Physics 2 Quiz: Interpreting Circuit Diagrams

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Interpreting Circuit DiagramsQuestion 1 of 2

A student reads a circuit diagram and identifies the following: a battery $\mathcal{E} = 20\ \text{V}$ (internal resistance negligible) drives current through an outer loop. The outer loop contains two nodes P and Q. Between P and Q there are three parallel branches: Branch 1 contains only resistor $R_1 = 10\ \Omega$ ; Branch 2 contains resistor $R_2 = 20\ \Omega$ in series with another resistor $R_3 = 20\ \Omega$ ; Branch 3 contains an ideal current source $I_s = 0.5\ \text{A}$ directed from Q to P (i.e., upward in the diagram). The battery is in a fourth branch, also between P and Q, with its positive terminal at P.

From the circuit diagram, a student must determine the conventional current through $R_1$ . Which of the following is correct?

The current through

R_1

is 2 A directed from P to Q, because the battery fixes the voltage across all parallel branches at 20 V, the ideal current source does not affect the terminal voltage since the battery is ideal, and

I_{R_1} = 20\ \text{V} / 10\ \Omega = 2\ \text{A}

The current through

R_1

is 1.5 A directed from P to Q, because the ideal current source

I_s = 0.5\ \text{A}

directed from Q to P reduces the net voltage across the parallel combination, effectively lowering the terminal voltage to

20 - 0.5 \times 10 = 15\ \text{V}

, so

I_{R_1} = 15\ \text{V} / 10\ \Omega = 1.5\ \text{A}

The current through

R_1

is 2 A directed from Q to P, because the ideal current source

I_s

directed from Q to P reverses the polarity of the voltage across all parallel branches, driving current upward (Q to P) through all resistive branches including

R_1

, and the magnitude is still

20\ \text{V} / 10\ \Omega = 2\ \text{A}

The current through

R_1

is 2.5 A directed from P to Q, because the ideal current source contributes additional current into node P, which superimposes on the battery-driven current; by superposition, the total current through

R_1

equals

20\ \text{V}/10\ \Omega + 0.5\ \text{A} \times (10\ \Omega / 10\ \Omega) = 2.5\ \text{A}

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Physics 2 Quiz

Physics 2 Quiz: Interpreting Circuit Diagrams

Practice Interpreting Circuit Diagrams in Physics 2 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 Interpreting Circuit Diagrams, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

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

From the circuit diagram, a student must determine the conventional current through $R_1$ . Which of the following is correct?

The current through $R_1$ is 2 A directed from P to Q, because the battery fixes the voltage across all parallel branches at 20 V, the ideal current source does not affect the terminal voltage since the battery is ideal, and $I_{R_1} = 20\ \text{V} / 10\ \Omega = 2\ \text{A}$ . (correct answer)
The current through $R_1$ is 1.5 A directed from P to Q, because the ideal current source $I_s = 0.5\ \text{A}$ directed from Q to P reduces the net voltage across the parallel combination, effectively lowering the terminal voltage to $20 - 0.5 \times 10 = 15\ \text{V}$ , so $I_{R_1} = 15\ \text{V} / 10\ \Omega = 1.5\ \text{A}$ .
The current through $R_1$ is 2 A directed from Q to P, because the ideal current source $I_s$ directed from Q to P reverses the polarity of the voltage across all parallel branches, driving current upward (Q to P) through all resistive branches including $R_1$ , and the magnitude is still $20\ \text{V} / 10\ \Omega = 2\ \text{A}$ .
The current through $R_1$ is 2.5 A directed from P to Q, because the ideal current source contributes additional current into node P, which superimposes on the battery-driven current; by superposition, the total current through $R_1$ equals $20\ \text{V}/10\ \Omega + 0.5\ \text{A} \times (10\ \Omega / 10\ \Omega) = 2.5\ \text{A}$ .

Explanation: When a circuit contains both a battery and an ideal current source connected between the same two nodes, the first thing to ask is: what controls the node voltage? An ideal voltage source (your battery) enforces a fixed potential difference between its terminals regardless of what else is connected in parallel — this is the defining property of an ideal voltage source. Since the battery fixes

V_{PQ} = 20\ \text{V}

, every parallel branch sees exactly that voltage, full stop. With

V_{PQ} = 20\ \text{V}

locked in, the current through

R_1

follows directly from Ohm's Law:

I_{R_1} = 20\ \text{V} / 10\ \Omega = 2\ \text{A}

, directed from the high-potential node P toward Q. That confirms A is correct. B is wrong because it treats the current source as if it creates a voltage drop that "subtracts" from the battery. An ideal current source in parallel with an ideal voltage source cannot alter the node voltage — the battery overrides it. The current source simply adjusts how much current the battery itself must supply, not the terminal voltage. C is wrong on two counts: the direction is reversed without justification, and it misunderstands which element controls polarity. The battery's positive terminal at P keeps P at higher potential; current through resistors flows P → Q. D misapplies superposition. When you activate the current source alone (battery replaced by a short), the node voltage becomes zero (both nodes shorted), so the current source drives no additional current through

R_1

. Superposition gives the same 2 A result as A, not 2.5 A. Study tip: Whenever an ideal voltage source and any other element share the same two nodes, the voltage source wins — node voltage is fixed, and you apply Ohm's Law directly.

Question 2

A student is analyzing a circuit diagram of a DC network with three nodes (A, B, C) and four branches. The branch currents and their defined positive directions are: $I_1$ from A to B through $R_1$ ; $I_2$ from B to C through $R_2$ ; $I_3$ from A to C through $R_3$ ; and $I_4$ from C to A through a battery $\mathcal{E}$ (positive terminal at A). The student writes KCL at all three nodes and obtains: Node A: $-I_1 - I_3 + I_4 = 0$ ; Node B: $I_1 - I_2 = 0$ ; Node C: $I_2 + I_3 - I_4 = 0$ .

Which of the following statements about the student's KCL equations is correct?

All three equations are correctly written and are independent, meaning they can be used simultaneously with KVL to solve for all four branch currents without any redundancy or inconsistency.
All three equations are correctly written, but they are not all independent — any one of the three equations can be derived from the other two, so only two of the three KCL equations can be used as independent constraints in solving the circuit. (correct answer)
The equation at Node C is incorrect because the student failed to account for the battery's EMF when writing KCL; batteries contribute to current balance at a node only if their internal resistance is nonzero, and since $\mathcal{E}$ is ideal, the $I_4$ term should not appear in the Node C equation.
The equation at Node A is incorrect because $I_4$ flows from C to A, meaning it enters Node A, so it should appear as $+I_4$ on the left side only when KCL is written as 'currents entering = currents leaving,' but the student's sign convention mixes entering and leaving currents inconsistently across the three equations.

Explanation: Whenever you see a question involving Kirchhoff's Current Law applied at every node of a circuit, you should immediately ask yourself: are all these equations truly independent? This is a fundamental property of KCL that's easy to overlook. Here's the key insight: in any circuit with

N

nodes, applying KCL at every node produces exactly

N

equations, but only

N-1

of them are independent. Why? Because every branch current appears exactly twice across all node equations — once leaving a node (negative) and once entering another (positive). This means if you add all the KCL equations together, every current cancels, giving

0 = 0

. That redundancy proves one equation is always derivable from the others. You can verify this directly: adding the student's three equations gives

(-I_1 - I_3 + I_4) + (I_1 - I_2) + (I_2 + I_3 - I_4) = 0,

which is

0 = 0

identically. So the Node C equation is just the negative sum of Nodes A and B — it carries no new information. Answer B is correct. Answer A is wrong because it claims all three equations are independent, which contradicts this fundamental property. Only two can serve as independent constraints. Answer C is wrong because KCL is purely a current-conservation law — it counts charges flowing in and out of a node, regardless of whether a branch contains a resistor, battery, or anything else. Batteries don't get special treatment in KCL. Answer D is wrong because the student's sign convention is actually consistent: currents leaving Node A are negative, and

I_4

enters Node A, so

+I_4

is correct. Study tip: For any circuit with

N

nodes, always use only

N-1

KCL equations. The last one is always redundant — don't waste it as an independent equation in your system.