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.

Two Kerala Engineering & Medical Entrance 2008 Questions on Resistive Networks

Questions involving resistive networks can be solved are often found in degree entrance examinations. Here is a reasonably good question which appeared in KEAM 2008 (Engineering) question paper:

In the circuit shown, if the resistance 5 Ω develops a heat of 42 J per second, the heat developed in 2 Ω must be about (in J s^–1)

(a) 25

(b) 20

(c) 30

(d) 35

(e) 40

The heat developed per second (power dissipation) in a resistance R is V²/R where V is the voltage across the resistance

[This expression has the forms I²R and VI where I is the current. Even though you can use these forms also to solve the problem, the form V²/R will be more convenient here]

Therefore we have

V²/5 = 42 watt

The power dissipated in the parallel branch containing the resistors 6 Ω and 9 Ω is V²/(6+9) = V²/15 and this must be 14 watt (since the denominator is 3 times)

The total power dissipated in the two parallel branches is thus 42+14 = 56 watt and the effective resistance of the two parallel branches is 5×15/(5+15) = 75/20 Ω.

If I is the current passing through the 2 Ω resistance, we have

I²×2 = W where W is the power dissipated in 2 Ω.

Since the same current passes through the effective resistance (75/20 Ω) of the parallel branches, we have

I²×75/20 = 56

From the above equations (on dividing),

40/75 = W/56 so that W = 40×56/75 = 30 watt nearly

The following MCQ which appeared in KEAM 2008 (Medical) question paper is simple:

In the Wheatstone’s network shown in the figure, the current I in the circuit is

(a) 1 A

(b) 2A

(c) 0.25 A

(d) 0.5 A

(e) 0.33 A

The bridge is balanced (since 2 Ω/4 Ω = 4 Ω/8 Ω). Therefore, the points B and D are equipotential points and there is no current through the diagonal 5 Ω resistor. The circuit thus reduces to 6 Ω and 12 Ω in parallel with the 2 V battery. The parallel combined value of 6 Ω and 12 Ω is 6×12/(6+12) = 4 Ω.

The current, I =2 V/4 Ω = 0.5 A.