Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics) — page 9.
Q35 (2026)
To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm. The least count of scale is 1 cm. The percentage error in the ratio of EMFs is ____
- $1.45$
- $1.65$
- $1.75$
- $1.55$
### Related Formula
$$\epsilon = \lambda l$$
$$\frac{\Delta y}{y} = \frac{\Delta l_1}{l_1} + \frac{\Delta l_2}{l_2}$$
### Core Logic
By potentiometer principle:
$\epsilon_1 = \lambda l_1$
$\epsilon_2 = \lambda l_2$
Ratio $y = \frac{\epsilon_1}{\epsilon_2} = \frac{l_1}{l_2}$
Maximum percentage error is the sum of fractional errors.
### Step 1: Calculate Percentage Error
$$\frac{\Delta y}{y} = \frac{1}{200} + \frac{1}{150}$$
$$\frac{\Delta y}{y} = \frac{3 + 4}{600} = \frac{7}{600}$$
Percentage error = $\frac{7}{600} \times 100\% = \frac{7}{6}\% \approx 1.16\%$
### Pattern Recognition
Note: The calculated exact percentage error is 1.16%, which does not match any of the provided options exactly. The closest official option provided was tracked as index (2) marking 1.65 by some keys, though technically a bonus question.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Class 11 Physics: Units and Measurements
Q30 (2026)
Two resistors of $100 \, \Omega$ each are connected in series with a 9V battery. A voltmeter of $400\Omega$ resistance is connected to measure the voltage drop across one of the resistors. The voltmeter reading is ____ V.
- $3$
- $4.5$
- $4$
- $2$
### Related Formula
$$R_{\text{eq(parallel)}} = \frac{R_1 R_2}{R_1 + R_2}$$
$V = I \cdot R$
### Core Logic
{{SOL_IMG1}}
The voltmeter is connected in parallel with one of the $100\, \Omega$ resistors.
The equivalent resistance of this parallel combination is:
$$R_{\text{parallel}} = \frac{400 \times 100}{400 + 100} = \frac{40000}{500} = 80\, \Omega$$
The total equivalent resistance of the circuit is:
$$R_{\text{eq}} = 100 + 80 = 180\, \Omega$$
### Step 1: Circuit Current and Voltmeter Reading
Current in the circuit:
$$I = \frac{E}{R_{\text{eq}}} = \frac{9}{180} = \frac{1}{20}\text{ A}$$
The reading of the voltmeter is the voltage across the parallel combination:
$$V = I \times 80 = \frac{1}{20} \times 80 = 4\text{ V}$$
### Pattern Recognition
When a real voltmeter is used, it draws current. Model it as a resistor in parallel with the test component to find the exact altered potential drop.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q37 (2026)
Two resistors $2\Omega$ and $3\Omega$ are connected in the gaps of bridge as shown in figure. The null point is obtained with the contact of jockey at some point on wire XY. When an unknown resistor is connected in parallel with $3\Omega$ resistor, the null point is shifted by 22.5 cm toward Y. The resistance of unknown resistor is ____ $\Omega$.
{{IMG1}}
- $3$
- $2$
- $4$
- $1$
### Related Formula
$$\frac{R_1}{R_2} = \frac{\ell}{100 - \ell}$$
### Core Logic
Initially, the meter bridge is balanced with $2\, \Omega$ and $3\, \Omega$:
$$\frac{2}{3} = \frac{x}{100 - x}$$
$$200 - 2x = 3x \implies 5x = 200 \implies x = 40 \text{ cm}$$
### Step 1: Shift after parallel connection
When an unknown resistor $R$ is connected in parallel with $3\, \Omega$, the new resistance in the right gap is $R' = \frac{3R}{3 + R}$.
The null point shifts by $22.5 \text{ cm}$ towards Y, so the new balancing length is $x' = 40 + 22.5 = 62.5 \text{ cm}$.
The new balance equation is:
$$\frac{2}{R'} = \frac{62.5}{100 - 62.5}$$
$$\frac{2}{\frac{3R}{3 + R}} = \frac{62.5}{37.5} = \frac{5}{3}$$
$$\frac{2(3 + R)}{3R} = \frac{5}{3}$$
$6 + 2R = 5R$
$$3R = 6 \implies R = 2\, \Omega$$
### Pattern Recognition
A shift 'towards Y' means the balance point moved right, which implies the right gap resistance decreased. Connecting in parallel correctly decreases resistance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q26 (2026)
The reading of the ammeter (A) in steady state in the following circuit (assuming negligible internal resistance of the ammeter) is ____ A.
{{IMG1}}
- $2$
- $1$
- $1 / 2$
- $0$
### Core Logic
In steady state, the capacitor acts as an open circuit. No current flows through the branch containing the $10\mu\text{F}$ capacitor.
{{SOL_IMG1}}
After removing the capacitor branch, we simplify the network of resistors.
### Step 1: Equivalent Resistance
$$\mathrm{I} = 2\mathrm{A}$$
{{SOL_IMG2}}
By further simplifying the parallel and series combinations, we determine the main current and the current branching into the ammeter.
### Step 2: Ammeter Reading
The current of $2\mathrm{A}$ splits equally through the symmetrical $8\Omega$ branches.
The ammeter reading is $1\mathrm{A}$.
### Pattern Recognition
Steady state DC + Capacitor = Open circuit. Always delete the capacitor branch first before calculating equivalent resistance.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Class 12 Physics: Electrostatics
Q29 (2026)
A regular hexagon is formed by six wires each of resistance r $\Omega$ and the corners are joined to centre by wires of same resistance. If the current enters at one corner and leaves at the opposite corner, the equivalent resistance of the hexagon between the two opposite corners will be
- $\frac{4}{5}r$
- $\frac{5}{8}r$
- $\frac{3}{4}r$
- $\frac{3}{5}r$
### Core Logic
By applying the principle of symmetry to the regular hexagonal resistor network, the circuit can be simplified across the axis of symmetry.
{{SOL_IMG1}}
{{SOL_IMG2}}
### Step 1: Network Reduction
Using equipotential nodes along the perpendicular bisector, we can resolve the network into parallel and series components.
{{SOL_IMG3}}
The equivalent resistance for each half is determined as follows:
$$\frac{R \times \frac{R}{2}}{\frac{3R}{2}} = \frac{R}{3}$$
### Step 2: Final Equivalent Resistance
Combining the simplified branches:
$$R_{\text{eq}} = \frac{2R \times \frac{4R}{3}}{2R + \frac{4R}{3}} = \frac{8R^2}{10R} = \frac{4}{5}R$$
### Pattern Recognition
For regular polygonal networks with central connections, always search for a line of equipotential symmetry perpendicular to the line joining the input and output nodes. Nodes on this axis can be detached or merged.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
