Kirchhoff’s Laws

PhysicsCurrent ElectricityFor JEE aspirants

Kirchhoff's laws are two fundamental rules for analysing electric circuits with multiple branches, loops, and sources. Kirchhoff's Current Law (junction rule) states that the algebraic sum of currents at any junction is zero - based on conservation of charge. Kirchhoff's Voltage Law (loop rule) states that the algebraic sum of potential changes around any closed loop is zero - based on conservation of energy. Together with Ohm's law, these enable systematic analysis of arbitrarily complex networks, including series/parallel resistor combinations and cell groupings tested extensively in JEE Physics and NEET Physics.

Key Formulas - Quick Reference

Junction rule (KCL): $\sum I_{in} = \sum I_{o u t}$ , i.e. $\sum I = 0$ at a node
Loop rule (KVL): $\sum Δ V = 0$ around any closed loop
Resistors in series: $R_{e q} = R_{1} + R_{2} + R_{3} + \dots$
Resistors in parallel: $\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}} + \dots$
$n$ identical cells in series: $I = \frac{n ε}{R + n r}$
$n$ identical cells in parallel: $I = \frac{n ε}{n R + r}$
Mixed grouping ( $m$ rows, $n$ per row): $I = \frac{mn ε}{m R + n r}$
Maximum current in mixed grouping: $R = \frac{n r}{m}$

1. Kirchhoff's Junction Rule (KCL)

Statement: At any junction in an electric circuit, the algebraic sum of currents is zero. Equivalently, the sum of currents entering a junction equals the sum of currents leaving it.

node \sum I = 0 or \sum I_{in} = \sum I_{out}

Basis: Conservation of electric charge. Charge cannot accumulate at a junction (no capacitor there), so what comes in must go out.

Figure 1: KCL - at any junction the sum of incoming currents equals the sum of outgoing currents. Here

i_{1} + i_{2} = i_{3} + i_{4}

2. Kirchhoff's Loop Rule (KVL)

Statement: The algebraic sum of potential differences (changes in potential) around any closed loop in a circuit is zero.

loop \sum Δ V = 0

Basis: Conservation of energy. A test charge returning to its starting point must have zero net work done on it by the electric field.

Figure 2: KVL - going once around the closed loop in the direction of the curved arrow, the algebraic sum of the potential changes across

ε_{1}

V_{1}

V_{2}

V_{3}

is zero.

Sign Conventions

Resistor traversed in direction of current: potential drops by $I R$ (write $- I R$ ).
Resistor traversed opposite to current: potential rises by $I R$ (write $+ I R$ ).
Battery traversed from $-$ to $+$ terminal: potential rises by $ε$ (write $+ ε$ ).
Battery traversed from $+$ to $-$ terminal: potential falls by $ε$ (write $- ε$ ).

Figure 3: Sign convention when traversing a loop. Battery: (a)

+

-

gives

- ε

; (b)

-

+

gives

+ ε

. Resistor: (c) traversed along the current

I

gives

- R I

; (d) traversed against the current gives

+ R I

Any consistent sign convention works - what matters is applying it uniformly around the loop.

Solved Example 1

Q: In the single-loop circuit shown, a 12 V cell and a 4 V cell are connected in opposition through a 4 Ω resistor and a 2 Ω resistor. Find (i) the current in the loop, and (ii) the potential difference across each resistor. Treat the cells as ideal (no internal resistance).

Circuit for Example 1: a single loop containing two cells (12 V and 4 V, both with positive terminals on top so they oppose each other around the loop) and two resistors (4 Ω on top, 2 Ω on bottom). Assumed current direction: clockwise.

Solution

Assume the current $I$ flows clockwise (A → B → bottom → back to A). Apply KVL, traversing the loop clockwise from A.

Contributions to the loop, in order:

Across the 4 Ω resistor (A → B), traversed with the current: $- 4 I$
Across the 4 V cell (top to bottom), traversed from $+$ to $-$ : $- 4$
Across the 2 Ω resistor (right to left along the bottom), traversed with the current: $- 2 I$
Across the 12 V cell (bottom to top), traversed from $-$ to $+$ : $+ 12$

KVL: sum equals zero.

- 4 I - 4 - 2 I + 12 = 0

6 I = 8

I = \frac{4}{3} A \approx 1.33 A (clockwise)

(ii) Potential difference across each resistor (Ohm's law):

V_{4Ω} = I R = \frac{4}{3} \times 4 = \frac{16}{3} V \approx 5.33 V

V_{2Ω} = I R = \frac{4}{3} \times 2 = \frac{8}{3} V \approx 2.67 V

Check: $V_{4Ω} + V_{2Ω} = \frac{16}{3} + \frac{8}{3} = 8$ V, which equals the net driving EMF $12 - 4 = 8$ V. KVL is satisfied.

3. Grouping of Resistances

Series Combination

Resistors carrying the same current are in series. If $V$ is the total potential difference and $I$ is the common current:

Figure 4: Resistors

R_{1}, R_{2}, \dots, R_{n}

connected in series - the same current

I

flows through each;

R_{e q} = R_{1} + R_{2} + \dots + R_{n}

V = I R_{1} + I R_{2} + I R_{3} + \dots = I (R_{1} + R_{2} + R_{3} + \dots)

R_{e q} = R_{1} + R_{2} + R_{3} + \dots

Key properties: Current is same through all resistors; total voltage is sum of individual voltages; equivalent resistance is greater than the largest individual resistance.

Parallel Combination

Resistors across the same potential difference are in parallel. If $V$ is the common voltage:

Figure 5: Resistors

R_{1}, R_{2}, \dots, R_{n}

connected in parallel - the same voltage

V

is applied across each;

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots + \frac{1}{R _{n}}

I = I_{1} + I_{2} + I_{3} = \frac{V}{R _{1}} + \frac{V}{R _{2}} + \frac{V}{R _{3}} = V (\frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}})

\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \frac{1}{R _{3}} + \dots

Key properties: Voltage is same across all resistors; total current is sum of branch currents; equivalent resistance is less than the smallest individual resistance.

For two resistors in parallel: $R_{e q} = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$ (product over sum).

Solved Example 2

Q: Find the equivalent resistance between A and B when the circuit is a cube of 12 identical resistors of $2 Ω$ each along a body diagonal AB.

Solution

By symmetry, the three vertices adjacent to A (call them C, O, D) are at the same potential; similarly, the three vertices adjacent to B are at the same potential.

So resistances AC, AO, AD are in parallel (3 resistors in parallel), and the middle "band" of 6 resistors is in parallel, and BC, BO, BD are in parallel.

R_{A B} = \frac{2}{3} + \frac{2}{6} + \frac{2}{3} = \frac{4}{6} + \frac{2}{6} + \frac{4}{6} = \frac{5}{3} Ω \approx 1.67 Ω

(Standard result for a cube of unit resistors along the body diagonal: $R_{A B} = \frac{5 R}{6}$ .)

4. Grouping of Identical Cells

Consider $N$ cells, each of EMF $ε$ and internal resistance $r$ , connected to an external resistance $R$ .

Series Grouping

All $n$ cells in a single line. Applying KVL:

n ε - I (n r) - I R = 0 \Rightarrow I = \frac{n ε}{R + n r}

When useful: If $R ≫ r$ (external much larger than internal), then $I \approx n ε / R$ . Series is best for high external resistance.

Parallel Grouping

All $n$ cells connected in parallel. Effective EMF is $ε$ and effective internal resistance is $r / n$ :

ε - I \cdot \frac{r}{n} - I R = 0 \Rightarrow I = \frac{n ε}{n R + r}

When useful: If $R ≪ r$ (external much smaller than internal), then $I \approx n ε / r$ . Parallel is best for low external resistance.

Mixed Grouping

$m$ rows in parallel, each row having $n$ cells in series. Total cells $= mn$ :

n ε - I \cdot \frac{n r}{m} - I R = 0 \Rightarrow I = \frac{mn ε}{m R + n r}

Condition for maximum current: Using AM-GM inequality on $(m R + n r)$ with product $mn r R$ fixed:

m R = n r \Leftrightarrow R = \frac{n r}{m}

i.e. maximum current is obtained when the external resistance equals the total internal resistance of one row divided by the number of rows.

Figure 6: Three cell-grouping configurations. Series:

I = n ε / (R + n r)

, best when

R ≫ r

. Parallel:

I = n ε / (n R + r)

, best when

R ≪ r

. Mixed:

I = mn ε / (m R + n r)

, maximum current when

R = n r / m

Solved Example 3

Q: A battery of 24 cells, each of EMF 1.5 V and internal resistance 0.5 $Ω$ , is to be connected in a mixed grouping ( $m$ rows, $n$ cells per row) to give maximum current through an external resistance of 3 $Ω$ . Find $m$ and $n$ .

Solution

Given $mn = 24$ and condition for max current: $R = n r / m$ .

3 = \frac{n \times 0.5}{m} \Rightarrow n = 6 m

Also $mn = 24 \Rightarrow m (6 m) = 24 \Rightarrow m^{2} = 4 \Rightarrow m = 2$ rows, $n = 12$ cells per row.

Maximum current: $I = \frac{mn ε}{m R + n r} = \frac{24 \times 1.5}{2 \times 3 + 12 \times 0.5} = \frac{36}{12} = 3$ A.

Frequently Asked Questions

How is Kirchhoff's junction rule related to conservation of charge? (JEE / NEET)

The junction rule follows directly from charge conservation. Since charge cannot be created, destroyed, or accumulated at an ideal junction (which has no capacitance), the total charge flowing in per unit time must equal the total charge flowing out. This gives

\sum I_{in} = \sum I_{o u t}

When should cells be connected in series versus parallel? (JEE Main / NEET)

Compare external resistance

R

with internal resistance

r

. If

R ≫ r

(large external), use series - each cell contributes fully to voltage, current

\approx n ε / R

. If

R ≪ r

(small external), use parallel - effective internal resistance drops to

r / n

, current

\approx n ε / r

. If

R

is comparable to

r

, use mixed grouping with

R = n r / m

Why does the equivalent parallel resistance become smaller than the smallest resistor? (JEE / NEET)

In parallel, all resistors share the same voltage but current splits. Adding another parallel path gives current an extra route, so total current rises for the same voltage. Since

R_{e q} = V / I_{t o t a l}

and

I_{t o t a l}

grows with each added resistor,

R_{e q}

shrinks - it must be less than any individual

R_{i}

because

1/ R_{e q} = \sum 1/ R_{i} > 1/ R_{i}

for any

i

In mixed grouping, why is the maximum current condition

m R = n r

? (JEE Main)

From

I = mn ε / (m R + n r)

, the numerator is fixed since

mn

= total number of cells is fixed. To maximise

I

, minimise the denominator

m R + n r

. With product

m R \cdot n r = mn r R

fixed (numerator times constants), the sum is minimised when the two terms are equal:

m R = n r

, giving

R = n r / m

Previous year questions on Kirchhoff’s Laws

13 questions from past papers, each with a step-by-step solution.

Show all 13 questions

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