Physics 2 Quiz: Kirchhoffs Current Law

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Kirchhoffs Current LawQuestion 1 of 8

In a resistive network, four branches meet at node P. Branch A carries $I_A = 4\text{ A}$ directed away from P. Branch B carries $I_B = 7\text{ A}$ directed toward P. Branch C carries $I_C$ directed toward P. Branch D carries $I_D = 2I_C$ directed away from P. All currents are steady-state DC values.

Which of the following correctly gives $I_C$ , the current in Branch C?

I_C = 1\text{ A}

, found by setting the total current into node P equal to the total current out of node P and solving the resulting linear equation in

I_C

I_C = -1\text{ A}

, found by treating all branch currents as positive quantities and subtracting outgoing from incoming currents, yielding a negative value that indicates Branch C actually flows away from P.

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.

I_C = 7\text{ A}

, found by equating Branch B's inward current directly to Branch C's inward current and ignoring the constraint imposed by Branches A and D on the node balance.

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Physics 2 Quiz

Physics 2 Quiz: Kirchhoffs Current Law

Practice Kirchhoffs Current Law 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 Kirchhoffs Current 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.

All questions

Question 1

Which of the following correctly gives $I_C$ , the current in Branch C?

$I_C = 1\text{ A}$ , found by setting the total current into node P equal to the total current out of node P and solving the resulting linear equation in $I_C$ .
$I_C = -1\text{ A}$ , found by treating all branch currents as positive quantities and subtracting outgoing from incoming currents, yielding a negative value that indicates Branch C actually flows away from P.
$I_C = 3\text{ A}$ , 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. (correct answer)
$I_C = 7\text{ A}$ , found by equating Branch B's inward current directly to Branch C's inward current and ignoring the constraint imposed by Branches A and D on the node balance.

Explanation: 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.

Question 2

In a planar resistive network, a student identifies two adjacent nodes, M and N, connected by a shared branch carrying current $I_{MN}$ . At node M, the KCL equation (inward positive) is: $8 - 3 + I_{MN} = 0$ . At node N, the KCL equation (inward positive) is: $-I_{MN} + 5 - 2 = 0$ .

Are both KCL equations self-consistent, and what is the value and direction of $I_{MN}$ ?

The equations are self-consistent. $I_{MN} = -5\text{ A}$ , meaning the current flows from N to M at a magnitude of $5\text{ A}$ . The sign flip between the two equations correctly reflects that the current enters one node and leaves the other through the shared branch.
The equations are inconsistent. Node M gives $I_{MN} = -5\text{ A}$ and node N gives $I_{MN} = -3\text{ A}$ ; since these differ, at least one branch current in the network must be incorrectly specified, indicating an error in the problem setup.
The equations are self-consistent. $I_{MN} = -5\text{ A}$ from node M and $I_{MN} = +3\text{ A}$ from node N, and the difference arises because the two nodes use opposite sign conventions, so both results describe the same physical current flowing from N to M.
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. (correct answer)

Explanation: 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.

Question 3

A student is analyzing a multi-loop DC circuit and identifies a node where three wires connect. The student writes the following KCL equation for that node: $I_1 + I_2 + I_3 = 0$ , where $I_1, I_2,$ and $I_3$ are signed currents with the sign convention that current flowing into the node is positive. After solving the system of equations, the student finds $I_1 = 5\text{ A}$ , $I_2 = -2\text{ A}$ , and $I_3 = -3\text{ A}$ .

Which of the following statements best describes the physical meaning of these results?

The results are physically inconsistent because KCL requires at least one current to flow into and at least one to flow out of every node; having two negative currents violates the requirement that the node cannot source or sink charge.
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 $5 = 2 + 3$ . (correct answer)
The results are physically valid only if the node is a supernode enclosing a voltage source, because ordinary nodes cannot have more outward currents than inward currents by the standard formulation of KCL.
The results are physically valid, but the student's KCL equation is incorrectly written. The correct form should be $I_1 = I_2 + I_3$ , which gives $5 = -2 + (-3) = -5\text{ A}$ , a contradiction showing the solution is wrong.

Explanation: 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.

Question 4

A DC network contains three nodes (A, B, C) and a ground reference. The following branch currents have been determined: $I_{AB} = 2\text{ A}$ (from A to B), $I_{BC} = 5\text{ A}$ (from B to C), $I_{CA} = 3\text{ A}$ (from C to A), $I_{Ag} = 1\text{ A}$ (from A to ground), $I_{Bg} = 0\text{ A}$ , and $I_{Cg} = 4\text{ A}$ (from C to ground). There are no independent sources.

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?

KCL fails only at node B, where the incoming current ( $I_{AB} = 2\text{ A}$ ) does not equal the outgoing current ( $I_{BC} = 5\text{ A}$ ), giving a discrepancy of $3\text{ A}$ that is not accounted for by $I_{Bg} = 0$ .
KCL fails only at the ground node, where the total current arriving at ground ( $I_{Ag} + I_{Bg} + I_{Cg} = 1 + 0 + 4 = 5\text{ A}$ ) has no return path in the absence of sources, so the error lies exclusively in the ground-branch currents rather than in the node-to-node branches.
KCL fails at all three nodes simultaneously, because in a network with no sources, the only self-consistent solution is zero current in every branch, and any nonzero branch currents necessarily violate KCL at every node.
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). (correct answer)

Explanation: 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.

Question 5

A student is analyzing a ladder network with the following properties: at node $k$ (for $k = 1, 2, 3, \ldots$ ), the current entering from the left series branch is $I_k$ , the current leaving through the shunt (parallel) branch to ground is $i_k$ , and the current exiting to the right through the next series branch is $I_{k+1}$ . The student writes the recurrence relation: $I_{k+1} = I_k - i_k$ .

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?

$I_5 = 0\text{ A}$ , 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. (correct answer)
$I_5 = 1\text{ A}$ , found by subtracting only $i_4$ from the previous series current $I_4$ , without accounting for the cumulative shunting from nodes 1 through 3, effectively applying KCL only at the final node.
$I_5 = 4\text{ A}$ , found by subtracting the total shunt current ( $3 + 4 + 2 + 1 = 10\text{ A}$ ) from the input and then dividing by the number of sections, treating each node as sharing the shunted current equally.
$I_5 = 10\text{ A}$ , found by assuming that the series current is conserved throughout the ladder and that the shunt currents $i_k$ are supplied independently by voltage sources at each node rather than drawn from the series current $I_k$ .

Explanation: 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.

Question 6

A transmission line junction has three conductors meeting at a single node. Conductor 1 carries a phasor current $\tilde{I}_1 = (4 + j3)\text{ A}$ into the node. Conductor 2 carries a phasor current $\tilde{I}_2 = (1 - j5)\text{ A}$ into the node. All currents are sinusoidal at the same frequency.

If KCL is applied to find the phasor current $\tilde{I}_3$ leaving the node through Conductor 3, what is $\tilde{I}_3$ ?

$\tilde{I}_3 = (5 + j8)\text{ A}$ , found by adding the real parts of $\tilde{I}_1$ and $\tilde{I}_2$ and separately adding their imaginary parts, then assigning the result as an outgoing phasor without sign adjustment.
$\tilde{I}_3 = (5 - j2)\text{ A}$ , found by applying KCL in phasor form: the sum of phasors entering equals the sum of phasors leaving, so $\tilde{I}_3 = \tilde{I}_1 + \tilde{I}_2 = (4+1) + j(3 + (-5)) = 5 - j2\text{ A}$ . (correct answer)
$\tilde{I}_3 = (3 + j8)\text{ A}$ , found by subtracting $\tilde{I}_2$ from $\tilde{I}_1$ since one conductor carries current with a lagging imaginary component, making it effectively the outgoing branch in the reactive sense.
$\tilde{I}_3 = 5\text{ A}$ , found by computing the magnitude of $\tilde{I}_1 + \tilde{I}_2$ using the Pythagorean theorem: $\sqrt{5^2 + (-2)^2} \approx 5.4\text{ A}$ , then rounding to the nearest integer for the outgoing real current.

Explanation: 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.

Question 7

A student is using the node-voltage method to analyze a circuit. At a particular node X, four branches are connected. The student correctly determines that the KCL equation for node X can be written as:

$\frac{V_X - V_A}{R_1} + \frac{V_X - V_B}{R_2} + \frac{V_X}{R_3} = I_S$

where $V_A = 12\text{ V}$ , $V_B = -6\text{ V}$ , $R_1 = 4\text{ Ω}$ , $R_2 = 3\text{ Ω}$ , $R_3 = 6\text{ Ω}$ , and $I_S = 2\text{ A}$ is a current source directed into node X. After solving, the student obtains $V_X = 6\text{ V}$ .

Using $V_X = 6\text{ V}$ , what is the current flowing through $R_2$ , and does it flow toward or away from node X?

The current through $R_2$ is $4\text{ A}$ , flowing toward node X from node B, because the imaginary part of the phasor current reverses the conventional direction, making node B the effective high-potential terminal in this branch.
The current through $R_2$ is approximately $1.3\text{ A}$ , flowing away from node X, found by computing the potential difference $V_X - V_B = 12\text{ V}$ but dividing by the series combination $R_2 + R_3 = 9\text{ Ω}$ , as if the two resistors share a single loop.
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. (correct answer)
The current through $R_2$ is $4\text{ A}$ , but its direction cannot be established from node voltages alone — Kirchhoff's Voltage Law (KVL) around the loop containing $R_2$ must be applied separately to confirm whether the current flows toward or away from node X.

Explanation: 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.

Question 8

A student is writing nodal equations for a circuit with nodes labeled 1, 2, 3, and a reference (ground). The student writes the following KCL equation for node 2, where all conductances are in siemens and all node voltages are measured with respect to ground:

$G_a(V_2 - V_1) + G_b(V_2 - V_3) + G_c V_2 = I_s$

Here, $I_s$ is a current source connected between node 2 and ground, directed into node 2. $G_a$ connects nodes 1 and 2; $G_b$ connects nodes 2 and 3; $G_c$ connects node 2 to ground.

Which of the following correctly identifies whether this KCL equation is written correctly, and if not, what error it contains?

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. (correct answer)
The equation is incorrect. The term $G_c V_2$ should be $G_c(V_2 - V_{\text{ground}})$ , and since $V_{\text{ground}} \neq 0$ in general, omitting the ground voltage introduces an error that violates KCL at node 2.
The equation is incorrect. The source current $I_s$ should appear on the left side with a negative sign, giving $G_a(V_2-V_1) + G_b(V_2-V_3) + G_c V_2 - I_s = 0$ , because KCL requires all terms to appear on the same side of the equation before it can be considered properly formulated.
The equation is incorrect. The term $G_a(V_2 - V_1)$ should be written as $G_a(V_1 - V_2)$ to represent current flowing into node 2 from node 1, and similarly $G_b(V_2 - V_3)$ should be $G_b(V_3 - V_2)$ , because KCL at a node must account for inward currents only.

Explanation: 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.