Kirchhoff's Voltage Law - Physics 2

1. Using KVL, the student wants to determine $R_{ext}$. Which of the following correctly applies KVL to find $R_{ext}$, and correctly identifies what the terminal voltage represents in the KVL equation?
   Answer: 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. When analyzing a battery circuit with internal resistance, the key framework is Kirchhoff's Voltage Law (KVL): the EMF must equal the sum of all voltage drops around the loop. The critical insight is understanding what "terminal voltage" physically means — it's the voltage measured at the battery's terminals, which equals the voltage drop across the external resistor, not the internal one. Starting from KVL: $\mathcal{E} - Ir - IR_{ext} = 0$, which you can rewrite as $\mathcal{E} - Ir - V_{\text{terminal}} = 0$ since $V_{\text{terminal}} = IR_{ext}$. Plugging in: $I = (\mathcal{E} - V_{\text{terminal}})/r = (10 - 8)/2 = 1\text{ A}$, giving $R_{ext} = 8/1 = 8\text{ }\Omega$. This is exactly what C does, and its conceptual description is correct — the terminal voltage equals the EMF minus the internal drop. A is wrong because it sets up KVL incorrectly with only two terms and misidentifies the terminal voltage as the driving source across the internal resistance — a fundamental conceptual reversal. B makes a critical error: it sets $V_{\text{terminal}} = Ir$, meaning it treats terminal voltage as the internal drop rather than the external drop. This is the most common misconception on this topic and leads to a nonsensical physical picture. D incorrectly adds $Ir$ to the EMF in the KVL equation (sign error), which would only apply if the battery were being charged. The negative current result is a red flag that the loop equation was set up wrong. Remember: terminal voltage always equals the voltage across what's outside the battery. When you see $V_{\text{terminal}} < \mathcal{E}$, the battery is discharging — never confuse this with charging.
2. Applying Kirchhoff's voltage law to this loop with an assumed clockwise current direction, which of the following equations correctly represents the loop, and what is the resulting current magnitude?
   Answer: $12 - 4I - 6 - 2I + 3 - 3I = 0$, 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. Whenever you see a Kirchhoff's Voltage Law (KVL) problem, your job is to apply a consistent sign convention as you traverse the loop. The standard rule: if you cross an EMF source from negative to positive terminal (i.e., positive terminal first in your traversal direction), you gain voltage — write +ε. If you cross it from positive to negative terminal, you lose voltage — write −ε. For resistors, if you traverse in the same direction as assumed current, you lose voltage — write −IR. Answer A applies these rules correctly. Traversing clockwise: $\mathcal{E}_1$ is entered at its positive terminal, giving $+12$. Then $R_1$ opposes the assumed current: $-4I$. $\mathcal{E}_2$ is entered at its negative terminal (a drop): $-6$. Then $-2I$. $\mathcal{E}_3$ is entered at its positive terminal: $+3$. Then $-3I$. This yields $12 - 4I - 6 - 2I + 3 - 3I = 0$, so $9 - 9I = 0$, giving $I = 1\text{ A}$ clockwise. ✓ Answer B is wrong because it adds resistor drops instead of subtracting them — the opposite of the correct sign convention for resistors traversed in the current direction. Answer C incorrectly treats $\mathcal{E}_2$ as $+6$ despite it being entered negative-terminal-first. The terminal orientation determines the sign, not whether the source "supplies energy." Answer D makes the opposite error with $\mathcal{E}_3$, treating it as $-3$ even though it's entered positive-terminal-first, which should give a voltage gain. Study tip: Draw a small $+/-$ label on each EMF as you traverse — the terminal you enter determines the sign. Positive terminal entry = voltage rise; negative terminal entry = voltage drop.
3. What is the correct voltage across the $10\text{ }\Omega$ resistor and the correct polarity of its terminals?
   Answer: 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. Whenever KVL gives you a negative current, that result is actually telling you something physically meaningful — don't just discard the sign. The negative sign means the actual current flows opposite to your assumed direction. If you assumed clockwise and got $I = -0.5\text{ A}$, the real current is $0.5\text{ A}$ counter-clockwise. This matters enormously for resistor polarity. The voltage magnitude across the $10\text{ }\Omega$ resistor is $|I| \times R = 0.5 \times 10 = 5\text{ V}$. But the high-potential terminal is determined by which side the actual current enters. Since actual current flows counter-clockwise, it enters the resistor from the right side — and by Ohm's law, current always enters a resistor at its higher-potential terminal. Therefore, the right terminal is at higher potential. That confirms B. Choice A is a classic trap: it treats polarity labels as permanently fixed to the assumed direction, ignoring what the negative sign physically reveals. Once you know current is actually counter-clockwise, the original polarity labels flip. Choice D makes the reasonable-sounding argument that the larger battery "wins" and drives clockwise current — but this is wrong because you must account for battery orientation, not just magnitude. The 4 V battery opposes the 9 V battery, and KVL already encoded all of that; the result is counter-clockwise current. Choice C incorrectly concludes the voltage is indeterminate. Two opposing batteries don't cancel to zero here — they produce a net $5\text{ V}$ drop across the resistor with a perfectly definite polarity. Study tip: Treat a negative KVL current as a direction correction, not an error. Flip the assumed direction, then re-assign all resistor polarities accordingly before interpreting the circuit.
4. Applying KVL to find the current in this loop, which analysis is correct?
   Answer: 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. When analyzing circuits with batteries in series, Kirchhoff's Voltage Law (KVL) requires you to track each EMF's sign based on direction, while every internal resistance always adds to the total resistance — regardless of battery orientation. This is the core concept being tested here. Walking through the correct analysis: assume current flows clockwise. The two clockwise batteries each contribute $+6\text{ V}$, while the counter-clockwise battery opposes the assumed direction, contributing $-6\text{ V}$. The net EMF is $6 + 6 - 6 = 6\text{ V}$. Crucially, all three internal resistances resist current flow regardless of orientation, so total resistance is $R + 3r = 3 + 3(1) = 6\text{ }\Omega$. This gives $I = 6/6 = 1\text{ A}$. Because current flows through the opposing battery in the direction that pushes charge against its EMF, that battery is being charged — not discharging. This confirms D. Choice A makes a critical error: it subtracts the opposing battery's internal resistance, as if orientation flips its contribution to resistance. It doesn't — internal resistance always dissipates energy. B ignores the opposing sign entirely, treating all three EMFs as aiding and dropping two internal resistances, producing a wildly inflated current. C gets the resistance right (implicitly $6\text{ }\Omega$) but miscounts the EMF signs, suggesting two batteries oppose the assumed direction when only one does, yielding the wrong magnitude and an incorrect claim about discharging. Your study tip: in any KVL problem, handle EMF signs and resistance separately. Signs depend on traversal direction; resistance is always positive and always sums.
5. When the student writes the KVL equation for Loop 2 only, which expression is correct?
   Answer: $(R_1 + R_2 + R_3)I_2 - R_1 I_1 = 0$, 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. Whenever you tackle mesh analysis (KVL with assigned loop currents), your job is to track every voltage drop around a given loop, paying close attention to how mesh currents interact on shared branches. For Loop 2, the resistors present are $R_1$ (shared with Loop 1), $R_2$, and $R_3$. Since both $I_1$ and $I_2$ are clockwise, they flow in opposite directions through the shared branch containing $R_1$. The net current through $R_1$ is therefore $(I_2 - I_1)$, making the voltage drop across it $R_1(I_2 - I_1)$. Adding the drops across $R_2$ and $R_3$, which carry only $I_2$, and setting the total equal to zero (no EMF in Loop 2) gives: $R_1(I_2 - I_1) + R_2 I_2 + R_3 I_2 = 0 \implies (R_1 + R_2 + R_3)I_2 - R_1 I_1 = 0$ This is exactly answer A, and the reasoning is sound: the $-R_1 I_1$ term reflects the opposing contribution of the adjacent mesh current. B is wrong because it adds $I_1$ and $I_2$ through $R_1$, which would only be correct if the currents flowed in the same direction through that branch — they don't. C incorrectly writes $+R_1 I_1$, implying the currents add through $R_1$, when they actually subtract because the mesh currents oppose each other in that branch. D is wrong because it excludes $I_2$ from the $R_1$ term entirely — $I_2$ absolutely passes through $R_1$ in the mesh formulation; that's the whole point of the shared branch. Your study tip: always draw arrows for each mesh current on the shared branch first. If arrows point opposite, subtract; if they point the same way, add. This single check prevents the most common mesh analysis mistakes.
6. What is the KVL equation for the loop containing Branches 1 and 2 (the loop with $I_1$ and $I_2$)?
   Answer: $12 - 6 - 4(I_1 - I_2) = 0$, 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. When applying mesh analysis, your job is to write KVL equations by tracing each loop and carefully accounting for every voltage source and resistor — including how shared branches carry contributions from multiple mesh currents. For Loop 1 (Branches 1 and 2), trace clockwise starting at Node A. Traveling through Branch 1 (A→B), you cross the 12 V battery from negative to positive terminal, giving +12 V. Returning through Branch 2 (B→A), you first cross the 6 V battery — its positive terminal is at B, so traveling B→A means entering the positive terminal and exiting the negative terminal, which is a voltage drop: −6 V. Then you cross the 4 Ω resistor. Since $I_1$ flows through Branch 2 in the same direction as your traversal, and $I_2$ flows opposite to your traversal through that shared branch, the net current is $(I_1 - I_2)$, giving a drop of $-4(I_1 - I_2)$. This yields: $12 - 6 - 4(I_1 - I_2) = 0$, confirming C is correct. A is wrong because the 8 Ω resistor in Branch 3 is not part of Loop 1 — it only appears in Loop 1's equation if the branch is shared by that loop, which it isn't. B fails on two counts: the batteries don't add (they oppose each other in this loop), and it ignores $I_2$'s contribution to the shared Branch 2 resistor. D incorrectly treats the 6 V battery as a rise rather than a drop. Entering the positive terminal during traversal is always a drop in KVL. A reliable strategy: always note which terminal you enter when crossing a battery — entering the positive terminal = voltage drop (−), entering the negative terminal = voltage rise (+). This single rule prevents most KVL sign errors.
7. Is the bridge balanced, and what does this imply for the galvanometer current and the application of KVL?
   Answer: 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. Whenever you encounter a Wheatstone bridge problem, your first move should always be to check the balance condition: $\frac{R_1}{R_3} = \frac{R_2}{R_4}$. If this ratio holds, the potentials at the two middle nodes are equal, and no current flows through the galvanometer — regardless of its resistance. Here, $\frac{R_1}{R_3} = \frac{5}{10} = 0.5$ and $\frac{R_2}{R_4} = \frac{15}{30} = 0.5$. The ratios match, confirming the bridge is balanced and $V_C = V_D$. With zero potential difference across the galvanometer, no current flows through it. You can therefore remove the galvanometer branch entirely and treat the circuit as two independent series combinations: the ACB branch ($5 + 10 = 15\ \Omega$) and the ADB branch ($15 + 30 = 45\ \Omega$) in parallel across the 20 V source. KVL on each outer loop is straightforward with no C–D branch to worry about. This is answer B. A introduces a fabricated condition — that $R_g$ must equal some geometric mean for balance. This is simply not a real requirement. The standard cross-product $R_1 R_4 = R_2 R_3$ (equivalent to the ratio condition) is both necessary and sufficient for balance; $R_g$ is irrelevant. C correctly identifies the balance condition but then contradicts itself by claiming current flows anyway. Current requires a potential difference — if $V_C = V_D$, then $\Delta V = 0$ and Ohm's law gives $I_g = 0$. D confuses branch resistance equality ($R_1 + R_3 = R_2 + R_4$) with the actual balance condition. Equal total branch resistances are not required; equal ratios within each branch are. Strategy tip: Memorize the ratio form $R_1/R_3 = R_2/R_4$ — it's faster than the cross-product and directly shows you why the middle potentials equalize.
8. At steady state, what does KVL predict for the voltage across the capacitor $V_C$?
   Answer: $V_C = 18\text{ V}$, 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. Whenever you see a capacitor in a DC circuit at steady state, the single most important fact to recall is that a fully charged capacitor acts as a perfect open circuit — zero current flows through it. Because the capacitor is in a single series loop with $R_A$ and $R_B$, blocking the capacitor means blocking current through the entire loop. With $I = 0\text{ A}$, Ohm's Law tells you the voltage drop across each resistor is $V = IR = (0)(R) = 0\text{ V}$. Applying KVL around the loop: $18\text{ V} - V_{R_A} - V_C - V_{R_B} = 0$, which simplifies to $18 - 0 - V_C - 0 = 0$, giving $V_C = 18\text{ V}$. Answer D is correct. Answer A is tempting but fatally flawed: it applies a voltage-divider formula assuming current flows, then assigns that divided voltage to the capacitor — a contradiction, since the capacitor is what stops the current. Answer B makes the same error, correctly identifying no current flows but then paradoxically using a current-dependent voltage-divider calculation anyway, arriving at $\frac{R_B}{R_A + R_B} \times 18 = 12\text{ V}$. Answer C invents a fictitious "equal energy storage" principle that has no basis in circuit analysis; energy storage in a capacitor and resistors are entirely different quantities and cannot be equated this way. Study tip: In any steady-state DC problem, identify capacitors first and replace them with open circuits. This immediately zeroes out current in any branch containing a capacitor, which cascades through your KVL/KCL analysis.

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