AP Physics B & C - Multiple Choice Practice Questions on Direct Current Circuits

Equations to be remembered for answering questions on direct current circuits and RC transients were posted on this site on 31^st March 2008. You can access them by clicking on the label ‘direct current circuit’ below this post or by trying a search using the search box at the top of this page. Today we will discuss some more multiple choice practice questions in this section:

(1) A uniform copper wire of radius r and length L has resistance R at temperature T. If this wire is carefully stretched so that its radius becomes kr where k is a constant, its resistance at the temperature T will be

(a) kR

(b) R/k²

(c) k²R

(d) k⁴R

(e) R/k⁴

If A is the area of cross section of the wire before the stretching, we have

R = ρL/A where ρ is the resistivity (specific resistance) of the material of the wire.

When the radius becomes kr, the area of cross section of the wire becomes k²A. Since the volume of the wire is unchanged, the length of the wire becomes L/k². The resistance of the wire after stretching therefore becomes R’ given by

R’ = ρ(L/k²)/k²A = ρL/ k⁴A = R/k⁴

[Note that the resistance of the wire after stretching is greater than R since k < 1].

(2) In the above question suppose the wire was stretched so that its length becomes 1.01 L. The resistance of the wire after stretching then becomes nearly

(a) 1.01R

(b) R/1.02

(c) 1.02R

(d) 1.04R

(e) R/1.04

Note that if a wire of resistance R is stretched so that its length becomes n times its original length, its area of cross section becomes (1/n) times its original area of cross section. The resistance of the wire therefore becomes n²R.

Since n = 1.01 the resistance after stretching is (1.01)²R =1.02R, nearly [Option (c)].

Here is a variation of question no.1:

(3) If the radius of a uniform metal wire of resistance R is changed by 1% by stretching it carefully under isothermal conditions, its resistance becomes

(a) 1.04R

(b) R/1.04

(c) 1.02R

(d) R/1.02

(e) 1.01R

We have R’ = R/k⁴ as shown in question no.1. Since the change (decrease) in the radius is 1%, k = 0.99.

Therefore, R’ = R/(0.99)⁴ = R/[1 – (0.01)]⁴ = R [1 – (0.01)]^–4

This gives R’ = 1.04R, nearly.

(4) The resistance of an ammeter of range 2 A is R. The shunt resistance required to make its range 6 A is

(a) R

(b) R/5

(c) R/6

(d) R/2

(e) R/3

Let S be the shunt resistance required. When a current of 6 A is passed through the ammeter- shunt combination, 2 A should pass through the ammeter and the remaining 4 A should pass through the shunt (fig.). Therefore we have

2×R = 4×S, on equating the potential difference across the ammeter resistance R to the potential difference across the shunt resistance S.

This gives S = R/2

The following questions are for AP Physics C aspirants (even though AP Physics B aspirants also can try them):

(5) A power supply having internal resistance r drives a current through a load of resistance P. The heat developed in the load resistance is H. When the same power supply drives a current through another load of resistance Q for the same time t, the heat developed in the load resistance is H itself. The internal resistance of the power supply is for a time

(a) √(PQ)

(b) √[(P² + Q²)/2]

(c) (P + Q)/2

(d) (P – Q)/2

(e) √[(P² – Q²)/2]

Since H = I²Rt where I is the current through the load resistance R, we have

H = [V/(P+r)]²Pt = [V/(Q+r)]²Qt where V is the emf of the power supply.

[V/(P+r) is the current in the load resistance P and V/(Q+r) is the current in the load resistance Q].

From the above equation we have

P/(P+r)² = Q/(Q+r)²

Or, P/(P² +r² +2Pr) = Q/(Q²+r² + 2Qr)

Or, PQ² + Pr² + 2PQr = QP² + Qr² + 2PQr from which

r²(P – Q) = QP² – PQ²

Or, r²(P – Q) = PQ(P – Q) so that r = √(PQ)

(6) A 12 V battery having internal resistance 0.2 Ω is being charged at a current of 2 A using the circuit shown in the figure. The generator which charges the battery has a terminal voltage of 16 V at this charging current. What is the value of the resistance R connected in the circuit?

(a) 0.2 Ω

(b) 0.8 Ω

(c) 1.8 Ω

(d) 2.6 Ω

(e) 3.4 Ω

The voltage between the terminals of the battery (terminal potential difference) while the battery is being charged (which is equal to the sum of the emf of the battery and the potential drop across its internal resistance) = 12 + 0.4 = 12.4 V.

[0.4 V is the voltage drop (2×0.2 volt) across the internal resistance of the battery].

Therefore, the voltage drop across the series resistance R is (16 – 12.4) V = 3.6 V.

Since the charging current is 2 A, we have 2 ×R = 3.6 from which R = 1.8 Ω.

You will find some useful multiple choice questions (with solution) in this section here.