MISCELLANEOUS SOLVED EXAMPLES

Example 1 :

The charge q flowing through a wire varies with time t as q = (0.1 + 0.2t + 0.3t²)C where t is in second. Find the current through the wire (i) initially and at (ii) t = 2s.

Sol. The instantaneous current is dq d 2

i (0.3t 0.2t 0.1)

dt dt

    or i = (0.6t + 0.2) C/s

(i) Initially, t = 0,  Initial current is i = 0.6 (0) + 0.2 A = 0.2 AA (ii) At t = 2s, the current is i = 0.6 (2) + 0.2 A = 1.4 A

Example 2 :

A potential difference applied to the ends of a wire made up of an alloy drives a current through it such that jr , where r is the distance of the point from the axis. If R be the radius of the wire, then find the total current through any cross section of the wire.

Sol. Let c be the centre of the cross-section considered at any section of the wire.

If r be the radius of a circular strip concentric with the cross-section considered and, dr be its thickness then (fig.) current through the circular strip of thickness dr will be

di j.dS   ( r) (2 rdr) cos 0 or _{di    2 ( r r ) dr}²

Cr dr R The total current i can be obtained by summing up

the currents flowing through individual circular strips, i.e.,

R

Three resistors of resistances 3, 5 and 1 and are joined in series. A potential difference of 24V is applied across the combination. Find the current through the resistors and voltage drops across each of them.

Sol. The effective resistance of the three resistances will be Reff      3 5 1 9

 The current through the resistors will be

eff subject to the condition that,

V₁ + V₂ + V₃ = 24V

The heat produced in 5resistor due to current flowing through it is 10 calorie/sec. What is heat dissipated in 4 resistance ? (4 ohm and 6 ohm are in series with 5 ohm in parallel)

Sol. The resistance 5 and 4 + 6 = 10 are in parallel so P =

A fuse wire of radius 8.0mm is replaced by a wire of radius 2.7 mm made up of the same substance. If the former wire has a maximum current capacity of 5.0A, then find the value for the second wire.

Sol. From eq.

Example 6 :

Three resistors of resistance R₁ 3 , R₂ 1 and R₃  are connected8 as shown in fig. across a cell fo 24V and negligible resistance. Find the current drawn from the cell, and the effective resistance across the cell.

R =3₁

+ – 24V

R =1₂

R =8₃ Sol. The current flowing through the branch containing the resistance R₁ and R₂ will be

1

i 24V 6A

(3 1 )

 

 

and the current flowing through the resistance R₃ will be 2

i 24V 3A

 8 



So, the total current drawn from the cell will be, ii₁i₂6A 3A 9A

If R be the effective resistance then, by definition 24V = (9A) R R = 2.67  Example 7 :

A standard 1 resistance and a low resistance r are connected in series in an electrical circuit. The p.d. across the combination is balanced at 6.30 m length of potentiometer, while the p.d. across 1 resistance is balanced at 6.00 m length. Find the value of the small resistance.

Sol. Let I is the current through the resistances, then I (1 + r) = x (6.30) and I (1) = x (6.00)

1 r 6.3 1 6.0

  or 1 + r = 1.05 or r = 0.05 ohm

Example 8 :

A potentiometer wire of length L and resistance 10 has a battery of 2.5V and a resistance in series in its primary circuit. The null point for a cell of emf 1 volt comes at L/2 distance from one end. If the series resistance in the primary circuit is doubled, then position of the new null point is –

(A) 0.5 L (B) 0.6 L (C) 1.0 L (D) None of the above

Sol. (B). For the first case ^{p W}

W

E R

x L R R



where R is the series resistance. Thus for the cell of 1 volt emf with balancing length L/2,

1 = xL/2 or 2.5 10 L

1 L 10 R2

 or 20 + 2R = 25 or R = 2.5 

For second case series resistance

R´ = 2R = 5  2.5 10

x L 10 5

  



1  x where the new balancing length 15L

1 0.6 L

  25 

 Example 9 :

It is desired to measure a small resistance r using the potentiometer. This resistance is connected to a big resistance R and a steady current is passed through it. When p.d. across R is balanced, the balancing length is found to be 320 cm. When p.d. across the series combination of the two is balanced, the balancing length is found to be 360cm. What is the ratio R : r –

(A) 8 : 1 (B) 9 : 8 (C) 1 : 8 (D) None of the above

Sol. (A).V₁IRx (320) and V₂ = I (R + r) = x (360)

Thus R r 360

R 320

  or r 9

1R 8 or r 9 1 R 8 1 8

Thus R 8

r 1

Example 10 :

In the network shown, each resistance is equivalent to R.

The equivalent resistance between points A and B is –

(A) R/3 (B) 2R/3

If a potential is applied across A and B, due to symmetry, the points C, O and D will have same potentials. We can therefore join all these points without affecting current in any branch. The network then re-duces to

Therefore, the equivalent resistance is

eq

R R 2R R  3  3  3 Example 11 :

A fuse with radius 1.0 mm blows at 15A. What is the radius of fuse made of same material which will blow at 30A.

Sol. For fuse ²  r³ s o

Two circular rings of identical radii and resistance of 36 each are placed in such a way that they cross each other's centre C₁ and C₂ as shown in figure.

Conducting joints are made at intersection points A and B of the rings. An ideal cell of emf 20 volts is connected across AB. The power delivered by cell is –

(A) 80 watt (B) 100 watt (C) 120 watt (D) 200 watt

C₁ C₂ A

Sol. (B).From the figure, AC₁ = AC₂ = C₁C₂ = radius B

 AC₁B = 120°

Hence the resistance of four sections are Hence equivalent resistance R across AB is

1 1 1 1 1

The thermo emf of a copper constantan thermocouple varies as E = T + T2

Example 14 :

Each of 3 resistors in figure has a resistance of 2 and can dissipate a maximum of 18 W without becoming excessively heated. Find the maximum power the circuit can dissipate.

Sol. _{Pmax (imax) R}²

2

18imax2

imax 3A Here Req = 3.

So maximum power of circuit = i²_maxR_eq 3² 3 27W Example 15 :

The circuit shown in the figure contains three resistors R₁ = 100, R₂ = 50 &

R₃ = 20and cells of emfs E₁ = 2V & E₂. The ammeter indicates a current of 50mA. Determine the currents in the resistors arid the emf of the second cell.

The internal resistance of the ammeter and of the cells should be neglected.

Sol. Applying KLV in upper loop

mA E₁

E₂

R₁ R₂

R₃ E₁ = (I + 0.05) R₁ + IR₂ I = – 20 mA

Current through R₁ = 30 mA towards right Current through R₂ = 20 mA towards left Applying KLV in lower loop

E₂ = (I + 0.05) 100 + (0.50) 20 = 4 volts Example 16 :

A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by T in a time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. Find the temperature of the wire is raised by the same amount T in the same time t the value of N.

Sol. In the first case, i ^3E

 R , where E is the emf of each cell and R (L) is the resistance of the wire

Also, i² Rt = m.s.T ... (1)

where m is the mass of L length of wire and S is the specific heat of the material of the wire.

In the second case, ^{i NE}

 R

  where _R2L

andi² R’t = m’s.T ... (2)

Dividing equation (2) by equation (1)

i R m

i . R m

  

   

   

    

N2 R m 9 .R m

 

 

N2 1

. 2

9 2   N = 6 Example 17 :

In a practical wheat stone bridge circuit as shown, when one more resistance of 100  is connected is parallel with unknown resistance ‘x’, then ratio   become ‘2’. ₁/ ₂ ₁ is balance length. AB is a uniform wire. Find the value of x.

100

G copper strips

A B

x

E r

copper strips

1 2

Sol.  Wheat stone bridge is in balanced condition 100

100 x

1 2

So

1 2

100x 100100x

   ¹

2

2

 

  x = 100 

Example 18 :

When a galvanometer is shunted with a 4 resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2 wire, the further reduction in the deflection will be (the main current remains the same).

(A) (8/13) of the deflection when shunted with 4 only (B) (5/14) of the deflection when shunted with 4 only (C) (3/4) of the deflection when shunted with 4 only (D) (3/13) of the deflection when shunted with 4only Sol. (A).

When shunted with 4

g

1 41

R 4

5  5  [Rg 16]

I/5 G

4I/5

R_g I

When further shunted with 2

I 16 (I I ) 4

   3

  I ¹

13



I' G R_g

I Further reduction in current = 1 1 8 1 I–I'

5 13 13 5

     

Hence further reduction in deflection = ⁸ 13 of the deflection when shunted with 4only.

Example 19 :

300 nos. of identical galvanic cells, each of internal resistance 9 are arranged a several in-series groups of cells connected in parallel. The arrangement has been laid out so that power output in an externally connected resistance of value 16 is maximum. If n number of cells are connected in every series group that form parallel combination, then find value of n.

R

E₁ E₂ E_n

Sol. Power developed in it is maximum when r_eq = R

nr R

N / n (N – total no. of cells) (n – cell in one series group) n RN

 r using values R = 16, N = 300, r = 9, n = 23.1

As n has to be integer and cofactor of 300, then nearest possible values are 20 and 25.

Power is maximum when current through it is maximum

2

I nE R n r

N





It is obvious that when n = 20, I is greater than when n =25. Therefore, n = 20 nos.

Example 20 :

Relation between current in conductor and time is shown in figure then determine.

(a) Total charge flow through the conductor (b) Write expression of current in terms of time

i₀ i

(c) If resistance of conductor is R then total heat dissipated across resistance R is Sol. (a) q



idt = area of given curve 0 0

q 1i t

2

(b) ⁰

Consider the potentiometer circuit arranged as in figure. The potenti-ometer wire is 600 cm long. (a) At what distance from the point. A should the jockey touch the wire to get zero deflection in the galvanom-eter ? (b) If the jockey touches the wire at a distance of 560 cm from A, what will be the current in the galvanometer ?

Sol. (a) When jockey is not connected E

I16r ... (1) N

r

15r A

Resistance per unit length 15r E/2 600 / cm

(b) Let potential at A is zero

Then apply Kirchoff’s law,

x E 0

If two bulbs rated at 600 W, 220 V are put in series in a 110 V line then what is power generated by each bulb ? Sol. In series voltage across each bulb V_A = R

RR V = V 2 = 110

2 = 55V Power consumed by each bulb P' =

2

The resistivity of a ferric-chromium-aluminium alloy is 51 × 10^–8

-m. A sheet of the material is 15 cm long, 6 cm wide and 0.014 cm thick. Determine resistance between (a) opposite ends and (b) opposite faces.

Sol. (a) As seen from figure (a) in this case,

 = 15 cm = 0.15 m

(b) As seen from figure (b) here

 = 0.014 cm = 14 × 10^–5 m A = 15 × 6 = 90 cm² = 9 × 10^–3 m²

 R = 51 × 10^–8 × 14 × 10^–5/9×10^–3 = 79.3 × 10^–10 

Example 24 :

Circuit for the measurement of resistance by potentiometer is shown.

The galvanometer is first connected at point A and zero deflection is observed at length PJ = 10 cm. In second case it is connect at point C and zero deflection is observed at a length 30 cm from P. Then the unknown resistance x is –

(A) 2R (B) R/2

In potentiometer wire potential difference is directly proportional to length.

Let potential drop unit length a potentiometer wire be K.

For zero deflection the current will flow independently in two circles IR = K × 10 ... (1)

All batteries are having emf 10 volt and internal resistance negligible. All resistors are in ohms. Calculate the current in the right most 2 resistor.

Sol. The simplified circuit is We have to find I.

Let potential of point P be O. Potential at other points P be O. Potential at other points are shown in the figure apply Kirchoff’s current law at x.

x 10 x 10 x 20 (x 10) 0

In the circuit diagram shown find the current through the 1 resistor.

Sol. ^{(v 10) 10 v 0 v 5} 0

Example 27 :

Three identical resistors are connected across a voltage source V so that one of them is in parallel with two others which are connected in series as shown.

The power dissipated through the first one, compared to the power dissipated by each of the other two, is approximately –

(A) the same (B) half as much

V

(C) twice as much (D) four times as much

Sol. (D).

The maximum current in a galvanometer can be 10 mA. It's resistance is 10. To convert it into an ammeter of 1 Amp. a resistor should be connected in –

(A) series, 0.1  (B) parallel, 0.1

(C) series, 100 (D) parallel, 100

Sol. (B).I_G = 10 mA

A hemispherical network of radius a is made by using a conducting wire of resistance per unit length ‘r’.

Find the equivalent resistance across OP.

A

D

B

C P

Sol. In given circuit A, B, C and D are at same potential by symmetry O

1 2

In the figure shown the current flowing through 2 R is – (A) from left to right (B) from right to left

(C) no current (D) none of these ^A

R 2R R B

Sol. (B).

In figure all resistance are connected in parallel.

A R 2R R B

So _eq 2R (R / 2) R 2R (R / 2)

 

 and current in all resistance flow from positive terminal of battery (means A end) to negative terminal of battery (means B end).

Example 31 :

The voltmeter shown in figure reads 18V across the 50 resistor. Find the resistance of the voltmeter.

Sol. i ^{30 18 12 1}A

A circuit consists of a battery, a resistor R and two light bulbs A and B as shown. If the filament in light bulb A burns out, then the following is true for light bulb B :

(A) It is turned off (B) Its brightness does not change

R

A B

(C) It gets dimmer (D) It gets brighter Sol. (D). So bulb B will become brighter.

Example 33 :

The given network is a part of a bigger network. Determine the value of current flowing in resistor R₆. Sol. The current I in R₆ can be calculated by applying

Kirchhoff’s current law. However, here it is not necessary to first find the currents through indi-vidual resistances, R₂, R₃, R₄ and R₅. Instead, we can treat the whole network inside the dotted box as a single junction. To this junction, currents I₁ and I₂ are entering the currents I₃ and I are leav-ing. Hence, according to KCL,

I₁ + I₂ – I₃ – I = 0

Example : 34

In the circuit shown, the charge on the 3 µF capacitor at steady state will be –

(A) 6 µC (B) 4 µC

2V

1V

2µF 3µF 1

2

(C) 2/3 µC (D) 3 µC

Sol. (B).

At steady state, there will be no current in the branches having capacitor only thus equivalent circuit diagram will be as shown in the figure.

AB AB

V 2

V 1 0

2

     AB

V 4V

 3

2V

1V

2µF 3µF 1

2 thus q = CV_AB = 4 µC

Example 35 :

Find current in the branch CD of the circuit (in ampere)

30V B

A D

C

2 2

2

3

Sol. R_eq = 3/2 ; 30

i 20 amp

3 / 2

 

From figure current through B D branch = 5 amp.

30V B

A D

2 2

2

A 3

D D

D

20A 10A

10A

5A

5A 2

2

2

3

D

Example 36 :

The efficiency of a cell when connected to a resistance R is 60%. What will be its efficiency if the external resistance is increased to six times.

Sol. Efficiency = output power input power

 



i R2

  Ei E

i R r

  ; R

R r

  

E r

R i

0.6 R 3R 3r 5R

In the circuit shown all five resistors have the same value 200 and each cell has an emf 3 volts. Find the open circuit voltage and the short circuit current for the terminals A and B.

A

Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0 × 10^-7 m² carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0 × 10³ kg/m³, and its atomic mass is 63.5 amu

Sol. The direction of drift velocity of conduction electrons is opposite to the electric field direction, i.e., electrons drift in the direction of increasing potential. The drift speed v_d is given by v_d = I/neA.

Now, e = 1.6 × 10^-19 C, A = 1.0 × 10^-7 m², I = 1.5 A. The density of conduction electrons, n is equal to the number of atoms per cubic metre (assuming one conduction electron per Cu atom as is reasonable from its valence electron count of one).

A cubic metre of copper has a mass of 9.0 × 10³ kg. Since 6.0 × 10²³ copper atoms have a mass of 63.5 g.

In document Current Electricity Notes JEE Main and Advanced (Page 38-49)