103 kg m–3 and atomic weight = 63)

(A) 0.3 mm/sec. (b) 0.1 mm/sec. (c) 0.2 mm/sec. (d) 0.2 cm/sec.

Q.4 In hydrogen atom, the electron makes 6.6 × 1015 revolutions per second around the nucleus in an orbit of radius 0.5 × 10–10 m. It is equivalent to a current nearly –

(A) 1A (b) 1 mA (c) 1 µA (d) 1.6 × 10–19 A

Q.5 In a neon discharge tube 2.9 × 1018 Ne+ ions move to the right each second while 1.2 × 1018 electrons move to the left per second. Electron charge is 1.6 × 10–19 C. The current in the discharge tube –

(A) 1A towards right (b) 0.66 A towards right (C) 0.66 A towards left (d) zero

Q.6 A non-conducting ring of radius R has two positive point charges lying diametrically opposite to each other, each of magnitude Q. The ring rotates with an angular velocity . If I is the equivalent current then,

(A) I = Q (B) I = 2Q/ (C) I = Q/ (D) i = zero

Q.7 A conductor of area of cross-section A having charge carriers, each having a charge q is subjected to a potential V. The number density of charge carriers in the conductor is n and the charge carriers along with their random motion are moving with average velocity v. A current I flows in the conductor. If j is the current density, then:

(A) | j | nqV, in the direction of charge flow.. (B) | j | nqV, in the direction opposite to charge flow.. (C) | j | nqV, in the direction perpendicular to charge flow.. (D) | j | nqV, in the direction of charge flow..

Q.8 The electron drift speed is small and the charge of the electron is also small but still, we obtain a large current in a conductor. This is due to:

(A) the conducting property of the conductor (B) the small resistance of the conductor

(C) the small electron number density of the conductor (D) the enormous electron number density of the conductor

ANSWERS

(1) (A) (2) (C) (3) (B) (4) (B)

(5) (B) (6) (C) (7) (D) (8) (D)

RESISTANCE :

1. When current flows in a conductor then conductor opposes the flow. Due to the applied electric field the flow of free electrons is obstructed due to temperature effect resulting in irregular velocity and because of collision with ions. This property of conductor is called its Resistance.

2. The value of resistance of a conductor is equal to the ratio of potential difference to the current.

of the conductor is : R V I     _{ }

4. Unit of resistance is ohm and Ohm = Volt@Ampere The symbol of ohm is .

5. If potential difference between the ends of the conductor is 1 volt and 1 ampere current flows through it then resistance of conductor is given by 1 ohm.

6. International ohm is defined has resistance of mercury column of length 106.3 cm, 1 mm square area of cross-section and mass 14.4521 gram at 0°C.

1 International ohm = 1.0048 ohm

7. 1 ohm= 1 volt / 1 ampere = 108 e.m.u. potential difference / 10–1 e.m.u. current or 1 ohm = 109 e.m.u. resistance

units and dimension of resistance are following &

ohm = volt / ampere= joule / coulomb/ ampere = (newton × metre) / (ampere × second) / ampere = kilogram × metre2 × second–3 ampere2

Dimension of resistance = ML2 T–3 A–2 8. Resistance of a conductor depends on

(a) Directly proportional to length (L) of the conductor, that is R  L

(b) Inversely proportional to the area of cross-section of the conductor, that is R  1

A or R  1 r2

 , where r is the radius of the conductor.

(c) Nature of material of the conductor.

(d) Temperature of the conductor. Resistance of metal conductor increases with increase in temperature.

9. Thus R  L A or R =  L A    

  where  is a constant called as 'specific resistance' or 'resistivity'.

  = RA L

 

 

 

If conductor is in the form of a wire of radius r, then  = R

2 r L _     

10. Specific resistance of a material is equal to the resistance of unit length of that material with unit area of cross-section and current perpendicular to the cross-section or specific resistance of a material is equal to the resistance of a cube of that material with a side of unit length and current perpendicular to the cross-section.

11. If R is the resistance of the conducting wire, R = V I = m ne A  2_ = 2 m ne      _ A       

If mean free path of electron is  and its root mean square velocity is v_rms , then  =

rms v   R = 2rms m v ne A 

Resistivity of the conductor  = RA  = 2

m

ne  and conductivity of the conductor

2 1 ne m     

12. For different substances their resistivity is also different e.g. _silver min imum1.6 10 8-mand

16 fused quartz maximum 10 m

    

insulator alloy semi conductor conductor

(Maximum for fused quartz) (Minimum for silver) 

      

13. On bending a straight wire from one or more points does not change its resistance, because the value of resistance of a wire depends on number of free electrons (n) and relaxation time () which does not change on bending the wire.

Effect of temperature on Resistance and Resistivity %

1. Resistance of a conductor depends on temperature. On increasing temperature random velocity of free electron increases. If the number of charge carrying electrons remain constant as in conductors then by increase in random velocity, resistivity increases.

In metals change in resistance with temperature is given by following relation : R_t = R₀ (1 + t + t2)

Where R_t and R₀ are resistances respectively at t°C and 0°C and &  are constants. Value of  is very small and usually it is

neglected. Thus R_t = R₀ (1 + t) or  = t 0

0

R R R x t



Constant  is called the temperature coefficient of resistance of the material. 2. If R₀ = 1 ohm] t = 1°C then  = (R_t – R₀)

Thus on increasing the temperature of a conductor of 1ohm resistance by 1°C increase in its resistance is equal to the temperature coefficient of resistance of the material of conductor. The value of temperature coefficient of resistance may be positive or negative.

3. On the base of calculations it is found that for most metals value of  is approximately 1/273 per °C.

Thus putting the value of in above equation

R_t = R₀ 1 273 

F

HG

I

KJ

F

HG

I

KJ

F

HG

I

KJ

Where T is the absolute temperature of the conductor. Thus R_t  T

Means resistance of a pure metal conductor is directly proportional to its absolute temperature. 4. Straight line graph is obtained between R_t and temperature t.

5. The value of resistivity or specific resistance changes with temperature. This change is due to change in resistance of the conductor with temperature. Dependence of resistivity on temperature is given by following equation : _t = ₀ (1 + t)

6. Resistivity or specific resistance of pure metal increases with increase in temperature and of insulator and semiconductor decreases with increase in temperature. For alloys it increases with increase in temperature but less than metals.

Variation of resistivity with temperature for metal, semiconductor and super conductor is shown in figure :

Temperature Metal  Temperature semiconductor  Temperature superconductor Tc 

7. For applying pressure on pure metals, their resistivity decreases but on applying tension force resistivity increases. Variation of resistance of some electrical material with temperature

Material Temp. coefficient resistance () Variation of resistance with temp. rise

Metals Positive Increases

Solid non-metal Zero Independent

Semi-conductor Negative Decreases

Electrolyte Negative Decreases

Ionised gases Negative Decreases

Alloys Small positive value Almost constant

Stretching of wire :

If a conducting wire stretches, it’s length increases, area of cross-section decreases so resistance increases but volume remain constant.

Suppose for a conducting wire before stretching it’s length =  , area of cross-section = A₁ A₁,

radius = r₁, diameter = d₁, and resistance 1 1 1

R A  

and resistance ₂ 2

2

R A  

Ratio of resistances before and after stretching

2 2 4 4 1 1 2 1 2 2 2 2 2 1 2 1 1 1 R A A r d R A A r d           _{ } _{ } _{ } _{ }          

  [Volume remain same V = AA11 = A22]

1. If length is given then

2 2 1 1 2 2 R R R     _{  }     

2. If radius is given then

4 1 2 4 2 1 R r 1 R R r r     _{  }   Colour coding of resistance

To know the value of resistance colour code is used. These code are printed in form of set of rings of strips. By reading the values of colour bands, we can estimate the value of resistance. The carbon resistance has normally four coloured rings or bands say A, B, C and D as shown in following figure.

A B C D

Colour band A and B : Indicate the first two significant figures of resistance in ohm.

Band C : Indicates the decimal multiplier i.e. the number of zeros that follows the two significant figures A and B.

D : Indicates the tolerance in percent about the indicated value or in other words it represents the percentage accuracy of the indicated value.

The tolerance in the case of gold is ±5% and in silver is ±10%. If only three bands are marked on carbon resistance, then it indicate a tolerance of 20%.

Colour code for carbon resistance :

Letters as an Colour figure Multiplier

aid to memory (A,B) C

B Black 0 100 B Brown 1 101 R Red 2 102 O Orange 3 103 Y Yellow 4 104 G Green 5 105 B Blue 6 106 V Violet 7 107 G Grey 8 108 W White 9 109

To remember the sequence of colour code following sentence should kept in memory. B B R O Y Great Britain Very Good Wife

OHM’S LAW

It the physical condition of the conductor (length, temperature, mechanical strain etc.) remains some, then the current flowing through the conductor is directly proportional to the potential difference across it’s two ends i.e. iVViR, where R is a proportionality constant, known as electric resistance.

1. Ohm’s law is not a universal law, the substances, which obey ohm’s law are known as ohmic substance. 2. Graph between V and i for a metallic conductor is a

straight line as shown. At different temperature V-i curves are different.

V i V i 1 2 1 2 T₁ T₂

Here tan > tan So R > R i.e., T > T

1 2

1 2

3. The device or substances which don’t obey ohm’s law e.g. gases, crys-tal rectifiers, thermionic valve, transistors etc. are known as non-ohmic or non-linear conductors. For these V-i curve is not linear.

V i Static resistance, st V 1 R i tan  

 , Dynamic resistance dyn

V 1 R i tan      Example 5 :

A conductor has its resistance of 100 ohm at 0°C, and its temperature coefficient of resistance is 0.008 per °C at 20°C. Then its resistance at 60°C is nearly –

(A) 132 (B) 148 (C) 157 (D) none of the above

Sol. (C). From R_T_T R₀₀ we get, ₂₀ 0 0 1 20       . This gives 0 0.0095 Thus R₆₀R (1₀   ₀ 60)100 (0.0095 60) 157 ohm Derivation of relation 0 T1 0 1 1 T      We know RT₁R (10  0 1T ) ... (1) T₂ 0 0 2 R R (1  T ) ... (2) Also RT₂ RT₁(1 T₁(T2T ))1 or RT₂RT₁ RT₁T₁(T2T )1 ... (3) From (1) and (2), T₂ T₁ 0 0 2 1 R R R  (T T ) ... (4) Using (3) and (4) we get, RT₁T₁R00

Now using (1) we get (1 0 1T )T₁ 0

Example 6 :

The current i through a non-ohmic conductor varies with voltage V as i   ( V V ) A2 , where  and  are positive constants and V is in volt. Find the dimensions of  and the resistance (dynamic) of the conductor for i = i₀.

Sol. From, the principle of homogeneity of dimensions, we have [i] [ V ]2 Here, V = potential difference =

2 2 3 1 Work [MLT ] [L] [ML T A ] Ch arg e [AT]      2 3 1 2 [A] [ML T A ]     or  [M2L T A ]4 6 3 Given, that current 2

i   V V di 2 V dV      At i = i₀ let V = V₀  i₀  V₀ V₀2 or 2 0 0 4 i V 2       

Since, negative V₀ is meaningless,  0





1

V 4 i

2

     

  The dynamic resistance of the non-ohmic conductor is

d

2

V V0 0

1 1 1

R

slope of i versus V graph (di / dV)  _{4 i}

  

Example 7 :

A wire has a resistance of 16.0 ohm. It is melted and drawn to a wire half its initial length. What will be the new resistance of the wire ?

Sol. Here, the factor by which the length is changed is n 1 2    



The new resistance R’ is given by

2 2 1 R R (n ) 16 4 2         _{ }   Example 8 :

The resistivity of silver at 0°C is 1.6 × 10–8 -m. If its temperature coefficient of resistance is 4.1 × 10–3 °C–1, find the resistivity of silver at 80°C.

Sol. Here, ₀ 1.6 × 10–8 _{-m, 4.1 × 10}–3 _°C–1 _{and t = 80°C}

Using   ₀[1  ( t)], the resistivity  at a temperature 80°C will be

8 3 1 8

1.6 10 m [1 (4.1 10 C ) (80 C)] 2.1 10 m

            

TRY IT YOURSELF

Q.1 If a 0.1% increase in length due to stretching, the percentage increase in its resistance will be –

(A) 0.2% (B) 2% (C) 1% (D) 0.1%

Q.2 The resistivity of iron is 1 × 10–7 ohm-m. The resistance of a iron wire of particular length and thickness is 1 ohm. If the length and the diameter of wire both are doubled, then the resistivity in ohm-m will be –

(A) 1 × 10–7 (B) 2 × 10–7 (C) 4 × 10–7 (D) 8 × 10–7

Q.3 The temperature coefficient of resistance for a wire is 0.00125/°C. At 300 K its resistance is 1ohm. The temperature at which the resistance becomes 2 ohm is –

(A) 1154 K (B) 1100 K (C) 1400 K (D) 1127 K

Q.4 The resistance of a wire is 20 ohms. It is so stretched that the length becomes three times, then the new resistance of the wire will be –

(A) 6.67 ohms (B) 60.0 ohms (C) 120 ohms (D) 180.0 ohms

Q.5 On increasing the temperature of a conductor, its resistance increases because –

(A) Relaxation time decreases (B) Mass of the electrons increases (C) Electron density decreases (D) None of the above

Q.6 The specific resistance of a wire is , its volume is 3m³ and its resistance is 3 ohms, then its length will be–

(A) 1 /  (B) 3 /  (C) 1_ 3 (D) 1

3 

Q.7 A piece of wire of resistance 4 ohms is bent through 180° at its mid point and the two halves are twisted together, then the resistance is –

(A) 8 ohms (B) 1 ohm (C) 2 ohms (D) 5 ohms

Q.8 Dimensions of a block are 1cm × 1cm × 100cm. If specific resistance of its material is 3 × 10–7 ohm-m, then the resistance between the opposite rectangular faces is –

(A) 3 × 10–9 ohm (B) 3 × 10–7 ohm (C) 3 × 10–5 ohm (D) 3 × 10–3 ohm Q.9 In the above question, the resistance between the square faces is –

(A) 3 × 10–9 ohm (B) 3 × 10–7 ohm (C) 3 × 10–5 ohm (D) 3 × 10–3 ohm

Q.10 Resistance of tungsten wire at 150°C is 133 . Its resistance temperature coefficient is 0.0045/°C. The resistance of the wire at 500°C will be –

(A) 180  (B) 225  (C) 258  (D) 317 

Q.11At what temperature will the resistance of a copper wire become three times its value at 0°C (Temperature coefficient of resistance for copper = 4 × 10–3 per °C)

(A) 400°C (B) 450°C (C) 500°C (D) 550°C

ANSWERS

(1) (A) (2) (A) (3) (D) (4) (B)

(5) (A) (6) (B) (7) (B) (8) (B)

POTENTIAL DIFFERENCE EMF and Batteries

To maintain a steady current in a conductor, we need a constant supply of electric energy. A device that supplies electrical energy is called a source of emf (The letters emf stand for electromotive force, Examples of emf sources are a battery, which converts chemical energy into electrical energy, and a generator, which converts mechanical energy into electrical energy. A source of emf does work on the charge passing through it, raising the potential energy of the charge. The work per unit charge is called the emf, E, of the source. The unit of emf is the volt, the same as the unit of potential difference. An ideal battery is a source of emf that maintains a constant potential difference between its two terminals, independent of the rate of flow of charge between them. The potential difference between the terminals of an ideal battery is equal in magnitude to the emf of the battery.

Figure shows a simple circuit consisting of a resistance R connected to an ideal battery. The straight lines indicate connecting wires of negligible resis-tance. The source of emf maintains a constant potential difference equal to E between points a and b, with point a being at the higher potential. There is negligible potential difference between points a and c or between points d and b because the connecting wire is assumed to have negligible resistance.

The potential difference from c and d is therefore equal in magnitude to the emf E, b d

a c + E – I R

and the current in the resistor is given by I = E/R . The direction of the current in this circuit s clockwise, as shown in the figure. Note that inside the source of emf, the charge flows from a region of low potential to a region of high potential, so it gains potential energy. When charge Q flows through the source of emf E, its potential energy is increased by amount QE . The charge then flows through the resistor, where this potential energy is converted into thermal energy. The rate at which energy is supplied by the source of emf is the power output : P QE EI

t 

 



In the simple circuit of figure, the power put out by the source of emf equals that dissipated in the resistor.

A source of emf can be thought of as a charge pump that pumps the charge from a region of low electrical potential energy to a region of high electrical potential energy. In a real battery, the potential difference across the battery terminals, called the terminal voltage, is not simply equal to the emf of the battery.

Consider the circuit consisting of a real battery and a resistor in fig. If the current is varied by varying the resistance R and the terminal voltage is mea-sured, the terminal voltage is found to decrease slightly as the current in-creases, just as if there were a small resistance within the battery.

Thus, we can consider a real battery of consist of an ideal battery of emf E plus

I V

E

a small resistance r, called the internal resistance of the battery. The internal resistance of cell depends on.

(1) Distance between electrodes (r  d) larger is the separation between electrodes more is the length of electrolyte through which ions have to move so more is internal resistance.

(2) Conductivity or nature of electrolyte (r  1/) (3) Concentration of electrolyte (r  c)

(4) Temperature of electrolyte (r  1/T)

(5) Nature and area of electrodes dipped in electrolyte (r  1/A) Terminal Potential Difference :

The potential difference between the two electrodes of a cell in a closed circuit i.e. when current is being drawn from the cell is called terminal potential difference.

(a) When cell is discharging :

When cell is discharging current inside the cell is from cathode to anode.

Current E r R    or E = R + r = V + r or V = E – r _R r I E I

When current is drawn from the cell potential difference is less than emf of cell. Greater is the current drawn from the cell smaller is the terminal voltage. When a large current is drawn from a cell its terminal voltage is reduced.

(b) When cell is charging :

When cells is charging current inside the cell is from anode to cathode.

Current V E r    or V = E + r V r I E I + –

(c) When cell is in open circuit : In open circuit R =   I E 0

R r

 

 So V = E

In open circuit terminal potential difference is equal to emf and is the maximum potential difference which a cell can provide. (d) When cell is short circuited :

In short circuit R = 0 so I E E

R r r

 

 and V = IR = 0

In short circuit current from cell is maximum and terminal potential difference is zero. (e) Power transferred to load by cell :

2 2 2 E R P R (r R)     s o P = P_max if dP 0 dR  r 4 E P 2 ma x r = R R P P = P_maxif r = R

Power transferred by cell to load is maximum when r = R and P_max =

2 2

E E

4r 4R KIRCHHOFF’S RULE

1. When any closed-circuit loop is traversed, the algebraic sum of the changes in potential must be equal to zero.

2. At any junction point in a circuit where the current can divide, the sum of the currents into the junction must equal the sum of the currents out of the junction.

Kirchhoff’s first rule, called the loop rule, follows directly from the conservation of energy. If we have a charge q at some point where the potential is V, the potential energy of the charge is qV. As the charge traverses a loop in a circuit, it loses or gains energy as it passes through resistors, batteries, or other devices, but when it arrives back at its starting point, its energy must again be qV. That is, the net change in the potential must be zero.

Kirchhoff’s second rule, called the junction rule, follows from the conserva-tion of charge. Figure shows the juncconserva-tion of three wires carrying currents I₁, I₂ and I₃. Since charge does not originate or accumulate at this point, the conser-vation of charge implies the junction rule, which for this cases gives

I₁ = I₂ + I₃ a I₁ I₂ I3 SINGLE-LOOP CIRCUITS

As an example of using Kirchhoff’s loop rule, consider the circuit shown in fig. containing two batteries with internal resistances r₁ and r₂ and three external resistors. We wish to find the current in terms of the emfs.

r2 + E2 – R₂ b – + c d + – e r1 + E1 – + – g f a _{+ –} I + – Battery 1 Battery 2 Changes in potential a b Drop IR b 1 2 2 2 3 1 c Drop IR c d Drop E d e Drop Ir e f Drop IR f g Increase E

Assume that I is clockwise, as indicated in figure, and apply Kirchhoff’s loop rule as we traverse the circuit in the assumed direction of the current, beginning at point a. The potential decreases and increases are given in the figure. Note that we encounter a potential difference as we traverse the source of emf between points c and d and a potential increase as we traverse the source of emf between f and g. Beginning at point a, we obtain from Kirchhoff’s loop rule.

1 12 2 2 3 1 1

IR IR E Ir IR E Ir 0

       

Note : You can use opposite sign convention also. Solving for the current I, we obtain

1 2 1 2 3 1 2 E E I R R R r r      

If E₂ is greater than E₁, we get a negative value for the current I, indicating that we have assumed the wrong direction for I. MULTI-LOOP CIRCUIT

To analyze circuits containing more than one loop, we need to use both of Kirchhoff’s rules, with Kirchhoff’s junction rule applied to points where the current splits into two or more parts.

The general methods for the analysis of multiloop circuits : 1. Draw a sketch of the circuit.

2. Choose a direction for the current in each branch of the circuit, and label the currents in the circuit diagram. Add plus and minus signs to indicate the high and low potential sides of each resistors, capacitor, or source of emf.

3. Replace any combination of resistors in series or parallel with its equivalent resistance. 4. Apply the junction rule to each junction where the current divides.

5. Apply the loop rule to each loop until you obtain as many equations as unknowns. 6. Solve the equations to obtain the values of the unknowns

7. Check your results by assigning a potential of zero to one point in the circuit and use the values of the currents found to determine the potentials at other points in the circuit.

Example 9 :

Suppose the elements in the circuit in figure have the values E₁ = 12V, E₂ = 4V, r₁ = r₂ = 1, R₁ = R₂ = 5, and R₃ = 4, as shown in figure. (a) Find the potentials at points a through g in the figure, assuming that the potential at point f is zero. (b) Find the power input and output in the circuit. 1 + 4V – 5 b – + c d + – 0V 1 + 12V _– + – g f a _{+ –} I + –

Sol. Picture the problem : To find the potential difference, we first need to find the current I in the circuit. The voltage drop across each resistor is then IR. To discuss the energy balance, we calculate the power into or out of each element using Equations, P = VI and P = V²/R

(a) 1. The current I in the circuit is found using equation

1 2 1 2 3 1 2 E E I R R R r r       12V 4V 8V I 0.5 A 5 5 4 1 1 16              

2. We now find the potential at each labeled point in the circuit :

g f 1 V V E  0 12V12V a g 1 V V Ir 12V (0.5A) (1 )  11.5V b a 1 V V IR 11.5V (0.5A) (5 )  9V c b 2 V V IR 9V (0.5A) (5 )  6.5V d c 2 V V E 6.5V4V2.5V e d 2 V V Ir 2.5V (0.5A) (1 )  2.0V f e 3 V V IR 2.0V (0.5A) (4 )  0

(b) 1. First, calculate the power delivered by the emf source E₁ :

E₁ 1

P E I(12V) (0.5A)6W

2. Part of this power is dissipated in the resistors, both internal and external :

2 2 2 2 2

R 1 2 3 1 2

2 R

P (0.5A) (5         5 4 1 1 ) 4.0 W 3. The remaining 2W of power goes into charging battery 2 :

E₂ 2

P E I(4V) (0.5A)2W Example 10 :

Calculate the currents I₁, I₂ and I₃ in the circuit shown in figure. Sol.

Junction rule at C yields

I₁ + I₂ – I₃ = 0 i.e., I₁ + I₂ = I₃ ....(1) while loop for meshes a and b yields respectively :

–14 – 4I₂ + 6I₁ – 10 = 0

i.e., 3I₁ – 2I₂ = 12 ....(2) and, 10 – 6I₁ – 2I₃ = 0

i.e., 3I₁ + I₃ = 5 ....(3) Substituting I₃ from Equation (1) in (3)

4I₁ + I₂ = 5

Solving equations (2) and (4) for I₁ and I₂, we find I₁ = 2A and I₂ = –3A And hence equation (1) yields, I₃ = –1A

The fact that I₂ and I₃ are negative implies that actual direction of I₂ and I₃ are opposite to that shown in the circuit.

You can solve above problems using nodal analysis. (Set zero potential reference of any junction, mark potential of other junctions and use Kirchhoff’s first rule.]

COMBINATION OF CELL

In series grouping of cell’s their emf’s are additive or subtractive while their internal resistances are always additive. If dissimilar plates of cells are connected together their emf’s are added to each other while if their similar plates are connected together their emf’s are subtractive.

(A) Series grouping

In series grouping anode of one cell is connected to cathode of other cell and so on. In identical cells are connected in series

R

r _E r _E r

E I

1. Equivalent emf of the combination E_eq = nE 2. Equivalent internal resistance r_eq = nr

3. Main current = Current from each cell = i nE R nr 



4. If out of n cells in series m cells be connected in an opposite manner, then the current will given by

(n m) m i j i 1 j 1 n i i 1 E E i r R               







5. Potential difference across external resistance V = iR 6. Potential difference across each cell V' = V/n 7. Power dissipated in the external circuit =

2 nE .R R nr       

8. Condition for maximum power R = nr and

2 max E P n 4r    _{ }   dP

0, Maximum power transfer theorem dR

 



 

 

9. Total power consumed in the circuit E²/(R + r) [and not E²R/(R + r)²] and will be maximum (= E²/r) when R = min = 0 with I = (E/r) = max. If R = r, P = (E²/2r) with I = (E/2r)

10. It is a common misconception that “current in the circuit will be maximum when power consumed by the load is maximum.” Actually current I = E/(R + r) is maximum (= E/r) when R = min = 0 with P_L = (E/r)2× 0 = 0 = min. while power consumed by the load E²R/(R + r)2 is maximum (E2/4r) when R = r and I = E/2r  max (= E/r)

(B) Parallel grouping

In parallel grouping all anodes are connected at one point and all cathode are connected together at other point. If n identical cells are connected in parallel. 1. Equivalent emf E_eq = E

2. Equivalent internal resistance R_eq = r/n 3. Main current i E R r / n   r r r R E E E I

4. Potential difference across external resistance = p.d. across each cell = V = iR 5. Current from each cell i i

n  

6. Power dissipated in the circuit

2 E P .R R r / n    _ _ 

7. Condition for max. power is R = r/n and

2 max E P n 4r    _{ }   8. This type of combination is used when nr >> R.

Dissimilar cells in parallel

Let two cells of emf E₁ and E₂ and internal resistances r₁ and r₂ be connected in parallel to an external resistance R.

Applying loop rule to loop d m n c d – IR –₁ r₁ + E₁ = 0 ...(1) Applying loop rule to loop a m n b a – IR –₂ r₂ + E₂ = 0 ...(2)

R m n 2E2 r2 r₁ E₁ 1 d c b a

Applying first rule at junction  I = I₁ + I₂.

Multiply equation 1 by r₂ and eqn. 2 by r₁ and put I₂ = I – I₁ we get –IR r₂ – I₁ r₁r₂ + E₁ r₂ = 0 and –IR r₁ – (I–I₁) r₂r₁ + E₂ r₁ = 0

On adding these we get, E₁ r₂ + E₂ r₁ = IR (r₁ + r₂) + Ir₁ r₂ = I (r₁ + r₂) 1 2

1 2 r r R r r         or 1 2 2 1 1 2 eq 1 2 eq 1 2 E (E r E r ) / (r r ) I r r R r R r r        equivalent cell r_eq Eeq R

Two dissimilar cells in parallel are equivalent to a single cell of internal resistance eq 1 2

1 2 r r r r r   and emf 1 2 2 1 1 2 1 2 eq 1 2 1 2 1 2 E r E r r r E E E r r r r r r      _  _   _{ } (C) Mixed Grouping

If n identical cell’s are connected in a row and such m row’s are connected in parallel as shown. 1. Equivalent emf of the combination E_eq = nE

2. Equivalent internal resistance of the combination r_eq = nr/m 3. Main current flowing through the load

nr mR mnE ) m / nr ( R nE i     4. Potential difference across load V = iR 5. Potential difference across each cell V' = V/n

E

r r r

E E

R I

6. Current from each cell i' = i/n

7. Condition for maximum power R nr m  and 2 max E P (mn) 4r 

8. In mixed grouping' as both current in the circuit and power transferred to the load are maximum under same condition it is preferred over series or parallel grouping of cells.

9. Total number of cell = mn

10. If N (= mn) and r/R = (m/n) are given then m = no. of rows and n = no. of cells contained in each row can be calculated for maximum current through R. It may be mentioned here, that the values of m and n should necessarily be positive integers. If, upon solving for m and n, the values come out to be fractions, then get the two set of integral values of m and n, one immediately lesser and the other greater than the obtained fractions. Next, check the value of i for these two set of m and n values, and then decide the values of m and n for i to be maximum.

Example 11 :

Figure shows a part of an electrical network. The current flows along the directions as shown in the diagram. Find the value of resistance R’. Sol. Let us consider the loop along the sense abcda.

Applying Kirchhoff’s second law to the loop abcda, yields

E=12V R' i =2.0A1 i =3A3 i =4.0A₂ a b c d R=5

R (4.0A) (2.0A)(0) 12V 5    (3A)0

or R 12 (15V) 0.75 4.0A



   



The value of the resistance is 0.75  and the direction of current flow is from b to a with a magnitude 4.0A. Example 12 :

Two cells P and Q connected in series have each an emf of 1.5 V and internal resistances 1.0 and 0.5 respectively. Find the current through them and the voltages across their terminals.

Sol. For a single closed loop, consisting of cells and resistors the current i flowing through it is given by

i i E i r r       i 1.5 1.5V 1.0 0.5     = 1.5A P i Q E =1.5V₁ E =1.5V2 r =1.01 r =0.52

The voltage across the cell P is

p 1 1

V E ir 1.5V 1.5 (1.0)V zero andQ is VQE2ir21.5V 1.5 (0.5)V 0.75V

Example 13 :

An accumulator is connected first to an external resistance R₁ and then to another external resistance R₂ for the same time. At what value of the internal resistance of the accumulator will the amount of heat dissipated in the external resistances be the same in the two cases ?

Sol. When an accumulator of emf E and internal resistance r is connected across a load resistance R, the heat dissipated in the external circuit 2 2 2 E Rt H I Rt (R r)    E as I (R r)        

According to given problem :

2 2 1 2 2 2 1 2 E R t E R t (R r) (R r) i.e., 2 2 1 1 2 (R R ) (r R R )0 And as R2R (given)1 r2R R1 20 Example 14 :

Twelve cells each having the same emf and negligible internal resistance are kept in a closed box. Some of the cells are connected in the reverse order. This battery is connected in series with an ammeter, an external resistance R and two cells of the same type as in the box. The current when they aid the battery is 3 ampere and when they oppose, it is 2 ampere. How many cells in the battery are connected in reverse order ?

Sol. Let n cells are connected in reverse order. Then emf of the battery is E’ = (12n – n)E – nE = (12 – 2n) E In case (i) I E 2E 3 R     or E' + 2E = 3R or (14 – 2n) E = 3R ... (1) In case (ii) I E 2E 2 R     or E' – 2E = 3R or (10 – 2n) E = 2R ... (2)

Dividing (1) and (2) 14 2n 3 10 2n 2

 

 or n = 1

One cell is connected in reverse order. Example 15 :

How will you connect 24 cells each of internal resistance 1 so as to get maximum power output across a load of 10? Sol. For maximum power output R r

n  m and n × m = p = 24 So 2 n pr R r m _m   _or 2 pr 24 1 m R 10    or _{m 2.41.56} (a) If m = 1 then n p 24 m   s o 2 2 2 1 _{2 2} RE 10E P 5E R r 10 1 n m 24                  (b) If m = 2 then n p 24 12 m 2    so 2 2 2 ₂ 10E P 5.6 E 10 1 12 2         

So to get maximum output power cells must be arranged in two rows having 12 cells in each row.

TRY IT YOURSELF

Q.1 By a cell a current of 0.9 A flows through 2 ohms resistor and 0.3A through 7 ohm resistor. The internal resistance of the cell is –

(A) 0.5 (B) 1.0 (C) 1.2 (D) 2.0

Q.2 A cell of emf 1.5V having a finite internal resistance is connected to a load resistance of 2 ohm. For maximum power transfer the internal resistance of the cell should be –

(A) 4 ohm (B) 0.5 ohm (C) 2 ohm (D) None of these

Q.3 The e.m.f. of a cell is E volts and internal resistance is r ohm. The resistance in external circuit is also r ohm. The p.d. across the cell will be –

(A) E/2 (B) 2E (C) 4E (D) E/4

Q.4 A cell of e.m.f. E is connected with an external resistance R, then p.d. across cell is V. The internal resistance of cell will be – (A) (E V) R E  (B) (E V) R V  (C) (V E) R V  (D) (V E) R E 

Q.5 When a resistance of 2ohm is connected across the terminals of a cell, the current is 0.5 amperes. When the resistance is increased to 5 ohm, the current is 0.25 amperes. The internal resistance of the cell is –

(A) 0.5 ohm (B) 1.0 ohm (C) 1.6 ohm (D) 2.0 ohm

Q.6 The terminal potential difference of a cell when short-circuited is (E = E.M.F. of the cell)

(A) E (B) E/2 (C) zero (D) E/3

Q.7 n identical cells each of e.m.f. E and internal resistance r are connected in series. An external resistance R is connected in series to this combination. The current through R is –

(A) nE Rnr (B) nE nRr (C) E Rnr (D) nE Rr

Q.8 A cell of internal resistance r is connected to an external resistance R. The current will be maximum in R, if –

(A) R = r (B) R < r (C) R > r (D) R = r/2

Q.9 The current flowing through the segment AB of the circuit shown in figure is (A) 1 amp from A to B

(B) 1 amp from B to A (C) 2 amp from A to B (D) 2 amp from B to A

Q.10 Two cells of the same emf  and different internal resistances r₁ and r₂ are connected in series to an external resistances R . The value of R for which the potential difference across the first cell is zero is given by (see figure) :

(A) R = 1 2 r r (B) R = r1 + r2 (C) R = r1 – r2 (D) R = r1 = r2

ANSWERS

Figure shows the fundamental diagram of wheatstone bridge. The bridge has four resistive arms, together with a source of emf (a battery) and a galvanometer. The current through the galvanometer depends on the potential difference between the point c and d. The bridge is said to be balanced when the potential difference across the galvanometer is 0 V so that there is no current through the galvanometer.

This condition occurs when the potential difference from point c to

G a b c _d R1 R2 R3 R4 I1 I2 I3 I4 Unknown Standard arm 

point a, equals the potential difference from point d to point a; or by referring to the other battery terminal, when the voltage from other point c to point b equals the voltage from point d to point b. Hence, the bridge is balanced when

I₁R₁ = I₁R₂ …(i)

if the galvanometer current is zero, the following conditions also exist:

3 1 3 1 R R ε I I    _…(ii) and 4 2 4 2 R R ε I I    _…(iii)

Combining Eqs. (i), (ii) and (iii) and simplifying, we obtain

4 2 2 3 1 1 R R R R R R    …(iv)

from which we get

R₁R₄ = R₂R₃ or 1 3

2 4

R R

R R …(v)

Equation (v) is the well known expression for balance of the wheatstone bridge. If three of the resistances have known values, the fourth may be determined from Equation (v). Hence, if R₄ is the unknown resistor, its resistance can be expressed in terms of remaining resistors 1 2 3 4 R R R R  _…(vi) COMBINATION OF RESISTANCE (a) Series combination

1. Resistances are connected in series as follows& 2. Same current pass through each resistance. 3. Potential difference between the ends of various

resistances depend on the value of resistance.

V₁ V₂ V₃

+ –

V

I R1 I R2 I R3

4. The sum of potential differences developed between the ends of resistances is equal to the total voltage applied in the circuit. Means, V = V₁ + V₂ + V₃

5. Equivalent resistance of the circuit is : R = R₁ + R₂ + R₃

Means equivalent resistance of resistances connected in series is equal to the sum of different resistances. 6. If n equal resistances of R' are connected in series then total resistance will be, R = nR'

(b) Parallel combination&

1. Resistances are connected in parallel as follows& R1

R2 R₃ I2 I3 I1 I V + –

2. Potential difference between the ends of each resistance is same. 3. Current through different resistances is inversely proportional to

the resistance.

4. The sum of currents through each resistance is equal to the total current flowing through the circuit.

Means I = I₁ + I₂ + I₃

5. If equivalent resistance of circuit is R then

1 2 3

1 1 1 1

R  R R R

That is for parallel combination of resistances reciprocal of equivalent resistances is equal to the sum of reciprocal of different resistances.

6. Total resistance of the circuit is less than the value of smallest resistance in the circuit.

7. If n resistances each of value R' are connected in parallel then equivalent resistances of the circuit will be : R = R'/n

8. On connecting resistances in parallel, the total resistances of the circuit decreases. Current divide inversely proportional to the resistance. I₁ = V R₁ & I2 = V R₂ and 1 2 1 1 1 R  R R  1 2 2 1 I R I  R , equivalent resistance R = R R R R 1 2 1 2

Thus V = IR from which

I₁ = IR R₁ = I R₁ R R R R 1 2 1 2

F

HG

I

KJ

Resistance R₃ is called the standard arm of the bridge and resistors R₂ and R₁ are called the ratio arms. Resistances in Mixed Combination

If resistance are arranged in series-parallel mixed grouping, we apply method of successive reduction to find equivalent resistance. Example 16 :

Find equivalent resistance between a and b.

If not able to reduce the given network in series and parallel through simple successive reduction follow the steps :

(1) Find same potential points in network and considering same potential point as a single point, redraw the circuit you will find resistance in series and parallel combination

Example 17 :

Find equivalent resistance between the points A and B? _{2R R}

3 2 2R 1 A B 4

Sol. Point 1 and 3 are at same potential. Similarly point 2 and 4 are at same potential. Joining resistance between 1 and 2, 2 and 3, 3 and 4 we find that they all are in parallel.

2,4 1,3 2R 2R R

So equivalent resistance

p

1 1 1 1 2

R  2R2RR R or Rp = R/2

Example : 18

Twelve equal resistances R are used to generate shape of a cube. Calculate equivalent resistance across the side of cube?

7 8 4 3 2 1 5 6 b a

Sol. By symmetry potential at point 4 and 5 is same. Similarly potential at point 3 and 6 is same. The equivalent circuits can be drawn as :

R R R R R R R R R R 8 7 3,6 2 1 4,5 b a R R  R R 2 2R 4,5 R/2 1 a b 2 3,6 R/2  R 7R 5 1 a b 2

The equivalent resistance between a and b is _eq

7 R R 7 5 R R 7 12 R R 5    

(2) If not able to see same potential point try to observe if network contains balanced wheatstone bridge. Example 19 :

Calculate the effective resistance between A and B in the following network.

D C 4 A C 6 B 3 2 7

Sol. The circuit can be redrawn as Here P R 2 Q S 3 so bridge is balanced Q=3 D C 7 _B A R=4 S=6 P=2 So the resistance between C and D is not useful.

Equivalent resistance = (P + Q) | | (R + S) eq (P Q)(R S) (2 3)(4 6) R P Q R S 2 3 4 6             5 10 10 15 3    

(3) Even if not able to observe balanced wheatstone bridge try to observe symmetry in network and use plane cutting method. Example 20 :

Seven resistors each of resistance R, are connected as shown in figure. The equivalent resistance between A and B is –

(A) 4R 3 (B) 3 R 2 (C) 7R (D) 8R 7 Sol. (D). R_AB = R_AC + R_CB Line of symmetry A B R/2 A C C B R/2 R _{R R} R R R

(4) If not able to see symmetry try to observe star-delta in the network. Star-Delta Conversions :

The combination of resistances shown in fig. 1 (a) is called a star connection and that shown in fig. 1 (b) is called a delta connection.

1 2 3 (a) 1 2 3 (b)

Fig. 1 (a) Star, (b) Delta connections

These two arrangements are electrically equivalent for the resistance measured between any pair of terminals. A star connection can be replaced by a delta, and a delta can be replaced by a star.

Star to Delta Conversion

If resistances R₁, R₂ and R₃ are known and connected in Star configuration (as in fig. 1 (a))then it can be replaced by a delta configuration with following resistances :

1 2 12 1 2 3 R R R R R R    _; 23 2 3 2 3 1 R R R R R R    _; 13 1 3 1 3 2 R R R R R R    Example 21 :

Consider the unbalanced wheatstone bridge shown in fig. 2.

Find its equivalent resistance between A and B (all values are in ohm) –

B A 1 X 3 3 Z ₁ 1 Y Fig. 2

Sol. Consider the star combination between XYZ. It can be reduced into a delta combination. Then the fig. 2 like fig.3, with

xy 1 3 R 1 3 7 1      xz 1 1 7 R 1 1 3 3      _A B Rxy X 3 Z ₁ Y Rxz Rzy Fig. 3 zy 1 3 R 1 3 7 1     

Thus the equivalent diagram now looks like fig. 4. The resistance between ends A and B is then easily determined. Its value is –

AB 5 R 3  7/3 X Y B A 7 7 3 1 Fig. 4

Thus the answer is R_AB = 5/3 ohm.

Note : For the circuit of fig. 5, memorize the following formula for equivalent resistance. (Try to

prove it by using star-delta conversion).

equivalent 3n 1 R r n 3     _ _  r nr nr r r Fig. 5

Delta to Star Conversion

Consider fig. 1 (a) and (b) again. If resistances R₁₂, R₂₃ and R₁₃ are known and connected in a delta configuration as in fig. 1 (b), then it can be replaced by a star connection with the following resistances –

12 13 1 12 13 23 R R R R R R    ; 23 21 2 12 13 23 R R R R R R    ; 31 32 3 12 13 23 R R R R R R    (note R13R , R31 23R , R32 12 R21) Example 22 :

Each resistance in the network is of r ohm. Then calculate the equivalent resistance between the terminals A and B. Sol. Consider the delta connection between points 123.

Converting it into a star connection, the fig. 6 now looks like fig. 7, with resistances –

B r r r r r r 3 r 1 2 Fig. 6 A 1 r r r R r r r 3      , 2 r R 3  , 3 r R 3 

Thus fig. 7, reduces to a balanced wheatstone bridge (fig. 8), whose equivalent resistance is

eq 8 R r 7  r 3 r r r 1 ₂ Fig. 7 A 4r/3 nr _r Fig. 8 B 4r/3 r/3

Note : The Star-delta conversion is useful if you do not wish to use Kirchhoff’s laws for solving equivalent resistance determina-tion problems. However, if numerical values are not given, at first sight, it leads to tedious algebra, when then leads to simpler expressions. If you have patience verify the following two results.

(i) The equivalent resistance for the circuit of fig. 9 is

r R R r r Fig. 9 eq r (3R r) R 3r R    2 4 6 8 12 4 2 2 B A Fig. 10

(ii) The equivalent resistance for the circuit of fig. 10 is (all resistances are in ohm), R_AB = 8 ohm

Example 23 :

Consider the unbalanced wheatstone bridge shown in fig. 11. Find the equiva-lent resistance between the points A and B. (all resistances are in ohm)

1 B A 3 5 ₇ 2 a d b c Fig. 11

Sol. Consider one of the delta combination, say abc. Then converting it into

equivalent star combination, we find fig. 12. A direct use of the conversion formula give –

1 1 2 1 R 1 2 5 4      B A 3 a d b c Fig. 12 R =1/41 R =5/43 R =5/82