Question 3.1: Question 3.1:
The storage battery of a car The storage battery of a car 0.4Ω, what is the maximum 0.4Ω, what is the maximum Answer
Answer
Emf of the battery,
Emf of the battery, E E = 12 V= 12 V In
Inteternrnal al rresesisistatancnce oe of tf the he babatttt Ma
Maxiximumum cum currrrent dent drarawn fwn froro According to Ohm’s law, According to Ohm’s law,
as an
as an emf emf of 1of 12 V. I2 V. If thf the intee internarnal rel resistsistancance of e of tt urrent that can be drawn from the
urrent that can be drawn from the battery?battery?
ery,
ery, r r = 0.4 Ω= 0.4 Ω the
the battery battery == I I
e
Current in the circuit,
Current in the circuit, I I = 0.5= 0.5 Resistance of the resistor = Resistance of the resistor = The
The relatrelation ion for cfor currenurrent usit usinn
Te
Termrmininal val vololtatage ge of of ththe ree resisiss According to Ohm’s law, According to Ohm’s law, V V == IR IR = 0.5 × 17 = 0.5 × 17 = 8.5 V = 8.5 V A A Ohm’s
Ohm’s law law is,is,
or or == V V
Current in the circuit,
Current in the circuit, I I = 0.5= 0.5 Resistance of the resistor = Resistance of the resistor = The
The relatrelation ion for cfor currenurrent usit usinn
Te
Termrmininal val vololtatage ge of of ththe ree resisiss According to Ohm’s law, According to Ohm’s law, V V == IR IR = 0.5 × 17 = 0.5 × 17 = 8.5 V = 8.5 V A A Ohm’s
Ohm’s law law is,is,
or or == V V
Total resistance = 1 + 2 + 3 Total resistance = 1 + 2 + 3 Current flowing through the Current flowing through the Emf of the battery,
Emf of the battery, E E = 12 V= 12 V Total re
Total resistance of thsistance of the circuite circuit The
The relatrelation ion for cfor currenurrent usit usinn
P
Pootteennttiial al ddrroop p aaccrroosss s 1 1 Ω Ω rreess From Ohm’s law, the value From Ohm’s law, the value V
V 11= 2 × 1= 2 V … (i)= 2 × 1= 2 V … (i)
P
Pootteennttiial al ddrroop p aaccrroosss s 2 2 Ω Ω rreess Again, from Ohm’s law, the Again, from Ohm’s law, the
6 Ω 6 Ω circuit circuit == I I
,, R R = 6 Ω= 6 Ω Ohm’s
Ohm’s law law is,is,
istor istor == V V 11
ff V V 11can be obtained ascan be obtained as
istor istor == V V 22
alue
There are three resistors of resistances,
R1= 2 Ω, R2= 4 Ω, and R3= 5 Ω
They are connected in parallel. Hence, total resistance ( R) of the combination is given by,
Therefore, total resistance of the combination is . Emf of the battery, V = 20 V
Therefore, the current throug total current is 19 A.
Question 3.5:
At room temperature (27.0 ° temperature of the element i temperature coefficient of th Answer
Room temperature, T = 27°C Resistance of the heating ele Let T i th i d t
h each resister is 10 A, 5 A, and 4 A respectivel
) the resistance of a heating element is 100 Ω. the resistance is found to be 117 Ω, given that
material of the resistor is
ent at T , R = 100 Ω rature of the filament.
y and the
hat is the he
A negligibly small current is section 6.0 × 10−7m2, and its the material at the temperatu Answer
Length of the wire, l =15 m Area of cross-section of the Resistance of the material of Resistivity of the material of Resistance is related with the
passed through a wire of length 15 m and unifo resistance is measured to be 5.0 Ω. What is the e of the experiment? ire, a = 6.0 × 10−7m2 the wire, R = 5.0 Ω the wire = ρ resistivity as m cross- resistivity of
Temperature, T 2= 100°C
Resistance of the silver wire Temperature coefficient of si It is related with temperature
Therefore, the temperature c
Question 3.8:
Aheating element using nich 3.2 A which settles after a fe temperature of the heating el coefficient of resistance of ni
at T 2, R2= 2.7 Ω
lver = α
and resistance as
efficient of silver is 0.0039°C−1.
ome connected to a 230 V supply draws an init seconds toa steady value of 2.8 A. What is th ement if the room temperature is 27.0 °C? Tem chrome averaged over the temperature range in
al current of steady erature olved is
Temperature co-efficient of Initial temperature of nichro Study state temperature reac T 2can be obtained by the rel
Therefore, the steady temper
Question 3.9:
ichrome,α= 1.70 × 10−4°C−1
e, T 1= 27.0°C
ed by nichrome = T 2
tion for α,
I 1= Current flowing through the outer circuit I 2= Current flowing through branch AB I 3= Current flowing through branch AD I 2− I 4= Current flowing through branch BC I 3+ I 4= Current flowing through branch CD I 4= Current flowing through branch BD
3 I 2+ 2 I 1− I 4= 2 … (3)
From equations (1) and (2), we obtain
I 3= 2(2 I 3+ 4 I 4) + I 4 I 3= 4 I 3+ 8 I 4+ I 4
3 I 3 = 9 I 4
3 I 4 = + I 3… (4)
Putting equation (4) in equation (1), we obtain
I 3= 2 I 2+ I 4
4 I 4 = 2 I 2 I 2= − 2 I 4… (5)
It is evident from the given figure that,
Therefore, current in branch
Answer
A metre bridge with resistors X andY is represented in the given figure.
When the galvanometer and galvanometer will show no d galvanometer.
Question 3.11:
A storage battery of emf 8.0 dc supply using a series resis during charging? What is the Answer
Emf of the storage battery, E Internal resistance of the batt
cell are interchanged at the balance point of the eflection. Hence, no current would flow throug
and internal resistance 0.5 Ω is being charged tor of 15.5 Ω. What is the terminal voltage of th purpose of having a series resistor in the chargi
= 8.0 V ery, r = 0.5 Ω bridge, the the by a 120 V e battery ng circuit?
A series resistor in a chargin The current will be extremel
Question 3.12:
In a potentiometer arrangem length of the wire. If the cell 63.0 cm, what is the emf of t Answer
Emf of the cell, E 1= 1.25 V
Balance point of the potentio The cell is replaced by anoth
circuit limits the current drawn from the exter high in its absence. This is very dangerous.
nt, a cell of emf 1.25 V gives a balance point at is replaced by another cell and the balance poin
e second cell? meter, l1= 35 cm er cell of emf E 2. al source. 35.0 cm t shifts to
Number density of free elect copper wire, l = 3.0 m
Area of cross-section of the Current carried by the wire, I= nAeV d
Where,
e = Electric charge = 1.6 × 1
V d = Drift velocity
ons in a copper conductor, n = 8.5 × 1028m−3L
ire, A = 2.0 × 10−6m2
= 3.0 A, which is given by the relation,
−19
C
Surface charge density of the Current over the entire globe Radius of the earth, r = 6.37 Surface area of the earth, A = 4πr 2
= 4π × (6.37 × 106)2 = 5.09 × 1014m2
Charge on the earth surface, q =σ × A
= 10−9× 5.09 × 1014 = 5.09 × 105C
Time taken to neutralize the
earth,σ = 10−9C m−2
, I = 1800 A × 106m
A secondary cell after long u Ω. What maximum current c motor of a car?
Answer
Number of secondary cells, Emf of each secondary cell, Internal resistance of each ce series resistor is connected t Resistance of the resistor, R Current drawn from the supp
se has an emf of 1.9 V and a large internal resis n be drawn from the cell? Could the cell drive
= 6 = 2.0 V
ll, r = 0.015 Ω
the combination of cells. 8.5 Ω
ly = I , which is given by the relation,
ance of 380 he starting
Question 3.16:
Two wires of equal length, one of aluminium and the other of copper have the same resistance. Which of the two wires is lighter? Hence explain why aluminium wires are preferred for overhead power cables. ( ρAl= 2.63 × 10−8Ω m, ρCu= 1.72 × 10−8Ω m,
Relative density of Al = 2.7, of Cu = 8.9.) Answer
Resistivity of aluminium, ρAl= 2.63 × 10−8Ω m
Relative density of aluminium, d 1 = 2.7
Letl1 be the length of aluminium wire and m1 be its mass.
And,
Mass of the aluminium wire, m1= Volume × Density
= A1l1 × d 1= A1l1d 1… (3)
Mass of the copper wire, m2= Volume × Density
Current A Volta V 0.2 3.94 0.4 7.87 0.6 11.8 0.8 15.7 1.0 19.7 2.0 39.4 Answer e Current A Voltage V 3.0 59.2 4.0 78.8 5.0 98.6 6.0 118.5 7.0 138.2 8.0 158.0
When a steady current flows current flowing through the speed are inversely proportio constant.
No, Ohm’s law is not univer semi-conductor is a non-oh According to Ohm’s law, the Voltage (V ) is directly propo Ris the internal resistance of
If V is low, then R must be v In order to prohibit the curre must have a very large intern current drawn can exceed th
in a metallic conductor of non-uniform cross-se onductor is constant. Current density, electric fi nal to the area of cross-section. Therefore, they
ally applicable for all conducting elements. Va ic conductor. Ohm’s law is not valid for it.
relation for the potential is V = IR tional to current ( I ).
the source.
ry low, so that high current can be drawn from t from exceeding the safety limit, a high tensio al resistance. If the internal resistance is not lar
safety limits in case of a short circuit.
ction, the eld, and drift are not
uum diode
the source.
supply
Alloys of metals usually hav Alloys usually have lower te The resistivity of the alloy, The resistivity of a typical in of 1022.
Question 3.20:
Given n resistors each of resi maximum (ii) minimum effe minimum resistance?
Given the resistances of 1 Ω, resistance of (i) (11/3) Ω (ii) Determine the equivalent res
greater resistivity than that of their constituent perature coefficients of resistance than pure m anganin, is nearly independent of increase of te sulator is greater than that of a metal by a factor
stance R, how will you combine them to get the tive resistance? What is the ratio of the maxim
2 Ω, 3 Ω, how will be combine them to get an (11/5) Ω, (iii) 6 Ω, (iv) (6/11) Ω?
stance of networks shown in Fig. 3.31.
metals. etals. mperature. of the order (i) m to quivalent
Hence, maximum resistance of the combination, R1=nR
Whenn resistors are connected in parallel, the effective resistance ( R2) is the minimum,
given by the ratio .
Hence, minimum resistance of the combination, R2=
The ratio of the maximum to the minimum resistance is,
The resistance of the given resistors is,
R1= 1 Ω, R2= 2 Ω, R3= 3 Ω2
Equivalent resistance,
Consider the series combination of the resistors, as shown in the given circuit.
Equivalent resistance of the circuit is given by the sum,
R’= 1 + 2 + 3 = 6 Ω
Equivalent resistance,
Consider the series combination of the resistors, as shown in the given circuit.
All the four resistors are con
Hence, equivalent resistance It can be observed from the connected in series.
Hence, equivalent resistance = 5 R
Question 3.21:
Determine the current drawn infinite network shown in Fi
ected in series.
of the given circuit is
iven circuit that five resistors of resistance R ea
of the circuit = R + R + R + R + R
from a 12 V supply with internal resistance 0.5 . 3.32. Each resistor has 1 Ω resistance.
ch are
Negative value of R’cannot
Internal resistance of the circ Hence, total resistance of the Supply voltage, V = 12 V
According to Ohm’s Law, c
e accepted. Hence, equivalent resistance,
uit, r = 0.5 Ω
given circuit = 2.73 + 0.5 = 3.23 Ω
What is the value ε?
What purpose does the high resistance of 600 kΩ have? Is the balance point affected by this high resistance?
Is the balance point affected by the internal resistance of the driver cell? Would the method work in the above situation if the driver cell of the potentiometer had an emf of 1.0 V instead of 2.0 V?
(f )Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of a thermo-couple)? If not, how will you modify the circuit?
Answer
The method would not work instead of 2.0 V. This is beca than the emf of the other cell The circuit would not work would be unstable, the balan large percentage of error. The given circuit can be mod The potential drop across A error would be small.
Question 3.23:
Figure 3.34 shows a potentio point with a standard resistor unknown resistance X is 68.5 failed to find a balance point
if the driver cell of the potentiometer had an e use if the emf of the driver cell of the potentio , then there would be no balance point on the w
ell for determining an extremely small emf. As e point would be close to end A. Hence, there
ified if a series resistance is connected with the is slightly greater than the emf measured. The
meter circuit for comparison of two resistances. R = 10.0 Ω is found to be 58.3 cm, while that
cm. Determine the value of X . What might you with the given cell of emf ε?
f of 1.0 V eter is less re. the circuit ould be a wire AB. ercentage The balance ith the do if you
Balance point for this resisto Hence, potential drop across The relation connecting emf
Therefore, the value of the u If we fail to find a balance p across R and X must be redu potential drop across R or X i
wire AB, a balance point is o
, l2= 68.5 cm
X , E 2= iX
nd balance point is,
known resistance, X , is 11.75 Ω.
int with the given cell of emf, ε, then the potent
ed by putting a resistance in series with it. Onl s smaller than the potential drop across the pote btained.
ial drop if the ntiometer
