Kirchhoff's junction rule is a reflection of (a) conservation of current density vector. (b) conservation of c

NCERT Exemplar · Q5

Q. Kirchhoff's junction rule is a reflection of

(a) conservation of current density vector.

(b) conservation of charge.

(c) the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction.

(d) the fact that there is no accumulation of charges at a junction.

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Concept understanding — Wheatstone Bridge Symmetry

Wheatstone Bridge Symmetry

Four resistors are arranged in a diamond. A battery sits across the top and bottom points; a sensitive galvanometer bridges the left and right points. The question is: when does no current flow through the galvanometer? The answer is symmetry.

The Intuition

Treat the bridge as two parallel voltage dividers — two resistors in series on the left, two in series on the right — with the galvanometer joining their midpoints. If both dividers produce the same midpoint voltage, there is no potential difference across the galvanometer and no current flows: the bridge is balanced. This happens when the resistance ratio on the left arm equals the ratio on the right arm.

Note

Symmetry here means equal ratios, not equal resistances. $1\ \Omega$ with $2\ \Omega$ on one side balances $100\ \Omega$ with $200\ \Omega$ on the other.

The Precise Statement

Label the four resistors: $R_1$ top-left, $R_2$ bottom-left, $R_3$ top-right, $R_4$ bottom-right. The battery connects the top junction (between $R_1$ , $R_3$ ) to the bottom junction (between $R_2$ , $R_4$ ); the galvanometer connects the two midpoints. The bridge is balanced when

\frac{R_1}{R_2} = \frac{R_3}{R_4}

Important

The balance condition is independent of the battery voltage and of the galvanometer's resistance — it depends only on the four resistor values.

Why This Matters

Unknown resistance: put the unknown in place of $R_4$ , adjust the others until balanced, then $R_4 = R_3\,\dfrac{R_2}{R_1}$ .
Strain gauges: stretching slightly changes a resistance, unbalancing the bridge; the deflection measures the strain.
Temperature sensors: a temperature-dependent resistor unbalances the bridge in proportion to the change. …

Why this formula?

Wheatstone Bridge Symmetry — Why the Formula Holds

The Wheatstone bridge is a circuit used to measure an unknown resistance by balancing two legs of a bridge. The key result is:

When the bridge is balanced, no current flows through the galvanometer, and:

$\frac{R_1}{R_2} = \frac{R_3}{R_4}$

Let's understand why this is true — not just memorize it.

1. The Circuit Setup

A Wheatstone bridge has four resistors arranged in a diamond:

$R_1$ and $R_2$ in series on the left branch
$R_3$ and $R_4$ in series on the right branch
A galvanometer (sensitive current detector) connects the midpoints of the two branches
A battery connects across the top and bottom

        A
       / \
      /   \
     R1    R3
     |     |
     G-----|   (G = galvanometer)
     |     |
     R2    R4
      \   /
       \ /
        B

2. The Condition for Balance — "No Current Through G"

Balance means the galvanometer shows zero deflection — no current flows through it. This implies points C (between $R_1$ and $R_2$ ) and D (between $R_3$ and $R_4$ ) are at the same electric potential. If $V_C = V_D$ , no potential difference exists across the galvanometer, so no current flows.

3. Deriving the Ratio — Step by Step

Step 1: Branch currents

With no current through G, the left branch carries a single current $I_1$ and the right branch a single current $I_2$ .

Step 2: Equal potentials at the midpoints

From A (common top) to the midpoints, since $V_C = V_D$ :

I_1 R_1 = I_2 R_3 \quad \text{(1)}

From the midpoints to B (common bottom), since $V_C = V_D$ :

I_1 R_2 = I_2 R_4 \quad \text{(2)}

Step 3: Divide (1) by (2)

\frac{I_1 R_1}{I_1 R_2} = \frac{I_2 R_3}{I_2 R_4}

The currents $I_1$ and $I_2$ cancel: …

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Kirchhoff's junction rule, $\sum I_{in} = \sum I_{out}$ , is a direct consequence of conservation of charge applied in steady state at a point — and 'charge is conserved at the junction' and 'no charge accumulates at the junction' are the same statement said two ways. Both (b) and (d) are correct; (a) and (c) are not.

Why the rule exists

A junction is just a point where several wires meet. In steady-state current flow, the charge $Q$ sitting at that point cannot be changing with time — if it were, the junction would be continuously gaining or losing charge forever, which never happens once currents settle down. So

\frac{dQ}{dt} = 0 \quad \text{at the junction (steady state).}

Since current is charge per unit time, $\dfrac{dQ}{dt} = \sum I_{in} - \sum I_{out}$ . Setting this to zero gives exactly Kirchhoff's junction rule:

\sum I_{in} = \sum I_{out}.

Evaluating each option

(a) Wrong. There is no general law that the current density vector $\vec{J}$ is conserved through a junction — different branches can have different cross-sectional areas and different $\vec J$ magnitudes/directions; it is the current (the integral of $\vec J$ over area, i.e. charge/time), not $\vec J$ itself, that balances.
(b) Correct. This is the underlying physical law: electric charge can neither be created nor destroyed, so it cannot pile up at a junction — hence $\sum I_{in}=\sum I_{out}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-M1 markMCQ
$1 \Omega$
›Reveal solutionSolution
The circuit is a symmetric delta network; using Kirchhoff’s laws and symmetry, the current $i_2$ through side PQ is found to be 1 A. The correct option is (C).

The key insight is that the three resistors are identical and form a closed loop (a delta network). The 3 A current enters at vertex P and leaves at vertex R. Because the resistors are all $1\,\Omega$ , the network is symmetric with respect to the line through P and the midpoint of QR — but here the exit is at R, not at the midpoint, so we cannot simply split the current equally. Instead, we must apply Kirchhoff’s current law (KCL) at the junctions and Kirchhoff’s voltage law (KVL) around the loop.

Let’s label the currents clearly:

$i_1$ flows from P to R (right side).

$i_2$ flows from P to Q (left side).

Let $i_3$ be the current from Q to R along the bottom side.

At vertex P, the incoming 3 A splits into $i_1$ and $i_2$ :
$i_1 + i_2 = 3 \quad \text{(1)}$
At vertex Q, current $i_2$ arrives from P, and then leaves via the bottom resistor toward R. So at Q:
$i_2 = i_3 \quad \text{(2)}$
At vertex R, current $i_1$ arrives from P, and current $i_3$ arrives from Q; both leave the network as the outgoing current (which must be 3 A, since current is conserved). So:
$i_1 + i_3 = 3 \quad \text{(3)}$
Now apply Kirchhoff’s voltage law around the triangle P→Q→R→P. Going clockwise:

From P to Q: voltage drop = $i_2 \times 1\,\Omega = i_2$ (positive if we go with the current direction).

From Q to R: voltage drop = $i_3 \times 1\,\Omega = i_3$ (direction from Q to R matches our assumed current).

From R back to P: voltage drop = $-i_1 \times 1\,\Omega = -i_1$ (because we are going opposite to the direction of $i_1$ ).

Sum of voltage drops around the closed loop must be zero:
$i_2 + i_3 - i_1 = 0 \quad \text{(4)}$
Now we have four equations, but only three unknowns. Solve them:

From (2): $i_3 = i_2$ .

Substitute into (1) and (3):

(1): $i_1 + i_2 = 3$

(3): $i_1 + i_2 = 3$ (same equation, so it’s consistent). …
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KCET 2026Set C21 markMCQ
$I_1$
›Reveal solutionSolution

Kirchhoff's Current Law (junction rule) applied successively at each of the 3 branch points gives $I_1$ , $I_2$ , and $I_3$ directly from the branch currents printed in the figure.

Step 1 — First junction

The figure shows $5\,\text{A}$ entering the first junction, with $1\,\text{A}$ branching off downward. By KCL, the current continuing along the main line is
$I_1 = 5\,\text{A} - 1\,\text{A} = 4\,\text{A}$
Step 2 — Second junction

At the second junction, the figure shows $1.5\,\text{A}$ branching off downward from the $I_1 = 4\,\text{A}$ arriving there. By KCL, the current continuing along the main line is
$I_2 = 4\,\text{A} - 1.5\,\text{A} = 2.5\,\text{A}$
Step 3 — Third junction …
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KCET 2025Set D-41 markMCQ
$E_1$
›Reveal solutionSolution

$E_{eq}$ is the weighted mean $(E_1 r_2 + E_2 r_1)/(r_1+r_2)$ , so it lies between the two emfs and is pulled towards the emf whose cell has the smaller internal resistance — here $E_1$ .

Step 1 — Derive the equivalent emf.

For two cells in parallel, the total current is the sum of branch currents. With terminal voltage $V$ :
$I = I_1 + I_2 = \frac{E_1 - V}{r_1} + \frac{E_2 - V}{r_2}$
Rearranging into the form $V = E_{eq} - I\,r_{eq}$ gives the standard results
$E_{eq} = \frac{E_1 r_2 + E_2 r_1}{r_1 + r_2}, \qquad r_{eq} = \frac{r_1 r_2}{r_1 + r_2}$
Step 2 — Note that it is a weighted average.

Write it as
$E_{eq} = \underbrace{\frac{r_2}{r_1+r_2}}_{w_1}E_1 + \underbrace{\frac{r_1}{r_1+r_2}}_{w_2}E_2, \qquad w_1 + w_2 = 1,\ w_1,w_2 > 0$
A convex combination of $E_1$ and $E_2$ must lie strictly between them, so with $E_2 > E_1$ :
$E_1 < E_{eq} < E_2$
This already eliminates (B) and (C).

Step 3 — Decide which emf it is nearer.

Crucially, $E_1$ carries the weight $w_1 = r_2/(r_1+r_2)$ — the other cell's resistance. Given $r_2 > r_1$ : …
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COMEDK 2024Set 2024-A1 markMCQ
$\mathrm{P}$
›Reveal solutionSolution

Imposing $V_B=V_P$ gives $x^2+2x-1=0$ , so the resistance of DP is $\sqrt{2}-1\ \Omega$ .

The square ABCD (total wire $4\,\Omega$ ) has each side $1\,\Omega$ : $AB=BC=CD=DA=1\,\Omega$ . Wire $AP=1\,\Omega$ joins A to P on DC. Let $DP=x$ , $PC=1-x$ . Apply p.d. between A and C; take $V_A=V$ , $V_C=0$ .

Potential of B (branch A–B–C, series $2\,\Omega$ ): $V_B=\dfrac{V}{2}$ .

Impose $V_P=\dfrac{V}{2}$ and use node equations at D and P.

At D: $\dfrac{V_D-V}{1}+\dfrac{V_D-V_P}{x}=0\Rightarrow V_D=\dfrac{V(x+\tfrac12)}{x+1}$ . …
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KCET 2023Set A-31 markMCQ
$5.5~\Omega$
›Reveal solutionSolution

Exploit self-similarity: an infinite ladder looks the same after you strip off its first section, which turns the problem into a quadratic equation for $R$ .

1. The self-similarity trick

Let $R$ be the resistance looking into the ladder at $A$ – $B$ . Strip off the first section: a $2\,\Omega$ resistor in the top rail, a $2\,\Omega$ resistor in the bottom rail, and the first $2\,\Omega$ shunt rung. What lies beyond that rung is again an infinite ladder identical to the original, so its resistance is also $R$ .

Hence the first shunt ( $2\,\Omega$ ) is in parallel with $R$ , and that combination is in series with the two $2\,\Omega$ rail resistors:
$R = 2 + 2 + \left(2 \parallel R\right) = 4 + \frac{2R}{2+R}$
2. Solve the equation

Multiply through by $(2+R)$ :
$R(2+R) = 4(2+R) + 2R$ $2R + R^2 = 8 + 4R + 2R$ $R^2 + 2R - 6R - 8 = 0$ $R^2 - 4R - 8 = 0$ $R = \frac{4 \pm \sqrt{16 + 32}}{2} = \frac{4 \pm \sqrt{48}}{2} = 2 \pm 2\sqrt{3}$
3. Choose the physical root …
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KCET 2020Set A-11 markMCQ
$1 \, \Omega$
›Reveal solutionSolution
By exploiting symmetry to identify equipotential points, the cube’s resistor network simplifies to a combination of series and parallel branches. The equivalent resistance between A and B is $\frac{5}{6} \, \Omega$ , which corresponds to option (A).

The key to solving a cubical resistor network is symmetry. When current enters at A and leaves at B, the geometry of the cube creates points that are at the same potential. You can then “short” those points (connect them without changing the circuit) and redraw the network into a much simpler one.

Let’s work through it step by step.

Identify the symmetry.

The cube has 12 edges, each a $1 \, \Omega$ resistor. Current enters at vertex A and exits at vertex B. Because the cube is symmetric about the line AB, the three vertices that are one edge away from A (call them C, D, E) are all at the same potential. Similarly, the three vertices that are one edge away from B (call them F, G, H) are also at the same potential. The remaining two vertices (the ones diagonally opposite on the faces not containing A or B) are also symmetric.

Tip

In a cube, the vertices that are equidistant from the entry and exit points are equipotential. Here, the three neighbours of A are all at one potential, and the three neighbours of B are at another.

Redraw the network using equipotential points.

Since the three neighbours of A are at the same potential, you can join them into a single node. Same for the three neighbours of B. This collapses the cube into a simpler circuit:

From A, three $1 \, \Omega$ resistors go to the “middle” node (the three neighbours of A).

From that middle node, six resistors go to the other middle node (the three neighbours of B) — but careful: each of the three neighbours of A is connected to two of the neighbours of B via the cube’s edges. That gives $3 \times 2 = 6$ resistors, each $1 \, \Omega$ , all in parallel between the two middle nodes.

Finally, from the second middle node, three $1 \, \Omega$ resistors go to B.

Calculate the equivalent resistance step by step.

From A to the first middle node: three $1 \, \Omega$ resistors in parallel. $R_1 = \frac{1}{3} \, \Omega$ …
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KCET 2019Set A-11 markMCQ
$Q. The readings of ammeter and voltmeter in the following circuit are respectively (A) 1.2 A, 120 V (B) 1.5 A, 100 V (C) 2.7 A, 220 V (D) 2.2 A, 220 V$
›Reveal solutionSolution

In an AC circuit the meters read RMS values. For this series R-L circuit the ammeter reads the RMS current drawn from the $220\ \text{V}$ source and the voltmeter reads $220\ \text{V}$ , so the correct option is (D): $2.2\ \text{A}$ and $220\ \text{V}$ .

In an AC circuit the ammeter and voltmeter measure RMS values. The resistor and inductor are in series across the $220\ \text{V}$ , $50\ \text{Hz}$ supply. The ammeter gives the total RMS current, and the voltmeter reads $220\ \text{V}$ .

1. Inductive reactance.

With $L = \dfrac{0.7}{\pi}\ \text{mH}$ and $f = 50\ \text{Hz}$ ,

$X_L = 2\pi f L = 2\pi \times 50 \times \frac{0.7}{\pi}\times 10^{-3} = 100 \times 0.7 \times 10^{-3} = 0.07\ \Omega$ …
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KCET 2018Set A-11 markMCQ
$\dfrac{1}{12}\ \Omega$
›Reveal solutionSolution
Between nodes M and N there are two parallel routes — the 6 Ω cross-bar and the 3 Ω + 3 Ω path through the apex R — which reduce to 3 Ω; then simply add the 4 Ω and 5 Ω legs in series.

Step 1 — Map the network onto its nodes

From the figure the connections are:

$P \to M$ : $4\,\Omega$

$M \to R$ (apex) : $3\,\Omega$

$R \to N$ : $3\,\Omega$

$M \to N$ (cross-bar) : $6\,\Omega$

$N \to Q$ : $5\,\Omega$

The key observation: the apex $R$ is a dead-end junction with only two resistors attached, so current entering it from M must all leave to N. The path $M \to R \to N$ is therefore a plain series chain.

Step 2 — Collapse the series pair through the apex
$R_{MRN} = 3 + 3 = 6\,\Omega$
Step 3 — Combine it with the cross-bar (parallel)

The 6 Ω chain $M{-}R{-}N$ and the 6 Ω cross-bar $M{-}N$ have the same two end nodes, M and N — that is precisely the definition of being in parallel:
$\frac{1}{R_{MN}} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \quad\Longrightarrow\quad R_{MN} = 3\,\Omega$
(For two equal resistors in parallel the result is simply half of one of them: $6/2 = 3\,\Omega$ .)

Step 4 — Add the two legs in series

Current entering at P has only one route: $P \to M$ (4 Ω), then through the M–N combination (3 Ω), then $N \to Q$ (5 Ω). Same current flows through all three in turn ⇒ series ⇒ resistances add: …
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