JEE Mains Physics · Current Electricity

Kirchhoff's Laws and the Wheatstone Bridge

Current entering a junction equals current leaving it, and the potential changes around any closed loop add to zero; dividers and the balanced Wheatstone bridge are the shortcuts these two laws give.

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Why this matters

Twenty-nine PYQs, sixteen of them multiple choice, and three from 2026; twenty-eight come with a figure. Ten are dividers: a current splitting between parallel branches, the voltage across one of several series resistors, and a bulb treated as a resistor. Nine are Wheatstone bridges: the unknown arm for balance, or the current once the middle arm drops out. Ten need Kirchhoff's laws in full: node potentials, circuits with two cells, and the current through a bridge that is not balanced.

Concept 1 of 3: Current and voltage dividers

Two resistors in series carry the same current, so the voltage splits in proportion to resistance: the bigger resistor takes the bigger share. Two resistors in parallel have the same voltage, so the current splits the other way: the smaller resistor takes the bigger share. Anything else connected across part of a divider, a bulb or a meter, is just another resistor in parallel with that part.

Definition

Series pair across V: $V_1 = V\dfrac{R_1}{R_1 + R_2}$ .
Parallel pair carrying I: $I_1 = I\dfrac{R_2}{R_1 + R_2}$ (the OTHER resistor on top).
A bulb rated $V_0$ , $P_0$ is a resistor $R = V_0^{2}/P_0$ . It does not have its rated voltage unless the circuit gives it that.
A load across part of a divider: combine it in parallel with that part first, then divide.
The potential difference between the midpoints of two dividers across the same supply is the difference of their two outputs.

Dividers

V_1 = V\frac{R_1}{R_1 + R_2}, \qquad I_1 = I\frac{R_2}{R_1 + R_2}

Worked example

A 12 V battery drives a 2 kΩ and a 4 kΩ resistor in series. Find the voltage across the 4 kΩ. Then a 4 kΩ load is connected across the 4 kΩ resistor; find the new voltage across it.

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The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 2 · Q13Moderate

Example 1 · Current Electricity · Kirchhoff's Laws and the Wheatstone Bridge

Two resistors of

200\Omega

and

400\Omega

are connected in series with a battery of 100 V . A bulb rated at

200\text{ }V,100\text{ }W

is connected across the

400\Omega

resistance. The potential drop across the bulb is

\_\_\_\_

Current divides inversely

In parallel, the larger share of current goes through the SMALLER resistor. Putting a resistor's own value on top of the fraction gives the other branch's current.

A rating is not the working voltage

A 200 V bulb in a 100 V circuit does not have 200 V across it. Use the rating only to find R, then let the circuit decide the voltage.

A load lowers the output

Anything connected across part of a divider lowers that part's resistance and so its voltage. Using the unloaded value is the trap option.

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Concept 2 of 3: The balanced Wheatstone bridge

A Wheatstone bridge is two dividers side by side with a resistor or galvanometer joining their middles. When both dividers split the voltage in the same ratio, their middles are at the same potential and nothing flows across. That middle arm can then be removed or replaced by a wire, and the network becomes two simple branches in parallel.

Definition

Arms P and Q in one branch, R and S in the other, with the middle arm joining P–Q to R–S: the bridge is balanced when $\dfrac{P}{Q} = \dfrac{R}{S}$ .
At balance the middle arm carries no current, whatever its resistance. Remove it: $R_{\text{eq}} = (P + Q) \parallel (R + S)$ .
An arm can itself be a combination; reduce it to one value first.
"The potential at A equals the potential at B" or "no current through the galvanometer" means balance: use the ratio to find the unknown.
Check the ratio before using balance. A bridge that is not balanced needs Kirchhoff's laws.
Heating one arm of a balanced bridge changes its resistance, so the bridge goes out of balance and a current flows in the middle arm.

Balance condition

\frac{P}{Q} = \frac{R}{S} \ \Rightarrow\ I_{\text{middle}} = 0, \qquad R_{\text{eq}} = \frac{(P + Q)(R + S)}{P + Q + R + S}

Worked example

A bridge has P = 3 Ω and Q = 6 Ω in one branch and R = 4 Ω and an unknown S in the other, with a 10 Ω resistor across the middle. It is balanced and driven by 12 V. Find S and the current from the battery.

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The same idea in a real exam question:

JEE Mains · 2025 · 28 January 2025 · Q99Moderate

Example 2 · Current Electricity · Kirchhoff's Laws and the Wheatstone Bridge

The value of current I in the electrical circuit as given below, when potential at A is equal to the potential at

B

, will be

\_\_\_\_\_\_

Pair the arms correctly

The ratio compares the two arms on the same side of the middle arm in each branch: P/Q = R/S, where P and R both touch the same input point. Cross-pairing gives a wrong unknown.

The middle arm's value is irrelevant at balance

At balance the middle arm carries nothing, so its resistance never enters the answer. A question that gives it is testing whether you notice.

Check before assuming balance

Five resistors in a diamond look like a bridge, but only equal ratios make it balanced. If the ratios differ, use Kirchhoff's laws.

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Concept 3 of 3: Kirchhoff's laws and node potentials

Charge does not pile up at a junction, so what flows in flows out: that is the junction law. Potential is like height: walk around any closed loop and you come back to where you started, so the rises and drops add to zero: that is the loop law. The quickest way to use both is to give each junction an unknown potential and write one junction equation for it.

Definition

Junction law: $\sum I_{\text{in}} = \sum I_{\text{out}}$ .
Loop law: $\sum \varepsilon = \sum IR$ around any closed loop.
Walking from one point to another: across a resistor in the direction of its current, the potential drops by IR; against the current, it rises by IR. Across a cell from − to +, it rises by $\varepsilon$ ; from + to −, it drops by $\varepsilon$ .
Node method: fix one point at 0 V. Give each other junction an unknown potential. At each, write $\sum \dfrac{V_x - V_i}{R_i} = 0$ over the branches leaving it. A cell in a branch shifts the far end's potential by its emf.
A current that comes out negative is flowing the other way; the size is still right.

Junction equation at a node x

\sum_i \frac{V_x - V_i}{R_i} = 0, \qquad \sum \varepsilon = \sum IR

Worked example

A junction X is joined to three points through 2 Ω, 3 Ω and 6 Ω. Those points are at 12 V, 6 V and 0 V. Find the potential of X and the current in each branch.

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The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q95Moderate

Example 3 · Current Electricity · Kirchhoff's Laws and the Wheatstone Bridge

Figure shows a part of an electric circuit. The potentials at points

a,b

and

c

are

30\text{ }V,12\text{ }V

and

2\text{ }V

respectively. The current through the

20\Omega

resistor will be.

The sign of a cell depends on the direction you cross it

Crossing a cell from − to + is a rise of ε, from + to − a drop. The direction of the current through the cell does not matter for this sign; it matters only for the IR term.

A negative current is not a mistake

If a current comes out negative, it flows opposite to the arrow you drew. Keep the size; flip the direction.

One equation per unknown node

With node potentials you need as many junction equations as unknown potentials, no more. Writing loop equations on top of them only adds algebra.

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Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Current and voltage dividers
Dividers
$V_1 = V\frac{R_1}{R_1 + R_2}, \qquad I_1 = I\frac{R_2}{R_1 + R_2}$
The balanced Wheatstone bridge
Balance condition
$\frac{P}{Q} = \frac{R}{S} \ \Rightarrow\ I_{\text{middle}} = 0, \qquad R_{\text{eq}} = \frac{(P + Q)(R + S)}{P + Q + R + S}$
Kirchhoff's laws and node potentials
Junction equation at a node x
$\sum_i \frac{V_x - V_i}{R_i} = 0, \qquad \sum \varepsilon = \sum IR$

Watch out for (9)

Current divides inversely
→ Current and voltage dividers
A rating is not the working voltage
→ Current and voltage dividers
A load lowers the output
→ Current and voltage dividers
Pair the arms correctly
→ The balanced Wheatstone bridge
The middle arm's value is irrelevant at balance
→ The balanced Wheatstone bridge
Check before assuming balance
→ The balanced Wheatstone bridge
The sign of a cell depends on the direction you cross it
→ Kirchhoff's laws and node potentials
A negative current is not a mistake
→ Kirchhoff's laws and node potentials
One equation per unknown node
→ Kirchhoff's laws and node potentials

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