Current Electricity Previous Year Questions — JEE Main Physics
50 past-year Current Electricity questions from JEE Main (Physics) — page 8.
Q40 (2026)
Two known resistance of $R\,\Omega$ and $2R\,\Omega$ and one unknown resistance $X\,\Omega$ are connected in a circuit as shown in the figure. If the equivalent resistance between points A and B in the circuit is $X\,\Omega$, then the value of X is ________ $\Omega$.
{{IMG1}}
- $(\sqrt{3}-1)R$
- $R$
- $2(\sqrt{3}-1)R$
- $(\sqrt{3}+1)R$
### Related Formula
For parallel combination: $R_{\text{eq}} = \frac{R_1 R_2}{R_1 + R_2}$
For series combination: $R_{\text{eq}} = R_1 + R_2$
### Core Logic
Based on the circuit topology, the $2R$ and $X$ resistors are in series. This combination is in parallel with the $R$ resistor.
The total equivalent resistance across A and B is given as $X$.
$$R_{\text{eq}} = \frac{(2R + X) \cdot (R)}{(2R + X) + R}$$
### Step 1: Setting up the Quadratic
Set the equivalent resistance equal to $X$:
$$X = \frac{(2R + X)R}{3R + X}$$
Cross-multiplying yields:
$$X(3R + X) = 2R^2 + XR$$
$$3RX + X^2 = 2R^2 + XR$$
$$X^2 + 2RX - 2R^2 = 0$$
### Step 2: Solving for X
Using the quadratic formula to solve for $X$:
$$X = \frac{-2R \pm \sqrt{(2R)^2 - 4(1)(-2R^2)}}{2}$$
$$X = \frac{-2R \pm \sqrt{4R^2 + 8R^2}}{2} = \frac{-2R \pm \sqrt{12R^2}}{2}$$
$$X = \frac{-2R \pm 2R\sqrt{3}}{2} = -R \pm R\sqrt{3}$$
Since resistance must be positive:
$$X = (\sqrt{3} - 1)R$$
### Pattern Recognition
Self-referential equivalent resistance problems always collapse into a quadratic equation $x^2 + ax + b = 0$. Solve strictly for the positive root.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q31 (2026)
A meter bridge with two resistances $R_{1}$ and $R_{2}$ as shown in figure was balanced (null point) at 40 cm from the point P. The null point changed to 50 cm from the point P, when $16~\Omega$ resistance is connected in parallel to $R_{2}$. The values of resistances $R_{1}$ and $R_{2}$ are \_\_\_\_.
{{IMG1}}
- $\mathrm{R}_2 = 16\Omega, \mathrm{R}_1 = \frac{16}{3}\Omega$
- $\mathrm{R}_2 = 4\Omega, \mathrm{R}_1 = \frac{4}{3}\Omega$
- $\mathrm{R}_2 = 8\Omega, \mathrm{R}_1 = \frac{16}{3}\Omega$
- $\mathrm{R}_2 = 12\Omega, \mathrm{R}_1 = \frac{12}{3}\Omega$
### Related Formula
$$\frac{R_1}{R_2} = \frac{l_1}{100 - l_1}$$
### Core Logic
{{SOL_IMG1}}
Initially:
$$\frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \implies R_1 = \frac{2}{3}R_2$$ ---- (1)
After connecting $16~\Omega$ in parallel with $R_2$:
$$\frac{R_1}{\left(\frac{R_2 \times 16}{R_2 + 16}\right)} = \frac{50}{50} \implies R_1 = \frac{16 R_2}{16 + R_2}$$ ---- (2)
Solving equations (1) and (2) yields:
$$R_2 = 8~\Omega, \quad R_1 = \frac{16}{3}~\Omega$$
### Pattern Recognition
Sees: Meter bridge balancing with parallel shunt resistance.
Shortcut: Set up length ratios for initial and modified states and solve linear simultaneous equations.
Check: Matches option (3). ✓
### Chapter Mix
Class 12 Physics: Current Electricity
Q36 (2026)
An electric power line having total resistance of $2 \Omega$, delivers 1 kW of power of 250 V. The percentage efficiency of transmission line is ____.
- $96.9$
- $86.5$
- $100$
- $92.5$
### Related Formula
$$P_{\text{out}} = V \cdot I$$
$$P_{\text{loss}} = I^2 R$$
$$\eta = \left(\frac{P_{\text{out}}}{P_{\text{net}}}\right) \times 100\%$$
### Core Logic
Calculating total current $I$:
$$1000 = 250 \times I \implies I = 4 \mathrm{~A}$$
Calculating power loss along the line $P_{\text{loss}}$:
$$P_{\text{loss}} = I^2 R = (4)^2 \times 2 = 32 \mathrm{~W}$$
Total input power supplied to line $P_{\text{net}}$:
$$P_{\text{net}} = P_{\text{out}} + P_{\text{loss}} = 1000 + 32 = 1032 \mathrm{~W}$$
Calculating transmission efficiency $\eta$:
$$\eta = \left(\frac{1000}{1032}\right) \times 100\% \approx 96.9\%$$
### Step 1: Final Conclusion
The percentage efficiency of the transmission line is $96.9\%$.
### Pattern Recognition
Efficiency formula: $\eta = \frac{P_{\text{out}}}{P_{\text{out}} + I^2 R} \times 100\%$.
Current $I = 1000/250 = 4\mathrm{~A}$. Loss $= 16 \times 2 = 32\mathrm{~W}$. $\eta = 1000/1032 = 96.9\%$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q49 (2026)
A cylindrical conductor of length 2m and area of cross-section $0.2 \mathrm{~mm}^2$ carries an electric current of 1.6 A when its ends are connected to a 2V battery. Mobility of electrons in the conductor is $\alpha \times 10^{-3} \mathrm{~m}^2/\mathrm{V}\cdot\mathrm{s}$. The value of $\alpha$ is :
(electron concentration = $5 \times 10^{28}/\mathrm{m}^3$ and electron charge = $1.6 \times 10^{-19}$ C)
### Related Formula
$I = n e A v_d$
$$v_d = \mu E = \mu \frac{V}{l}$$
$$\mu = \frac{I l}{n e A V}$$
### Core Logic
Combining current density and mobility equations:
$$I = n e A \left(\mu \frac{V}{l}\right) \implies \mu = \frac{I \cdot l}{n \cdot e \cdot A \cdot V}$$
Substituting given values $I = 1.6 \mathrm{~A}, l = 2 \mathrm{~m}, n = 5 \times 10^{28} /\mathrm{m}^3, e = 1.6 \times 10^{-19} \mathrm{~C}, A = 0.2 \times 10^{-6} \mathrm{~m}^2, V = 2 \mathrm{~V}$:
$$\mu = \frac{1.6 \times 2}{(5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (0.2 \times 10^{-6}) \times 2}$$
$$\mu = \frac{3.2}{1.6 \times 10^3 \times 2} = \frac{3.2}{3200} = 1.0 \times 10^{-3} \mathrm{~m}^2/\mathrm{V}\cdot\mathrm{s}$$
Comparing with $\alpha \times 10^{-3} \implies \alpha = 1$.
### Step 1: Final Conclusion
The value of $\alpha$ is 1.
### Pattern Recognition
Mobility formula: $\mu = I l / (n e A V)$. Direct parameter plug-in yields $\alpha = 1$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
Q34 (2026)
A wire of uniform resistance $\lambda\Omega/m$ is bent into a circle of radius r and another piece of wire with length 2r is connected between points A and B (AOB) as shown in figure. The equivalent resistance between points A and B is ____ $\Omega$.
{{IMG1}}
- $\frac{3\pi\lambda r}{8}$
- $(\pi + 1)2r\lambda$
- $\frac{6\pi\lambda r}{3\pi + 16}$
- $2\pi \lambda r$
### Related Formula
$R = \rho L$
For parallel resistors:
$$\frac{1}{R_{\text{eq}}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$$
### Core Logic
The system represents three resistors connected in parallel between nodes A and B: the upper arc, the lower arc, and the straight diameter wire. Using $R = \lambda L$ where $\lambda$ is resistance per unit length.
### Step 1: Assign Resistances
Length of the upper arc = $\frac{\pi r}{2}$
Resistance $R_{1} = \lambda \cdot \frac{\pi r}{2}$
Length of the straight wire AB = $2r$
Resistance $R_{2} = \lambda \cdot 2r$
Length of the remaining larger arc = $2\pi r - \frac{\pi r}{2} = \frac{3\pi r}{2}$
Resistance $R_{3} = \lambda \cdot \frac{3\pi r}{2}$
### Step 2: Equivalent Resistance Calculation
$$\frac{1}{R_{AB}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$$
$$\frac{1}{R_{AB}} = \frac{2}{\lambda \pi r} + \frac{1}{2\lambda r} + \frac{2}{3\lambda \pi r}$$
$$\frac{1}{R_{AB}} = \frac{1}{\lambda r}\left[\frac{2}{\pi} + \frac{1}{2} + \frac{2}{3\pi}\right]$$
$$\frac{1}{R_{AB}} = \frac{1}{\lambda r}\left(\frac{12 + 3\pi + 4}{6\pi}\right) = \frac{1}{\lambda r}\left(\frac{16 + 3\pi}{6\pi}\right)$$
### Step 3: Final Inversion
$$R_{AB} = \lambda r \left(\frac{6\pi}{16 + 3\pi}\right)$$
### Pattern Recognition
Sees: "Uniform resistance wire bent into shape" → immediately break the shape down into parallel/series segments defined purely by their arc/line lengths multiplied by the linear density $\lambda$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Current Electricity
