College Physics Quiz: Compound Direct Current Circuits

4 questions · exam conditions

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Compound Direct Current CircuitsQuestion 1 of 4

Two identical batteries, each with EMF $12\text{ V}$ and internal resistance $2\,\Omega$ , are connected in parallel to a load resistor $R_L$ . If the power delivered to the load is maximum, what is the efficiency of power transfer from the batteries to the load?

25\%

33\%

50\%

67\%

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College Physics Quiz

College Physics Quiz: Compound Direct Current Circuits

Practice Compound Direct Current Circuits in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Compound Direct Current Circuits, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$25\%$
$33\%$
$50\%$ (correct answer)
$67\%$

Explanation: Two identical batteries in parallel have equivalent EMF

12\text{ V}

and equivalent internal resistance

2\,\Omega/2 = 1\,\Omega

. Maximum power transfer occurs when the load resistance equals the internal resistance, so

R_L = 1\,\Omega

. The total circuit resistance is

1 + 1 = 2\,\Omega

, giving current

I = 12/2 = 6\text{ A}

. Power delivered to load:

P_L = I^2 R_L = 36 \times 1 = 36\text{ W}

. Power lost in internal resistance:

P_{internal} = I^2 \times 1 = 36\text{ W}

. Total power:

P_{total} = 36 + 36 = 72\text{ W}

. Efficiency =

P_L/P_{total} = 36/72 = 50\%

. This is a general result: maximum power transfer always occurs at 50% efficiency.

Question 2

A complex circuit network contains 5 nodes and multiple resistors. Using Kirchhoff's current law (KCL) at each node, a system of equations is set up. If the currents $I_1 = 3\text{ A}$ , $I_2 = -1\text{ A}$ , $I_3 = 2\text{ A}$ , and $I_4 = -2\text{ A}$ are known, and the equation for node E is $I_3 + I_5 - I_6 + I_7 = 0$ , what additional constraint is needed to solve for $I_5$ , $I_6$ , and $I_7$ ?

One additional KCL equation from another node that involves these currents
Two additional KVL equations around independent loops that include these currents
The values of all resistances in branches containing $I_5$ , $I_6$ , and $I_7$
Both KVL equations and resistance values, since KCL alone provides insufficient constraints (correct answer)

Explanation: The single KCL equation

I_3 + I_5 - I_6 + I_7 = 0

provides only one constraint for three unknowns (

I_5

I_6

I_7

). With

I_3 = 2\text{ A}

, we get

2 + I_5 - I_6 + I_7 = 0

, or

I_5 - I_6 + I_7 = -2

. This is one equation with three unknowns, which is insufficient. We need two additional independent equations. These must come from Kirchhoff's voltage law (KVL) around loops that contain these current-carrying branches, combined with Ohm's law (

V = IR

) which requires knowing the resistance values. KCL equations alone cannot provide the additional constraints needed because they only relate currents at nodes, not the voltage drops that determine the current magnitudes through specific resistances.

Question 3

A circuit contains three identical resistors, each with resistance $R$ , connected to form a triangle. A battery with EMF $\mathcal{E}$ and internal resistance $r$ is connected between two vertices of the triangle. What is the current delivered by the battery?

$\frac{\mathcal{E}}{R + r}$
$\frac{2\mathcal{E}}{3R + 2r}$
$\frac{3\mathcal{E}}{2R + 3r}$ (correct answer)
$\frac{\mathcal{E}}{\frac{2R}{3} + r}$

Explanation: The three resistors form a triangle, and the battery is connected between two vertices. From the battery's perspective, one resistor (

R

) is in parallel with the series combination of the other two resistors (

2R

). The equivalent resistance of this parallel combination is

\frac{R \cdot 2R}{R + 2R} = \frac{2R^2}{3R} = \frac{2R}{3}

. The total resistance in the circuit is the internal resistance plus this equivalent resistance:

r + \frac{2R}{3}

. Therefore, the current is

I = \frac{\mathcal{E}}{r + \frac{2R}{3}} = \frac{3\mathcal{E}}{3r + 2R} = \frac{3\mathcal{E}}{2R + 3r}

Question 4

A student builds a circuit to measure an unknown resistance $R_x$ using a Wheatstone bridge configuration. The bridge uses known resistors: $R_1 = 100\,\Omega$ , $R_2 = 150\,\Omega$ , and $R_3 = 200\,\Omega$ . When the bridge is balanced (zero current through the galvanometer), what is $R_x$ , and what happens to the bridge balance if $R_1$ is increased to $120\,\Omega$ ?

$R_x = 300\,\Omega$ ; current flows from junction A to junction B through the galvanometer (correct answer)
$R_x = 300\,\Omega$ ; current flows from junction B to junction A through the galvanometer
$R_x = 133\,\Omega$ ; current flows from junction A to junction B through the galvanometer
$R_x = 133\,\Omega$ ; current flows from junction B to junction A through the galvanometer

Explanation: For a balanced Wheatstone bridge,

\frac{R_1}{R_2} = \frac{R_3}{R_x}

. Substituting:

\frac{100}{150} = \frac{200}{R_x}

, which gives

R_x = \frac{200 \times 150}{100} = 300\,\Omega

. When

R_1

increases to

120\,\Omega

, the ratio

\frac{R_1}{R_2} = \frac{120}{150} = 0.8

becomes greater than

\frac{R_3}{R_x} = \frac{200}{300} = 0.667

. This means the voltage at junction A (between

R_1

and

R_2

) becomes higher than the voltage at junction B (between

R_3

and

R_x

). Therefore, current flows from the higher potential junction A to the lower potential junction B through the galvanometer.