Q: (a) Three resistors 1Ω,2Ω and 3Ω are combined in series. What is the total resistance of the combination?(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

(a) Equivalent resistance is Req=R1+R2+R3=(1+2+3)Ω=6Ω(b)Current in the circuit is given by I=REI=612=2ΩPotential drop across 1Ω, V1=IR1=2×1=2VPotential drop across 2Ω,V2=IR2=2×2=4VPotential drop across 3Ω,V3=IR3=2×3=6V

Q: (a) Three resistors 2Ω,4Ω and 5Ω are combined in parallel. What is the total resistance of the combination?(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, detere the current through each resistor, and the total current drawn from the battery.

(a) For Parallel connection,Rp1=R11+ R21+ R31Rp1=21+ 41+ 51Rp1=2010+5+4Rp=1920Ω(b) R1=2Ω,R2=4Ω,R3=5ΩI1=R1V=20/2=10AI2=R2V=20/4=5AI3=R3V=20/5=4ATotal current = I1+I2+I3=19A

Q: A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0×10−7m2, and its resistance is measured to be 5.0Ω. What is the resistivity of the material at the temperature of the experiment?

Let resistivity coefficient be ρR=ρAlρ=lRAρ=2×10−7Ωm

Q: The four arms of a Wheatstone bridge (Figure) have the following resistances:AB=100Ω,BC=10Ω,CD=5Ω, and DA=60Ω. A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Considering the mesh BADB, we have 100I1+15Ig−60I2=0 or 20I1+3Ig−12I2=0 [(a)] Considering the mesh BCDB, we have 10(I1−Ig)−15Ig−5(I2+Ig)=0 10I1−30Ig−5I2=0 2I1−6Ig−I2=0 [(b)] Considering the mesh ADCEA, 60I2+5(I2+Ig)=10 65I2+5Ig=10 13I2+Ig=2 [(c)] Multiplying Eq. (b) by 10 20I1−60Ig−10I2=0 [(d)] From Eqs. (d) and (a) we have 63Ig−2I2=0 I2=31.5Ig [(e)] Substituting the value of I2 into Eq. [(c)], we get 13(31.5Ig)+Ig=2 410.5Ig=2 Ig=4.87 mA

Question 1

The storage battery of a car has an emf of 12 V. If the internal resistance of the battery is 0.4 Ω, what is the maximum current that can be drawn from the battery?

Accepted Answer

The maximum current is given by I=Internal resistanceVoltage​=0.412​=30A

Question 2

A battery of emf 10 V and internal resistance 3 W is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the teral voltage of the battery when the circuit is closed?

Accepted Answer

A battery of emf 10 V and internal resistance 3 W is connected to a resistor. If the current in the circuit is 0.5 A, what is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Question 3

(a) Three resistors 1Ω,2Ω  and 3Ω are combined in series. What is the total resistance of the combination?(b) If the combination is connected to a battery of emf 12 V and negligible internal resistance, obtain the potential drop across each resistor.

Accepted Answer

(a) Equivalent resistance is Req​=R1​+R2​+R3​=(1+2+3)Ω=6Ω(b)Current in the circuit is given by I=RE​I=612​=2ΩPotential drop across 1Ω, V1​=IR1​=2×1=2VPotential drop across 2Ω,V2​=IR2​=2×2=4VPotential drop across 3Ω,V3​=IR3​=2×3=6V

Question 4

(a) Three resistors 2Ω,4Ω and 5Ω are combined in parallel. What is the total resistance of the combination?(b) If the combination is connected to a battery of emf 20 V and negligible internal resistance, detere the current through each resistor, and the total current drawn from the battery.

Accepted Answer

(a) For Parallel connection,Rp​1​=R1​1​+ R2​1​+ R3​1​Rp​1​=21​+ 41​+ 51​Rp​1​=2010+5+4​Rp​=1920​Ω(b) R1​=2Ω,R2​=4Ω,R3​=5ΩI1​=R1​V​=20/2=10AI2​=R2​V​=20/4=5AI3​=R3​V​=20/5=4ATotal current = I1​+I2​+I3​=19A

Question 5

A negligibly small current is passed through a wire of length 15 m and uniform cross-section 6.0×10−7m2, and  its  resistance is  measured  to  be  5.0Ω.  What is the resistivity of the material at the temperature of the experiment?

Accepted Answer

Let resistivity coefficient be ρR=ρAl​ρ=lRA​ρ=2×10−7Ωm

Question 6

Determine the current in each branch of the network shown in Fig.

Accepted Answer

Each branch of the network is assigned an unknown current to be determined by the application of Kirchhoff's rules. To reduce the number of unknowns at the outset, the first rule of Kirchhoff is used at every junction to assign the unknown current in each branch. We then have three unknowns I1​,I2​ and I3​ which can be found by applying the second rule of Kirchhoff to three different closed loops. Kirchhoff's second rule for the closed loop ADCA gives,
10−4(I1​−I2​)+2(I2​+I3​−I1​)−I1​=0
[(a)]
that is, 7I1​−6I2​−2I3​=10
For the closed loop ABCA, we get
10−4I2​−2(I2​+I3​)−I1​=0
that is, I1​+6I2​+2I3​=10
[( b)]
For the closed loop BCDEB, we get
5−2(I2​+I3​)−2(I2​+I3​−I1​)=0
that is, 2I1​−4I2​−4I3​=−5 Equations (a,b,c) are three simultaneous equations in three unknowns. These can be solved by the usual method to give
I1​=2.5 A,I2​=85​ A,I3​=187​ A
The currents in the various branches of the network are
AB:85​ A,CA:221​ A,DEB:187​ A
AD:187​ A,CD:0 A,BC:221​ A
It is easily verified that Kirchhoff's second rule applied to the remaining closed loops does not provide any additional independent equation, that is, the above values of currents satisfy the second rule for every closed loop of the network. For example, the total voltage drop over the closed loop BADEB
5 V+(85​×4)V−(815​×4)V
equal to zero, as required by Kirchhoff's second rule.

Question 7

A network of resistors is connected to a 16 V battery with internal resistance of 1Ω, as shown in Fig. 3.19: (a) Compute the equivalent resistance of the network. (b) Obtain the current in each resistor. (c) Obtain the voltage drops VAB​,VBC​ and VCD​

Accepted Answer

(a) The network is a simple series and parallel combination of resistors. First the two 4Ω resistors in parallel are equivalent to a resistor =[(4×4)/(4+4)]Ω=2Ω
In the same way, the 12Ω and 6Ω resistors in parallel are equivalent to a resistor of
[(12×6]/(12+6)]Ω=4Ω
The equivalent resistance R of the network is obtained by combining these resistors (2 Ω and 4Ω with 1Ω in series, that is,
R=2Ω+4Ω+1Ω=7Ω
(b) The total current I in the circuit is
I=R+rε​=(7+1)Ω16V​=2 A
Consider the resistors between A and B. If I1​ is the current in one of the 4Ω resistors and I2​ the current in the other,
I1​×4=I2​×4
that is, I1​=I2​, which is otherwise obwious from the symmetry of the two arms. But I1​+I2​=I=2 A. Thus,
I1​=I2​=1 A
that is, current in each 4Ω resistor is 1 A. Current in 1Ω resistor between B and C would be 2 A.
Now, consider the resistances between C and D. If I3​ is the current in the 12Ω resistor, and I4​ in the 6Ω resistor,
I3​×12=I4​×6, lie.. I4​=2I3​
But, I3​+I4​=I=2 A
Thus, I3​=(32​)A,I4​=(34​)A
that is, the current in the 12Ω resistor is (2/3)A, while the current in the 6Ω resistor is (4/3)A.
(c) The voltage drop across AB is
VAB​=I1​×4=1 A×4Ω=4 V, 
This can also be obtained by multiplying the total current between A and B by the equivalent resistance between A and B. that is,
VAB​=2 A×2Ω=4 V
The voltage drop across BC is
VBC​=2 A×1Ω=2 V
Finally, the voltage drop across CD is
VCD​=12Ω×I3​=12Ω×(32​)A=8 V. 
This can alternately be obtained by multiplying total current between C and D by the equivalent resistance between C and D. that is,
VCD​=2 A×4Ω=8 V
Note that the total voltage drop across AD is 4 V+2 V+8 V=14 V. Thus, the terminal voltage of the battery is 14 V, while its emf is 16 V. The loss of the voltage (=2 V) is accounted for by the internal resistance 1Ω of the battery [2 A ×1Ω=2 V].

Question 8

The four arms of a Wheatstone bridge (Figure) have the following resistances:AB=100Ω,BC=10Ω,CD=5Ω, and DA=60Ω. A galvanometer of 15Ω resistance is connected across BD. Calculate the current through the galvanometer when a potential difference of 10 V is maintained across AC.

Accepted Answer

Considering the mesh BADB, we have
100I1​+15Ig​−60I2​=0
 or 20I1​+3Ig​−12I2​=0
[(a)]
Considering the mesh BCDB, we have
10(I1​−Ig​)−15Ig​−5(I2​+Ig​)=0
10I1​−30Ig​−5I2​=0
2I1​−6Ig​−I2​=0
[(b)]
Considering the mesh ADCEA,
60I2​+5(I2​+Ig​)=10
65I2​+5Ig​=10
13I2​+Ig​=2
[(c)]
Multiplying Eq. (b) by 10
20I1​−60Ig​−10I2​=0
[(d)]
From Eqs. (d) and (a) we have
63Ig​−2I2​=0
I2​=31.5Ig​
[(e)]
Substituting the value of I2​ into Eq. [(c)], we get
13(31.5Ig​)+Ig​=2
410.5Ig​=2
Ig​=4.87 mA

Question 9

A resistance of RΩ draws current from a potentiometer. The potentiometer has a total resistance R0​Ω (Fig. 3.29). A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

Accepted Answer

While the slide is in the middle of the potentiometer only half of its resistance (R0​/2) will be between the points A and B. Hence, the total resistance between A and B, say, R1​, will be given by the following expression:
R1​1​=R1​+(R0​/2)1​
R1​=R0​+2RR0​R​
The total resistance between A and C will be sum of resistance between A and B and B and C, i.e., R1​+R0​/2
∴ The current flowing through the potentiometer will be
I=R1​+R0​/2V​=2R1​+R0​2V​
The voltage V1​ taken from the potentiometer will be the product of current I and resistance R1​,
V1​=IR1​=(2R1​+R0​2V​)×R1​
Substituting for R1​, we have a
V1​=2(R0​+2RR0​×R​)+R0​2V​×R0​+2RR0​×R​
V1​=2R+R0​+2R2VR​
or V1​=R0​+4R2VR​.

Chapter 3 : Current Electricity