Physics 2 Quiz: Kirchhoffs Voltage Law

What this quiz covers

This quiz focuses on Kirchhoffs Voltage Law, 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.

Question 1

1. $V_C = 6\text{ V}$, because in steady state the capacitor acts as an open circuit, and the 18 V divides proportionally between the two resistors; since $R_B$ is twice $R_A$, the capacitor charges to the voltage across the smaller resistor.
2. $V_C = 12\text{ V}$, because in steady state the current is determined by the series resistance, and the capacitor voltage is found by applying the voltage-divider rule using only $R_B$ out of the total resistance.
3. $V_C = 9\text{ V}$, because in steady state the capacitor blocks DC and the voltage divides equally between $R_A$ and $R_B$ due to equal energy storage in each element, leaving half the battery voltage for the capacitor.
4. $V_C = 18\text{ V}$, because in steady state no current flows through the series loop, so there are no resistive voltage drops, and KVL requires the capacitor to account for the entire EMF. (correct answer)

Question 2

1. KVL gives $V_{\text{terminal}} - Ir = 0$, so $I = V_{\text{terminal}}/r = 4\text{ A}$, and $R_{ext} = (\mathcal{E} - V_{\text{terminal}})/I = 0.5\text{ }\Omega$. The terminal voltage serves as the effective source voltage driving current through the internal resistance once the external load is connected.
2. KVL gives $\mathcal{E} - I(r + R_{ext}) = 0$, so $I = \mathcal{E}/(r + R_{ext})$. Setting $V_{\text{terminal}} = Ir$ (the drop across the internal resistance) gives $I = 8/2 = 4\text{ A}$, and $R_{ext} = (\mathcal{E} - Ir)/I = 4\text{ }\Omega$. The terminal voltage represents the portion of the EMF dissipated internally.
3. KVL gives $\mathcal{E} - Ir - V_{\text{terminal}} = 0$, where $V_{\text{terminal}} = IR_{ext}$. Solving: $I = (\mathcal{E} - V_{\text{terminal}})/r = 1\text{ A}$, so $R_{ext} = V_{\text{terminal}}/I = 8\text{ }\Omega$. The terminal voltage equals the voltage across the external resistor and equals the EMF minus the internal voltage drop. (correct answer)
4. KVL gives $\mathcal{E} + Ir - V_{\text{terminal}} = 0$, so $I = (V_{\text{terminal}} - \mathcal{E})/r = -1\text{ A}$, giving $R_{ext} = V_{\text{terminal}}/|I| = 8\text{ }\Omega$. The terminal voltage being less than the EMF indicates the battery is being charged, and the internal drop is subtracted from the terminal voltage to recover the EMF.

Question 3

1. $(R_1 + R_2 + R_3)I_2 - R_1 I_1 = 0$, because the shared branch carries net current $(I_2 - I_1)$ and the voltage across $R_1$ due to $I_1$ opposes the $I_2$ contribution, while no EMF appears in Loop 2. (correct answer)
2. $(R_2 + R_3)I_2 + R_1(I_1 + I_2) = 0$, because mesh currents flowing through a shared resistor always add regardless of their relative directions, and no EMF source is present in Loop 2.
3. $(R_1 + R_2 + R_3)I_2 + R_1 I_1 = 0$, because the two mesh currents flow in opposite directions through $R_1$, so their contributions to the voltage drop across $R_1$ must be summed to get the total drop in Loop 2.
4. $(R_2 + R_3)I_2 - R_1 I_1 = 0$, because $R_1$ is a shared element and its voltage drop in Loop 2 is determined solely by $I_1$, the current from the adjacent loop, while $I_2$ does not pass through $R_1$ in the mesh formulation.

Question 4

1. $12 - 4I - 6 - 2I + 3 - 3I = 0$, giving $I = 1\text{ A}$ clockwise, because each EMF traversed positive-terminal-first contributes a positive term and each resistor drop opposes the assumed current direction. (correct answer)
2. $12 + 4I - 6 + 2I + 3 + 3I = 0$, giving $I = -\frac{9}{9} = -1\text{ A}$, because resistor voltage drops are added when traversing in the direction of assumed current and EMFs are treated as fixed positive quantities.
3. $12 - 4I + 6 - 2I + 3 - 3I = 0$, giving $I = \frac{21}{9} \approx 2.3\text{ A}$ clockwise, because the second EMF, though encountered negative-terminal-first, still supplies energy to the loop and must be added.
4. $12 - 4I - 6 - 2I - 3 - 3I = 0$, giving $I = \frac{3}{9} = 0.33\text{ A}$ clockwise, because all EMF sources encountered while traversing the loop in the current direction produce voltage drops regardless of terminal orientation.

Question 5

1. $12 - 6 - 4I_1 - 8I_2 = 0$, because the loop traversal must account for the voltage drop across the $8\text{ }\Omega$ resistor in Branch 3, which is shared between both mesh loops and therefore appears in both KVL equations.
2. $12 + 6 - 4I_1 = 0$, because both batteries drive current in the same direction around the Loop 1 mesh when the clockwise direction is defined consistently, and only $I_1$ flows through the resistor in Branch 2.
3. $12 - 6 - 4(I_1 - I_2) = 0$, because traversing from A to B through Branch 1 gives +12 V, and returning B to A through Branch 2 gives −6 V (opposing EMF) and a resistor drop that accounts for both mesh currents sharing that branch. (correct answer)
4. $12 + 6 - 4(I_1 - I_2) = 0$, because the 6 V battery in Branch 2 is oriented with its positive terminal at B, and traversing B to A through Branch 2 in the clockwise direction means entering the positive terminal, which registers as a voltage rise in the mesh equation.

Question 6

1. The magnitude is $5\text{ V}$, and the left terminal is at higher potential, because KVL polarity assignments are locked to the assumed current direction; a negative result changes only the current's magnitude interpretation, not the polarity labels assigned during the initial traversal.
2. The magnitude is $5\text{ V}$, and the right terminal is at higher potential, because the negative current result means actual current flows counter-clockwise, entering the resistor from the right side, and current enters a resistor at its high-potential terminal. (correct answer)
3. The magnitude is $5\text{ V}$, and neither terminal is definitively at higher potential, because a negative KVL current indicates the two batteries oppose each other with equal effect, leaving the resistor in an indeterminate voltage state that requires additional information to resolve.
4. The magnitude is $5\text{ V}$, and the left terminal is at higher potential, because the 9 V battery is larger than the 4 V battery and dominates the loop, driving current clockwise from its positive terminal on the left side through the resistor, making the left terminal the entry point for conventional current.

Question 7

1. The net EMF is $6 + 6 - 6 = 6\text{ V}$ but internal resistances cancel for the opposing battery, giving total resistance $3 + 2(1) = 5\text{ }\Omega$ and $I = 1.2\text{ A}$, because the reversed battery's internal resistance subtracts from the circuit resistance rather than adding to it.
2. The net EMF driving the loop is $6 + 6 + 6 = 18\text{ V}$ and the total resistance is $3 + 1 = 4\text{ }\Omega$, giving $I = 4.5\text{ A}$, because all three EMF sources contribute positively to current flow regardless of orientation.
3. The net EMF driving the loop is $6 - 6 - 6 = -6\text{ V}$, meaning $I = -1\text{ A}$; the negative sign means the assumed direction was wrong, so the magnitude is $1\text{ A}$ but all three batteries are discharging equally in the corrected direction.
4. The net EMF driving the loop is $6 + 6 - 6 = 6\text{ V}$ and the total resistance is $3 + 3(1) = 6\text{ }\Omega$, giving $I = 1\text{ A}$. The opposing battery is being charged by the current from the other two. (correct answer)

Question 8

1. The bridge is unbalanced: although the cross-product check gives $R_1 R_4 = 5 \times 30 = 150$ and $R_2 R_3 = 15 \times 10 = 150$, these values being equal is a necessary but not sufficient condition for balance — the galvanometer resistance $R_g$ must also equal the geometric mean of the bridge resistors for zero galvanometer current, which is not satisfied here.
2. The bridge is balanced because $R_1/R_3 = R_2/R_4$ (both equal $1/2$), so no current flows through the galvanometer; the galvanometer branch can be removed, and KVL applied to each outer loop yields a consistent current in each branch without needing to account for the C–D branch. (correct answer)
3. The bridge is balanced because $R_1/R_3 = 5/10 = 0.5$ and $R_2/R_4 = 15/30 = 0.5$, confirming $V_C = V_D$; however, the galvanometer current is nonzero because current always flows through any resistive path that exists between two nodes, regardless of the potential difference across it.
4. The bridge is unbalanced because the correct balance condition requires equal total resistance in each parallel branch: since $R_1 + R_3 = 15\text{ }\Omega \neq R_2 + R_4 = 45\text{ }\Omega$, the potentials at C and D differ, driving a nonzero galvanometer current that must be included in all KVL loop equations.