Kirchhoff's Current Law - Physics 2

1. A student checks KCL at each node and at the ground node. At which node(s) does KCL fail, indicating an error in the stated branch currents?
   Answer: KCL fails at both node B and node C. At B: in $= 2\text{ A}$, out $= 5\text{ A}$, discrepancy $= 3\text{ A}$. At C: in $= 5\text{ A}$, out $= 3 + 4 = 7\text{ A}$, discrepancy $= 2\text{ A}$. Node A satisfies KCL independently ($3\text{ A}$ in $= 2 + 1 = 3\text{ A}$ out). Whenever you're asked to verify KCL in a multi-node network, your job is systematic: at every node, sum all currents entering and subtract all currents leaving — the result must be zero. Apply this to each node and the ground independently, and let the math reveal where consistency breaks down. Start with node A: currents in = $I_{CA} = 3\text{ A}$; currents out = $I_{AB} + I_{Ag} = 2 + 1 = 3\text{ A}$. KCL holds — no discrepancy. Now node B: currents in = $I_{AB} = 2\text{ A}$; currents out = $I_{BC} + I_{Bg} = 5 + 0 = 5\text{ A}$. Discrepancy = $3\text{ A}$ — KCL fails. Now node C: currents in = $I_{BC} = 5\text{ A}$; currents out = $I_{CA} + I_{Cg} = 3 + 4 = 7\text{ A}$. Discrepancy = $2\text{ A}$ — KCL fails. This confirms D is correct: KCL fails at B and C, but node A checks out on its own. A is wrong because it identifies only node B, ignoring the equally real violation at node C. B is wrong because it misattributes the entire problem to the ground node — the errors originate at the internal nodes, not exclusively in the ground branches. C is wrong because a network without sources can still carry consistent nonzero currents (e.g., in a loop); the claim that all currents must be zero is false. Strategy tip: Always check every node individually rather than stopping at the first failure. On circuit problems, errors are often distributed across multiple nodes, and missing one costs you the correct answer.
2. Suppose it is found that $I_1 = 10\text{ A}$, $i_1 = 3\text{ A}$, $i_2 = 4\text{ A}$, $i_3 = 2\text{ A}$, and $i_4 = 1\text{ A}$. What is $I_5$, the current entering the fifth series section?
   Answer: $I_5 = 0\text{ A}$, found by applying the recurrence four times: $I_2 = 7$, $I_3 = 3$, $I_4 = 1$, $I_5 = 0\text{ A}$, indicating that all input current has been shunted to ground by the fourth node. Ladder networks are a classic application of Kirchhoff's Current Law (KCL), and the key insight is that the recurrence relation $I_{k+1} = I_k - i_k$ must be applied sequentially and cumulatively — each step depends on the result of the previous one, not on the original input current alone. Starting with $I_1 = 10\text{ A}$, apply the recurrence at each node in order: $I_2 = I_1 - i_1 = 10 - 3 = 7\text{ A}$ $I_3 = I_2 - i_2 = 7 - 4 = 3\text{ A}$ $I_4 = I_3 - i_3 = 3 - 2 = 1\text{ A}$ $I_5 = I_4 - i_4 = 1 - 1 = 0\text{ A}$ This confirms that answer A is correct: all input current has been shunted to ground by the fourth node, leaving nothing to enter the fifth series section. Answer B is wrong because it only subtracts $i_4$ from the original $I_1$, skipping three intermediate steps. KCL must be applied at every node, not just the last one. Answer C introduces a fictitious averaging operation — dividing by the number of sections has no basis in KCL or circuit theory whatsoever. Answer D fundamentally misunderstands the circuit topology: the shunt currents are drawn from the series branch, not supplied by independent sources, so the series current must decrease at every node. Your strategy here: whenever you see a recurrence relation in a circuit problem, always chain each result into the next step. Treat it like a running balance — each node "spends" some current, and you carry the remainder forward.
3. Which of the following correctly gives $I_C$, the current in Branch C?
   Answer: $I_C = 3\text{ A}$, found by noting that $I_D = 2I_C$ means Branch D carries twice the current of Branch C, so the two branches together contribute a net outward current of $I_C$ that must be balanced by Branches A and B alone. Whenever you see a node in a DC circuit, your first instinct should be Kirchhoff's Current Law (KCL): the sum of all currents entering a node equals the sum of all currents leaving it. This is simply conservation of charge — no charge can accumulate at a node in steady state. Setting up KCL for node P, currents in are $I_B = 7\text{ A}$ and $I_C$, while currents out are $I_A = 4\text{ A}$ and $I_D = 2I_C$. Writing in = out: $7 + I_C = 4 + 2I_C$ Solving: $3 = I_C$, so $I_C = 3\text{ A}$. This confirms C is correct. Notice how the relationship $I_D = 2I_C$ means the two branches together produce a net outward contribution of $2I_C - I_C = I_C$, which must be balanced by the net inward contribution from Branches A and B ($7 - 4 = 3\text{ A}$). Both ways of seeing it give the same answer. A is wrong because it claims $I_C = 1\text{ A}$, which doesn't satisfy the equation — plugging in gives $8 \neq 6$. It describes the right method but arrives at a wrong number, likely from an algebra error. B is wrong on two levels: the value is incorrect, and a negative result from a properly set-up KCL equation would signal a direction error, not confirm a physical negative current here. D is wrong because it ignores Branches A and D entirely, violating KCL — you must account for every branch at the node. Study tip: Always write KCL explicitly before solving. Listing every branch with its direction prevents you from accidentally omitting a term — the most common KCL mistake on circuit problems.
4. Are both KCL equations self-consistent, and what is the value and direction of $I_{MN}$?
   Answer: The equations are inconsistent. Node M gives $I_{MN} = -5\text{ A}$ while node N gives $I_{MN} = +3\text{ A}$; because these magnitudes differ ($5 \neq 3$), the network violates charge conservation and no valid steady-state solution exists. Whenever you encounter KCL problems involving a shared branch between two nodes, your job is to check whether the branch current is mathematically consistent across both equations — not just whether each equation looks correct in isolation. Start by solving each KCL equation for $I_{MN}$. At node M: $8 - 3 + I_{MN} = 0 \Rightarrow I_{MN} = -5\text{ A}$. At node N: $-I_{MN} + 5 - 2 = 0 \Rightarrow I_{MN} = +3\text{ A}$. These two results disagree — $-5 \neq +3$ — which means the network as stated cannot satisfy charge conservation simultaneously at both nodes. Answer D is correct: the equations are inconsistent, and no valid steady-state solution exists. Now let's address the traps. Answer A misreads node N's equation entirely — it claims both nodes yield $I_{MN} = -5\text{ A}$, but solving $-I_{MN} + 3 = 0$ from node N actually gives $+3$, not $-5$. Answer B correctly identifies an inconsistency but invents a second wrong value ($-3$) for node N — the actual result from node N is $+3$, so B's arithmetic is flawed. Answer C is the most seductive distractor: it acknowledges the two different values ($-5$ and $+3$) but wrongly attributes the difference to "opposite sign conventions." If the conventions are properly applied — which they are here, since $I_{MN}$ appears with opposite signs in the two equations — a consistent system would give equal magnitudes. A magnitude mismatch ($5 \neq 3$) signals a real error, not a convention difference. Study tip: When a shared branch current is solved from two nodes and the magnitudes differ, that's always a red flag for an inconsistent network — sign flips are expected and fine, but magnitude mismatches are not.
5. Which of the following correctly identifies whether this KCL equation is written correctly, and if not, what error it contains?
   Answer: The equation is correct. Each conductance term represents the current leaving node 2 through that branch (since current flows from higher to lower potential), and setting the sum of outgoing currents equal to the incoming source current $I_s$ is a valid application of KCL. When writing nodal equations, you have flexibility in how you apply KCL — the key is consistency. You can sum currents leaving a node and set them equal to currents entering, or move everything to one side and set the sum to zero. Both formulations are algebraically equivalent and physically valid. The equation given sums all currents leaving node 2 through the resistive branches: $G_a(V_2 - V_1)$ is the current leaving toward node 1, $G_b(V_2 - V_3)$ is the current leaving toward node 3, and $G_c V_2$ is the current leaving to ground. Setting this equal to $I_s$ (the current entering from the source) perfectly satisfies KCL: total current out equals total current in. A is correct. B is wrong because ground is the reference node by definition, so $V_{\text{ground}} = 0$ always. Writing $G_c(V_2 - V_{\text{ground}})$ is identical to $G_c V_2$. There is no omission or error here — this is a common misconception about what "reference node" means. C is wrong because KCL has no requirement about which side of the equation terms appear on. Moving $I_s$ to the left and writing $-I_s = 0$ is mathematically identical to the original — it's the same equation rearranged. "Proper formulation" doesn't dictate sign placement. D is wrong because it confuses the sign convention. If you choose to sum outgoing currents, then $G_a(V_2 - V_1)$ is correct — positive when current flows away from node 2. Forcing all terms to represent inward currents is one valid choice, but it's not the only valid choice. Study tip: Always decide upfront whether you're summing currents in or currents out, then apply that convention consistently across all branches — either approach gives the correct nodal equation.
6. Which of the following statements best describes the physical meaning of these results?
   Answer: The results are physically valid. $I_1 = 5\text{ A}$ flows into the node, while $I_2 = 2\text{ A}$ and $I_3 = 3\text{ A}$ flow out of the node, and the magnitudes satisfy charge conservation since $5 = 2 + 3$. Whenever you see a question involving KCL and signed currents, your first job is to correctly interpret what the signs mean — they encode direction relative to your chosen convention, not some absolute property of the node. Here, the student defined into the node as positive. So a positive result means current physically flows in, and a negative result means current physically flows out. With $I_1 = 5\text{ A}$, $I_2 = -2\text{ A}$, and $I_3 = -3\text{ A}$: one branch carries 5 A into the node, while the other two carry 2 A and 3 A out. Checking conservation: $5 = 2 + 3$ ✓. The node neither creates nor destroys charge — this is exactly what KCL demands. B is correct. A is wrong because it invents a rule that doesn't exist. KCL only requires that the algebraic sum of currents equals zero — there is no constraint on how many currents flow in versus out. A node with one inflow and five outflows is perfectly valid. C is wrong for a similar reason. Supernodes are a technique for handling voltage sources between two non-reference nodes, not a loophole needed to "allow" multiple outward currents. Ordinary nodes handle any current distribution just fine. D is wrong on two counts. First, $I_1 = I_2 + I_3$ is equivalent to $I_1 + I_2 + I_3 = 0$ only if you flip the signs of $I_2$ and $I_3$ — but the student's variables are already signed, so the original equation is correctly written. Second, plugging the solution into the original equation gives $5 + (-2) + (-3) = 0$ ✓, confirming consistency. Study tip: Always anchor sign interpretation to the stated convention before evaluating results. On circuits questions, negative values are information, not errors — they simply mean the assumed direction was opposite to the physical flow.
7. If KCL is applied to find the phasor current $\tilde{I}_3$ leaving the node through Conductor 3, what is $\tilde{I}_3$?
   Answer: $\tilde{I}_3 = (5 - j2)\text{ A}$, found by applying KCL in phasor form: the sum of phasors entering equals the sum of phasors leaving, so $\tilde{I}_3 = \tilde{I}_1 + \tilde{I}_2 = (4+1) + j(3 + (-5)) = 5 - j2\text{ A}$. When you encounter phasor circuit analysis, the key principle to remember is that KCL applies to phasors exactly as it does to real-valued currents — conservation of current holds at every node, just with complex arithmetic. At any node, the sum of phasor currents entering equals the sum of phasor currents leaving. Here, both $\tilde{I}_1$ and $\tilde{I}_2$ enter the node, so $\tilde{I}_3$ (leaving) must equal their phasor sum: $\tilde{I}_3 = \tilde{I}_1 + \tilde{I}_2 = (4+1) + j(3+(-5)) = 5 - j2 \text{ A}$. This is answer B, and the process is straightforward — add real parts separately, add imaginary parts separately. A reaches the numerically correct phasor $(5 - j2)$ through the right arithmetic but then incorrectly claims "no sign adjustment" is needed as if there's ambiguity — the result is actually correct as stated, making A's explanation misleading and self-contradictory. The answer it displays, $(5 + j8)$, doesn't match its own arithmetic, revealing it as a distractor built on careless addition errors. C subtracts $\tilde{I}_2$ from $\tilde{I}_1$ based on a false premise — that a lagging imaginary component changes which branch is "effectively outgoing." KCL doesn't work this way; current direction is defined by the problem setup, not by the sign of the imaginary part. D commits a critical conceptual error: converting the phasor to its magnitude before assigning it as the answer. The magnitude tells you the peak amplitude of the sinusoid, not the phasor itself. Phasors must remain in complex form until you deliberately convert to time domain. Study tip: Always keep phasors in complex rectangular form throughout KCL/KVL calculations. Only convert to magnitude (or time domain) as a final step when the problem explicitly asks for it.
8. Using $V_X = 6\text{ V}$, what is the current flowing through $R_2$, and does it flow toward or away from node X?
   Answer: The current through $R_2$ is $4\text{ A}$, flowing away from node X toward node B, because $V_X = 6\text{ V} > V_B = -6\text{ V}$, so the potential difference drives conventional current from the higher-potential node X through $R_2$ to the lower-potential node B. When you see a question about the node-voltage method, your core tool for finding branch currents is simple: the current through any resistor equals the voltage difference across it divided by its resistance, and the sign tells you direction. For $R_2$, the branch connects node X to node B. The potential difference is $V_X - V_B = 6 - (-6) = 12 \text{ V}$. Dividing by $R_2 = 3 \text{ Ω}$ gives $I = \frac{12}{3} = 4 \text{ A}$. Because $V_X > V_B$, conventional current flows from the higher-potential side (node X) through $R_2$ toward the lower-potential side (node B) — meaning it flows away from node X. That's exactly what C describes, making it correct. A is wrong on two levels: while it arrives at 4 A, it invents a fictional "phasor reversal" concept that doesn't apply here — this is a DC resistive circuit, and phasors are irrelevant. The direction it concludes is also backwards. B uses the right voltage difference ($V_X - V_B = 12 \text{ V}$) but then incorrectly treats $R_2$ and $R_3$ as a series combination, dividing by $R_2 + R_3 = 9 \text{ Ω}$. In the node-voltage method, each branch is analyzed independently — $R_3$ connects node X to ground in its own separate branch. D is wrong because node voltages alone are entirely sufficient to determine current direction. KVL is not needed separately; direction is encoded directly in the sign of $V_X - V_B$. Study tip: In node-voltage analysis, always write branch current as $\frac{V_{\text{from}} - V_{\text{to}}}{R}$. If the result is positive, current truly flows in that assumed direction — no extra KVL loop required.

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