Physics 2 Quiz: Multi Loop Circuits

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Multi Loop CircuitsQuestion 1 of 4

Two batteries with EMFs $\mathcal{E}_1 = 10 \text{ V}$ and $\mathcal{E}_2 = 4 \text{ V}$ and internal resistances $r_1 = 1 \text{ } \Omega$ and $r_2 = 2 \text{ } \Omega$ respectively are connected in a loop with an external resistor $R = 3 \text{ } \Omega$ . Battery 1's positive terminal connects to battery 2's positive terminal through $R$ , and both batteries' negative terminals are connected together. Applying KVL around the single loop, what is the current in the circuit and the rate at which battery 2 is charging or discharging?

I = 1 \text{ A}

clockwise (from

\mathcal{E}_1

's positive terminal through

R

\mathcal{E}_2

's positive terminal); battery 2 is charging at a rate of

4 \text{ W}

because current enters its positive terminal, indicating it absorbs energy from the circuit.

I = 1 \text{ A}

clockwise; battery 2 is discharging at a rate of

4 \text{ W}

because the current flows out of its positive terminal in the defined direction, indicating it delivers energy to the circuit just as battery 1 does.

I = 2.33 \text{ A}

clockwise, found by dividing the larger EMF by the total resistance

(10/4.3)

; battery 2 is charging because the larger battery always forces the smaller one to charge when they are connected in opposition.

I = 1 \text{ A}

clockwise; battery 2 is neither charging nor discharging because the two batteries' EMFs drive currents in opposite directions that exactly cancel in battery 2's branch, resulting in zero net power transfer to battery 2.

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Physics 2 Quiz

Physics 2 Quiz: Multi Loop Circuits

Practice Multi Loop Circuits 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 Multi Loop Circuits, 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

$I = 1 \text{ A}$ clockwise (from $\mathcal{E}_1$ 's positive terminal through $R$ to $\mathcal{E}_2$ 's positive terminal); battery 2 is charging at a rate of $4 \text{ W}$ because current enters its positive terminal, indicating it absorbs energy from the circuit. (correct answer)
$I = 1 \text{ A}$ clockwise; battery 2 is discharging at a rate of $4 \text{ W}$ because the current flows out of its positive terminal in the defined direction, indicating it delivers energy to the circuit just as battery 1 does.
$I = 2.33 \text{ A}$ clockwise, found by dividing the larger EMF by the total resistance $(10/4.3)$ ; battery 2 is charging because the larger battery always forces the smaller one to charge when they are connected in opposition.
$I = 1 \text{ A}$ clockwise; battery 2 is neither charging nor discharging because the two batteries' EMFs drive currents in opposite directions that exactly cancel in battery 2's branch, resulting in zero net power transfer to battery 2.

Explanation: When two batteries oppose each other in a loop, your first instinct should be to apply Kirchhoff's Voltage Law (KVL): the algebraic sum of all EMFs and voltage drops around the loop equals zero. The key skill here is correctly identifying which battery "wins" and what that means for the other one. Choose a clockwise current direction. Traversing the loop,

\mathcal{E}_1

drives current clockwise while

\mathcal{E}_2

opposes it. KVL gives:

\mathcal{E}_1 - \mathcal{E}_2 - I(r_1 + r_2 + R) = 0

10 - 4 - I(1 + 2 + 3) = 0 \implies I = \frac{6}{6} = 1 \text{ A clockwise}

Now, the critical insight: current flows into battery 2's positive terminal (since

\mathcal{E}_1

dominates and pushes current against

\mathcal{E}_2

's natural direction). When current enters a battery's positive terminal, the battery is being charged — it absorbs power at a rate of

P = \mathcal{E}_2 \cdot I = 4 \times 1 = 4 \text{ W}

. This confirms answer A is correct. B is wrong because it claims current exits battery 2's positive terminal, making it a source — the opposite of what happens. Current actually enters that terminal, so battery 2 absorbs, not delivers, energy. C uses an incorrect formula; you cannot simply divide one EMF by total resistance — you must use the net EMF (

\mathcal{E}_1 - \mathcal{E}_2

) when batteries oppose each other. D is wrong because power transfer to battery 2 is nonzero; zero net power would require

I = 0

, which only happens if both EMFs were equal. Study tip: Always check which terminal current enters on each battery. Entering the positive terminal = charging (absorbing power); exiting the positive terminal = discharging (supplying power). This single rule resolves most battery-opposition problems.

Question 2

In a two-loop circuit, Kirchhoff's current law (KCL) is applied at a junction node, yielding $I_1 + I_2 = I_3$ . After applying KVL to both loops, the system of equations produces $I_1 = 2 \text{ A}$ and $I_3 = 5 \text{ A}$ . A student claims that the power delivered by a $10 \text{ V}$ battery (in the branch carrying $I_2$ , with current flowing out of its positive terminal) is 30 W. Which of the following best evaluates this claim?

The claim is correct, because $I_2 = I_3 - I_1 = 3 \text{ A}$ , and the power delivered by the battery is $P = \mathcal{E} I_2 = 10 \times 3 = 30 \text{ W}$ , confirming that the battery acts as a source supplying energy to the circuit. (correct answer)
The claim is incorrect, because the power delivered by the battery must account for internal resistance; without knowing the internal resistance, the delivered power cannot be determined from EMF and branch current alone.
The claim is incorrect, because $I_2 = I_1 + I_3 = 7 \text{ A}$ , and the correct power is $P = 10 \times 7 = 70 \text{ W}$ , since KCL requires all currents entering a node to sum to the current leaving it.
The claim is incorrect, because $I_2 = I_3 - I_1 = 3 \text{ A}$ is right, but the power delivered is $P = \mathcal{E}^2 / R_{\text{total}}$ , which requires knowledge of the circuit's total resistance rather than just the branch current.

Explanation: When tackling multi-loop circuit problems, your workflow should always follow the same sequence: apply KCL at junctions first, then use KVL to find individual currents, and finally calculate power. Keeping these steps distinct prevents the most common errors. Here, KCL at the junction gives

I_1 + I_2 = I_3

. Solving for the unknown branch current:

I_2 = I_3 - I_1 = 5 - 2 = 3 \text{ A}

. Since the problem states current flows out of the positive terminal of the 10 V battery, the battery is acting as a source — it is delivering energy to the circuit. The power delivered by a source is simply

P = \mathcal{E} \cdot I

, giving

P = 10 \times 3 = 30 \text{ W}

. The student's claim is correct, making A the right answer. B introduces a real concept — internal resistance does reduce terminal voltage — but the question specifies the EMF source directly and asks for power delivered by the battery, meaning we use the given EMF and branch current as stated. No additional information is needed here. C misapplies KCL by adding

I_1

and

I_3

to get

I_2

. This gets the junction equation backwards. KCL states that currents entering a node equal currents leaving; since

I_1

and

I_2

both enter while

I_3

leaves,

I_2 = I_3 - I_1

, not

I_1 + I_3

. D correctly finds

I_2 = 3 \text{ A}

but then abandons the valid

P = \mathcal{E} I

formula in favor of

P = \mathcal{E}^2 / R_{\text{total}}

, which is unnecessary and inapplicable when the branch current is already known. A useful rule of thumb: once you know a source's EMF and the current through it,

P = \mathcal{E} I

is always sufficient — don't overcomplicate it by reaching for resistance-based formulas.

Question 3

In a two-mesh circuit, the mesh equations are $10I_1 - 4I_2 = 20$ and $-4I_1 + 8I_2 = -8$ . After solving, a student correctly finds $I_1 = 2 \text{ A}$ and $I_2 = 0 \text{ A}$ . The student then claims: "Since $I_2 = 0$ , the right-hand loop carries no current, so the branch shared between the two meshes carries only $I_1$ , and the right-hand loop is effectively an open circuit." Which statement best assesses this claim?

The claim is fully correct: a mesh current of zero means no current flows anywhere in that loop, the shared branch carries only $I_1 = 2 \text{ A}$ , and the right-hand loop's elements can be removed without affecting any currents in the left-hand loop.
The claim is partially correct but misleading: $I_2 = 0$ does mean the shared branch carries only $I_1 = 2 \text{ A}$ , which is correct, but the branches exclusive to the right loop also carry $I_2 = 0 \text{ A}$ , so those branches dissipate no power — yet calling the loop an 'open circuit' incorrectly implies a physical break rather than a balanced zero-current condition.
The claim is incorrect: $I_2 = 0$ means the mesh current assigned to the right loop is zero, but the shared branch still carries $I_1 = 2 \text{ A}$ , so current does flow through that element. Branches exclusively in the right loop carry zero current, but this is a balanced electrical condition — not an open circuit — and the right loop's elements are still physically present and connected. (correct answer)
The claim is incorrect because $I_2 = 0$ is a mathematical artifact of the chosen mesh directions; reversing the direction of $I_2$ would yield $I_2 = -2 \text{ A}$ , which shows the right loop does carry current and is not an open circuit.

Explanation: Whenever you work with mesh analysis, keep this distinction sharp: mesh currents are mathematical tools, not necessarily the physical current in any single branch. The actual current through a branch is found by combining whatever mesh currents share that branch. Here,

I_2 = 0

tells you the right-loop mesh current is zero — nothing more, nothing less. The shared branch between the two meshes carries

I_1 - I_2 = 2 - 0 = 2 \text{ A}

, which is physically real current flowing through a physically connected element. Meanwhile, branches exclusive to the right loop carry only

I_2 = 0 \text{ A}

, meaning no current passes through them — but that's because of how the circuit voltages and resistances balance out, not because those elements are missing or broken. This is why C is correct: the right loop is electrically balanced at zero mesh current, not an open circuit. A is wrong because it goes too far — claiming you can remove the right loop's elements without affecting the left loop ignores that those elements are still part of the network and their presence shapes the mesh equations themselves. B is tempting but ultimately misplaces the error. It accepts "open circuit" language as merely a poor word choice, when in fact the deeper conceptual mistake is conflating a zero mesh current with a physical disconnection — making the claim incorrect, not just misleading. D is wrong because mesh current direction is a chosen convention; reversing

I_2

's direction would give

I_2 = 0 \text{ A}

regardless, since zero has no sign ambiguity. No "artifact" is hiding a nonzero current. Study tip: Always distinguish between a mesh current and a branch current — on circuit analysis questions, the trap is almost always treating one as the other.

Question 4

In a multi-loop circuit, Kirchhoff's voltage law is applied around a loop containing two resistors $R_1 = 10 \text{ } \Omega$ and $R_2 = 15 \text{ } \Omega$ in series with a single battery $\mathcal{E} = 25 \text{ V}$ . However, $R_1$ is shared with an adjacent loop that carries an additional mesh current $I_2 = 1 \text{ A}$ in the opposite direction to the primary mesh current $I_1$ . If the loop equation for the primary mesh gives $I_1 = 1.5 \text{ A}$ , what is the actual current through $R_1$ and the power it dissipates?

The actual current through $R_1$ is $I_1 + I_2 = 2.5 \text{ A}$ (since the adjacent mesh current flows opposite to $I_1$ , the two currents add in the branch), and the power dissipated is $P = (2.5)^2(10) = 62.5 \text{ W}$ .
The actual current through $R_1$ is $I_1 - I_2 = 0.5 \text{ A}$ (since the adjacent mesh opposes $I_1$ through the shared branch, the net current is reduced), and the power dissipated is $P = (0.5)^2(10) = 2.5 \text{ W}$ . (correct answer)
The actual current through $R_1$ is $I_1 = 1.5 \text{ A}$ (mesh currents are defined per loop, so $R_1$ carries only the primary mesh current regardless of adjacent loops), and the power dissipated is $P = (1.5)^2(10) = 22.5 \text{ W}$ .
The actual current through $R_1$ is $\sqrt{I_1^2 + I_2^2} = \sqrt{3.25} \approx 1.80 \text{ A}$ (since the two mesh currents are in orthogonal loops, their contributions add in quadrature), and the power dissipated is $P \approx (1.80)^2(10) \approx 32.5 \text{ W}$ .

Explanation: Whenever you see a multi-loop circuit problem using mesh analysis, your first instinct should be to ask: what is the actual branch current? Mesh currents are mathematical tools — the real current in any shared branch is the algebraic sum of all mesh currents flowing through it, with careful attention to direction. In this problem,

R_1

sits in a shared branch. The primary mesh current

I_1 = 1.5 \text{ A}

flows through it in one direction, while the adjacent mesh current

I_2 = 1 \text{ A}

flows through it in the opposite direction. When two currents oppose each other in the same branch, you subtract:

I_{R_1} = I_1 - I_2 = 1.5 - 1.0 = 0.5 \text{ A}

. The power dissipated is then

P = I^2 R = (0.5)^2(10) = 2.5 \text{ W}

, confirming B is correct. A gets the arithmetic backwards. When two mesh currents flow in opposite directions through a branch, they partially cancel — they don't add. Adding them would only be correct if both currents flowed in the same direction through

R_1

. C makes the critical mistake of treating mesh currents as if they were real, isolated branch currents. Mesh currents are fictitious loop variables; you must always combine them to find the true physical current in any shared element. D introduces a completely fabricated rule. Mesh currents are not vectors in perpendicular spatial directions — they are scalar loop variables in a circuit. Quadrature addition has no place here. Study tip: For any shared branch in mesh analysis, always write

I_{\text{branch}} = I_{\text{mesh,A}} \pm I_{\text{mesh,B}}

, where the sign depends on whether the adjacent mesh current aids or opposes the primary one. Never use a mesh current alone as a branch current without checking for sharing.