(a) Explain the statement: "Current is a scalar although we represent current with an arrow". (b) Use Kirchhof

Question

Q. (a) Explain the statement: "Current is a scalar although we represent current with an arrow".

(b) Use Kirchhoff's rules to find the current through the

3\ \Omega

resistor in the circuit shown in the figure.

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Current is a scalar because it obeys algebraic addition at junctions, not vector addition. Using the node-voltage method (a direct application of Kirchhoff's rules), the current through the $3\ \Omega$ resistor is $\frac{7}{13}\ \text{A} \approx 0.54\ \text{A}$ , flowing upward from E to B.

Part (a): Why current is a scalar despite the arrow

When we draw an arrow next to a current $I$ , it is tempting to think of current as a vector. But here is the crucial distinction: a vector quantity must obey the parallelogram law of addition. If you have two currents meeting at a junction, you simply add them algebraically — $I_1 + I_2 = I_3$ — not by constructing a parallelogram. The arrow on a circuit diagram is merely a sign convention: it tells you which direction we call positive for that branch. The actual magnitude is a number, and the direction is captured by the sign of that number in the equations.

Watch out

A common mistake is to treat current as a vector and try to resolve it into components. In a wire, current has only two possible directions (forward or backward), so a single signed number suffices — this is the hallmark of a scalar.

Part (b): Finding the current through the $3\ \Omega$ resistor

We will use the node-voltage method, which is Kirchhoff's current law (KCL) applied at a single node. This is often faster than writing multiple loop equations.

1. Label the nodes and choose a reference.

The circuit has six labelled nodes: A, B, C (top row) and F, E, D (bottom row). Notice that A–B is a plain wire, so $V_A = V_B$ . Similarly, E–D is a plain wire, so $V_E = V_D$ . Let us take node F as the reference (ground): $V_F = 0$ .

2. Determine the voltages fixed by the batteries.

The $3\ \text{V}$ cell in the left branch has its positive plate at A (toward A) and negative plate at F. Since $V_F = 0$ , this means $V_A = +3\ \text{V}$ . And because A–B is a wire, $V_B = 3\ \text{V}$ as well.

The $5\ \text{V}$ cell in the top branch has its negative plate at B and positive plate at C. So $V_C = V_B + 5 = 3 + 5 = 8\ \text{V}$ .

Tip

Always check the orientation of the battery: the positive terminal is the longer line in the symbol. The voltage rises from the negative to the positive terminal.

3. Identify the unknown node voltage.

The only node whose voltage we do not yet know is E (and D, since they are connected by a wire). Let $V_E = V_D = x$ .

4. Apply Kirchhoff's current law at node E.

Three branches meet at E:

The $3\ \Omega$ resistor from B to E: current $I_1 = \frac{V_B - x}{3}$ (flowing from B to E if $V_B > x$ ).
The $2\ \Omega$ resistor from C to D: current $I_2 = \frac{V_C - x}{2}$ (flowing from C to D if $V_C > x$ ). …

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

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Kirchhoff's junction rule for electric circuits is based on which of the following physical quantities?$
(i) Mass
(ii) Energy
(iii) Electric charge
(iv) None of these
›Reveal solutionSolution

Kirchhoff's junction (current) rule is just conservation of charge at a node.

Kirchhoff's junction rule states that the algebraic sum of currents meeting at a junction is zero ( $\sum I=0$ ). Since current is charge flow per unit time, this rule is a direct statement that charge can neither be cr …

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CBSE 2026Set ANNUAL1 mark
$Q. The value of current 'I' in the given electric circuit will be ______ A.$
›Reveal solutionSolution
By Kirchhoff's junction rule (conservation of charge), the total current entering a junction equals the total leaving it. Here 4.2 + 3.1 + 1.2 A flow in and 5.1 A + I flow out, so I = 8.5 − 5.1 = 3.4 A.

Kirchhoff's junction (current) rule states that the algebraic sum of currents at a junction is zero — equivalently, total current in = total current out, because charge cannot pile up.

From the figure, list the currents at the junction network:

Entering: 4.2 A (from the left), 3.1 A (diagonally, from the upper-left), and 1.2 A (diagonally, from the upper-right).

Leaving: 5.1 A (diagonally, toward the lower-right) and the unknown current I (to the right). …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Kirchhoff's first law of electricity follow which conservation law?$
(a) Mass
(b) Momentum
(c) Energy
(d) Charge
›Reveal solutionSolution

Kirchhoff's Current Law (junction rule) is a direct consequence of conservation of electric charge.

Kirchhoff's first law states that at any junction in a circuit, the algebraic sum of currents meeting at that junction is zero ( $\sum I_{in} = \sum I_{out}$ ). Charge cannot accumulate at a junction in steady state, so whatever charge flows in per second must flow out per second — this is …

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CBSE 2025Set A1 markMCQ
$Q. First law of Kirchhoff of current distribution follows —$
(i) Law of energy conservation
(ii) Law of charge conservation
(iii) Law of conservation of momentum
(iv) Law of mass conservation
›Reveal solutionSolution

Kirchhoff's Current Law (junction rule) is a direct consequence of conservation of electric charge.

Kirchhoff's first law, also called the junction rule or current law, states that the sum of currents entering a junction in an electrical circuit equals the sum of currents leaving that junction (ΣI_in = ΣI_out), i.e. the algebraic sum of all currents at a junction is zero. Since current is the rate of flow of charge, and charge can neither be created nor destroyed at a junction (charge does not accumulate there in steady state …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The algebraic sum of all the currents at any point in an electrical circuit is$
(a) positive
(b) negative
(c) zero
(d) infinite
›Reveal solutionSolution

Kirchhoff's junction rule is a statement of conservation of charge: at any point (junction) in a circuit, charge cannot accumulate, so current flowing in must equal current flowing out.

If currents entering a junction are taken as positive and currents leaving as negative (or vice versa), then since no charge can pile up at a point in steady state:

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Kirchhoff's junction rule is reflection of:$
(a) conservation of current density vector.
(b) the fact that the momentum with which a charged particle approaches a junction is unchanged as the charge particle leaves the junction.
(c) conservation of charge and the fact that there is no accumulation of charges at a junction.
(d) None of the above.
›Reveal solutionSolution

Kirchhoff's junction (current) rule follows directly from conservation of electric charge.

Kirchhoff's junction rule states that the algebraic sum of currents meeting at a junction in an electrical circuit is zero, i.e. the sum of currents entering a junction equals the sum of currents leaving it.

…

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CBSE 2024Set A1 markMCQ
$Q. Kirchhoff's second law of electricity is related to (A) conservation of mass (B) conservation of charge (C) conservation of energy (D) conservation of momentum$
›Reveal solutionSolution

Kirchhoff's voltage (loop) law: sum of EMFs = sum of IR drops around a loop — a statement of energy conservation.

Kirchhoff's second law (the loop or voltage law) states that the algebraic sum of potential changes around any closed loop is zero: $\sum \varepsilon = \sum IR$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The current I in the following circuit given below is$
(a) 1.7 A
(b) 3.7 A
(c) 1.3 A
(d) 1A
›Reveal solutionSolution

Kirchhoff's current (junction) law: total current into a junction equals total current out; applying it at each junction of the circuit gives I = 1.7 A.

At the left junction, two branches each carrying 2 A flow in, so by charge conservation the single wire leaving that junction toward the right junction carries their sum: $2A+2A=4A$ .

…

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CBSE 2023Set MODEL1 markMCQ
$Q. Kirchhoff's second law relates to:$
(a) Conservation of mass
(b) Conservation of charge
(c) Conservation of energy
(d) Conservation of momentum
›Reveal solutionSolution

Kirchhoff's voltage (second) law is a statement of energy conservation in a closed loop.

Kirchhoff's second law (the loop rule) states that the algebraic sum of changes in potential around any closed loop in a circuit is zero. Since potential difference is energy per unit charge, this is equivalent to saying that the total energy gained by a charg …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Kirchhoff's loop rule is a direct consequence of law of conservation of$
(1) Charge
(2) Momentum
(3) Angular momentum
(4) Energy
›Reveal solutionSolution

The loop rule says the algebraic sum of potential changes around any closed loop is zero — this is just a statement that the electric field is conservative, so energy is conserved.

Going around a closed loop and summing all emfs and IR drops, $\sum \Delta V = 0$ . If this were not true, a charge taken around the loop would gain or lose energy for free, violating conservation of ene …

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CBSE 2023Set ANNUAL1 mark
$Q. Are Kirchhoff's rules applicable to both a. c. and d. c. ?$
›Reveal solutionSolution

Kirchhoff's rules follow from conservation of charge and energy, which hold at every instant, so they apply equally to a.c. and d.c. circuits.

Kirchhoff's current rule (junction rule) is a statement of conservation of electric charge, and Kirchhoff's voltage rule (loop rule) is a statement of conservation of energy. Both conservation laws hold at every instant of time, regardless of whether the currents/voltages are steady (d.c.) or varying with time (a.c.). Hence, at any given instant, the algebraic sum of currents at a junct …

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CBSE 2020Set ANNUAL1 markMCQ
$Q. Kirchhoff's point rule is a direct consequence of law of conservation of$
(a) Energy
(b) Linear momentum
(c) Angular momentum
(d) Charge
›Reveal solutionSolution

Kirchhoff's two rules each restate a conservation law: the junction rule restates conservation of charge, and the loop rule restates conservation of energy.

Kirchhoff's point rule (also called the junction rule or current law) says that at any junction in a circuit, the sum of currents entering the junction equals the sum of currents leaving it:

sum of I_in = sum of I_out

…

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Q.(a) Explain the statement: "Current is a scalar although we represent current with an arrow".

Concept understanding — Kirchhoff's Current and Voltage Rules

(a) Current is a scalar although we represent current with an arrow

(b) Current through the $3\ \Omega$ resistor

Part (a): Why current is a scalar despite the arrow

Part (b): Finding the current through the $3\ \Omega$ resistor

Q.(a) Explain the statement: "Current is a scalar although we represent current with an arrow".

Part (a): Why current is a scalar despite the arrow

Part (b): Finding the current through the 3 Ω3\ \Omega3 Ω resistor

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Part (b): Finding the current through the $3\ \Omega$ resistor

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