Physics B & C - Multiple Choice Practice Questions on Electric Circuits

"Nearly every man who develops an idea works it up to the point where it looks impossible, and then he gets discouraged. That's not the place to become discouraged."

– Thomas A. Edison

Today we will discuss a few multiple choice practice questions on  electric circuits. Questions in this section were discussed on earlier occasions. You may click on the label ‘electric circuits’ below this post to access all posts in this section. Here are the questions with their solution:

(1) The voltmeter in the shown shown has a resistance of. The internal resistance of the 10 V battery is negligible. What is the reading of the voltmeter?

(a) 2 V

(b) 3.3 V

(c) 4 V

(d) 5V

(e) 6 V

The voltmeter of resistance 100 KΩ is connected across a 100 KΩ resistor and these two give a parallel combined resistance value of 50 KΩ. The total resistance in series with the battery is 100 KΩ + 100 KΩ + 50 KΩ = 250 KΩ.

The current in the series circuit is 10 V/250 KΩ and hence the potential difference across the parallel combination of the voltmeter and the 100 KΩ resistor is (10 V/250 KΩ) × 50 KΩ = 2 V [Option (a)].

[We retained the resistances in KΩ itself in the above expressions so that we could obtain the potential difference in volts in a convenient manner].

 

(2) The adjoining figure shows junctions J₁ and J₂ of an electric circuit. Currents at these junctions sufficient for arriving at the answer are indicated in the figure. What is the value of the current I?

(a) 1 A

(b) 3 A

(c) 4 A

(d) 8 A

(e) 10 A

In accordance with Kirchoff’s current law (KCL), the current I₁ flowing outwards from junction J₁ towards junction J₂ is 5 A – 2 A = 3 A.

[Algebraic sum of currents at a junction is zero according to KCL:

             5 A – 2 A + I₁ = 0

Therefore, I₁ = – 3 A, the negative sign showing that the current flows outwards from the junction].

Total current flowing towards junction J₂ is 3 A + 5 A = 8 A.

Since a current of 4 A is shown as flowing out from junction J₂, the remaining current I₂ that has to flow out must be 4 A [Option (c)].

(3) A 2 μF capacitor is connected in series with a 1 μF capacitor and a 60 V power supply. The potential difference across the 2 μF capacitor is

(a) 60 V

(b) 40 V

(c) 30 V

(d) 20 V

(e) zero

The effective capacitance connected across the power supply is the series combined value of the 2 μF and 1 μF capacitors which is (²/₃) μF

[If capacitors C₁ and C₂ are  connected in series, their effective capacitance C is given by the reciprocal relation, 1/C = 1/ C₁ + 1/C₂ so that C = C₁C₂/( C₁ + C₂)]

The charge Q supplied by the power supply is given by

             Q = CV = (²/₃) × 60 = 40 μC

[We obtain the charge in micro coulomb since the capacitance is in μF].

The capacitors hold the same charge in series connection. Therefore, the voltage across the 2 μF

capacitor is Q/C₁ = 40 μC/2 μF = 20 volt [Option (d)].

[You will be able to work out this problem in no time if you remember that the voltages are distributed among series connected capacitors in inverse proportion to their capacitances].

The following questions are specifically meant for AP Physics C aspirants:

(4) The resistance R of any uniform wire of length L and cross section area A is related to the resistivity (or, specific resistance) ρ of its material as

             R = ρL/A

Suppose you have n wires of the same length L and the same area of cross section A made of materials of resistivities ρ, 2ρ, 3ρ, 4ρ, ……nρ respectively. If they are connected in series, what will be the resistivity of the material of the combined wire?

(a) n ρ

(b) n(n+1)ρ

(c) n ρ/2

(d) (n+1)ρ/2

(e) n(n+1)ρ/2

The resistances of the 1^st, 2^nd, 3^rd, 4^th, ……n^th wires are respectively given by R₁ = ρL/A, R₂ = 2ρL/A, R₃ = 3ρL/A, R₄ = 4ρL/A, ……. R_n = nρL/A.

The total resistance R of the series combination is given by

             R = R₁ + R₂ + R₃ + R₄ + ……+ R_n

Or, R = (ρL/A)(1 + 2 + 3 + 4 + ……n), on substituting for R₁, R₂, R₃ etc.

This gives R = (ρL/A)[n(n + 1)/2] = n(n + 1) ρL/2A

The length of the compound wire is nL and its area of cross section is A. If the resistivity is ρ its resistance can be written as

             R = ρnL/A

Therefore ρ = RA/nL = [n(n + 1) ρL/2A](A/nL) = (n+1)ρ/2

[Even without writing all the above steps you could have written the answer, arguing that the resistivity of the combined wire must be the average value of the resitivities of the individual wires, since the increments in resistivity occur in a regular manner. Therefore ρ = (ρ+2ρ+ 3+4ρ+ ……+nρ)/n = (n+1)ρ/2]

(5) In the circuit shown in the adjoining figure, the ammeter reading is zero when the switch S is closed. If the batteries are of zero internal resistance, the value of resistance X is

(a) 100 Ω

(b) 200 Ω

(c) 300 Ω

(d) 400 Ω

(e) 600 Ω

Since the current through the ammeter is zero, the voltage drop across the series combination of X and 600 Ω must be 4 V.

[The battery emf of 4 V and the p.d. across the series combination of X and 600 Ω have tu be equal and and in opposition to attain the condition of zero current through the ammeter].

The 6 V battery drives a current through the three resistors and 4 V is dropped across the series combination of X and 600 Ω. The remaining 2 V is dropped across the 400 Ω resistor.

Since the potential drop across the series combination of X and 600 Ω is 4V, we have

              X + 600 Ω = 400 Ω

Therefore X = 200 Ω

[If you want to use Kirchoff’s laws and write down mathematical steps for solving the above problem, here is how you will proceed:

             For the loop containing the 4 V battery, ammeter, X and 600 Ω, we have

             (I₁ + I₂)(X+600) = 4 where I₁ and I₂ are the currents supplied by the 4 volt and 6 volt batteries respectively.

Since I₁ = 0 we have

             I₂(X+600) = 4 …………..(i)

             For the loop containing the 6 V battery, 400 Ω, X and 600 Ω, we have

             I₂(400+X+600) = 6 ……..(ii)

Dividing Eq. (i) by Eq. (ii), we have

             (X+600) /(400+X+600) = 2/3

This gives X = 200 Ω]