Multiple Choice Practice Questions involving Power in Direct Current Circuits for AP Physics B Exam.

If you grasp fundamental principles thoroughly, you will be able to answer complicated questions without much difficulty. The following questions meant for AP Physics aspirants are relatively simple but they will definitely help you to test your understanding of fundamental principles.

(1) For transferring 50 coulomb of charge through a 10 Ω resistor, the work required is 150 J. The potential difference across the resistor is

(a) 500 V

(b) 50 V

(c) 10 V

(d) 5 V

(e) 3 V

We have W = VQ where W is the work done in transferring Q coulombs of charge through a potential difference of V volt.

Therefore, V = W/Q = 150/50 volt = 3 V.

(2) Four resistors R₁, R₂, R₃ and R₄ having resistances 12 Ω, 16 Ω, 26 Ω and 36 Ω respectively are connected in parallel with a battery of negligible internal resistance. The current in the 12 Ω resistor is 0.5 A. What is the electric power dissipated in the 36 Ω resistor?

(a) 0.5 W

(b) 1 W

(c) 6 W

(d) 18 W

(e) 36 W

Since the resistors are in parallel, the potential difference across them will be the same. (This is true even if the battery has internal resistance).

The potential difference across the 12 Ω resistor is 12×0.5 volt = 6 V.

The potential difference across the 36 Ω resistor also is 6 V.

Therefore, the power dissipated in the 36 Ω resistor = V²/R = 6²/36 = 1 W.

(3) Two 24 V, 16 W electric lamps are connected in series and this series combination is connected across a 12 V battery of negligible internal resistance. The power consumed by each lamp is

(a) 1 W

(b) 2 W

(c) 4 W

(d) 8 W

(e) 16 W

If the resistance of each lamp is R, we have (from power, P = V²/R)

16 = 24²/R ………..(i)

When the series combination of the lamps is connected across a 12 V battery, each lamp has 6 volt across it. The power (X) consumed by each lamp is therefore given by

X = 6²/R …………(ii)

From equations (i) and (ii) X = 1 W.

(4) A 12 V battery of internal resistance 3 Ω is connected in series with a 1 Ω resistor and a variable resistor. The power dissipated in the variable resistor will be maximum when the current through it is

(a) 0.5 A

(b) 1 A

(c) 2 A

(d) 3 A

(e) 4 A

According to maximum power transfer theorem maximum power transfer occurs when the external resistance is equal to the internal resistance of the battery. Therefore, the variable resistor and the fixed 1 Ω resistor together must be 3 Ω for maximum power transfer. The current in the circuit then is 12 V/6 Ω = 2 A [Option (c)].

You will find similar useful questions at AP Physics Resources and at physicsplus.