Current Electricity Previous Year Questions (JEE Main 2026)

The full-scale deflection current $I_g$ is fixed by the meter, so it is the same before and after the modification. Adding a series resistor $R_s$ raises the range.

Before: the full-scale current is set by the $20\ \text{V}$ range across $x\ \Omega$:

$$I_g = \frac{20}{x}$$

After: for a $30\ \text{V}$ range the same $I_g$ flows through $(x + R_s)$:

$$I_g = \frac{30}{x + R_s}$$

Equate the two expressions:

$$\frac{20}{x} = \frac{30}{x + R_s} \implies 20(x + R_s) = 30x$$

$$x + R_s = 1.5x \implies R_s = \frac{x}{2}$$

A resistor of $\dfrac{x}{2}\ \Omega$ must be added in series (series resistors always extend a voltmeter’s range).

Answer: A — connect a resistor of $\frac{x}{2}\ \Omega$ in series with the voltmeter.

Net reactance:

$$X = X_L - X_C = 20.0 - 80.0 = -60\ \Omega$$

Impedance:

$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{80^2 + (-60)^2} = \sqrt{6400 + 3600} = \sqrt{10000} = 100\ \Omega$$

Power factor:

$$\cos\phi = \frac{R}{Z} = \frac{80}{100} = 0.8$$

(The driving voltage and frequency are not needed for the power factor.)

Answer: A — $0.8$

Step 1 — Bulb resistance from its rating:

$$R_{\text{bulb}} = \frac{V^2}{P} = \frac{(200)^2}{100} = 400\ \Omega$$

Step 2 — Bulb parallel with the $400\ \Omega$ resistor:

$$R_p = \frac{400 \times 400}{400 + 400} = 200\ \Omega$$

Step 3 — Total resistance (this $R_p$ in series with the $200\ \Omega$ resistor):

$$R_{\text{total}} = 200 + 200 = 400\ \Omega$$

Step 4 — Battery current:

$$I = \frac{100}{400} = 0.25\ \text{A}$$

Step 5 — Voltage across the parallel section (= across the bulb):

$$V_{\text{bulb}} = I \times R_p = 0.25 \times 200 = 50\ \text{V}$$

Answer: B — $50$

Let the coil resistance be $G$ and the full-scale (galvanometer) current be $I_g$; $I_g$ is the same in both experiments.

Series case ($470\ \Omega$ in series, $10\ \text{V}$ full scale):

$$I_g = \frac{10}{G + 470} \quad\text{(1)}$$

Shunt case ($S = 2\ \Omega$ across the coil, total current $I = 0.5\ \text{A}$): The shunt carries $(I - I_g)$ and the voltage across both is equal:

$$I_g\,G = (I - I_g)\,S \implies I_g\,G = (0.5 - I_g)(2)$$

$$I_g\,G = 1 - 2 I_g \implies I_g(G + 2) = 1 \implies I_g = \frac{1}{G + 2} \quad\text{(2)}$$

Equate (1) and (2):

$$\frac{10}{G + 470} = \frac{1}{G + 2}$$

$$10(G + 2) = G + 470 \implies 10G + 20 = G + 470$$

$$9G = 450 \implies G = 50\ \Omega$$

(Check: $I_g = \tfrac{1}{52} \approx 19.2\ \text{mA}$ from both equations.)

Answer: 50

For two cells in parallel, use the equivalent-cell (Thevenin) method.

Equivalent EMF and internal resistance (both cells aiding):

$$E_{eq} = \frac{\dfrac{E_1}{r_1} + \dfrac{E_2}{r_2}}{\dfrac{1}{r_1} + \dfrac{1}{r_2}} = \frac{\dfrac{1}{2} + \dfrac{2}{1}}{\dfrac{1}{2} + 1} = \frac{2.5}{1.5} = \frac{5}{3}\ \text{V}$$

$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 \times 1}{3} = \frac{2}{3}\ \Omega$$

Find $R$ from the given $1\ \text{A}$:

$$I = \frac{E_{eq}}{r_{eq} + R} = 1 \implies r_{eq} + R = E_{eq} = \frac{5}{3} \implies R = \frac{5}{3} - \frac{2}{3} = 1\ \Omega$$

Reverse one cell’s polarity (say the $1\ \text{V}$ cell $\to -1\ \text{V}$); $r_{eq}$ is unchanged:

$$E_{eq}' = \frac{-\dfrac{1}{2} + \dfrac{2}{1}}{1.5} = \frac{1.5}{1.5} = 1\ \text{V}$$

New current:

$$I' = \frac{E_{eq}'}{r_{eq} + R} = \frac{1}{\tfrac{2}{3} + 1} = \frac{1}{\tfrac{5}{3}} = \frac{3}{5}\ \text{A}$$

So $\dfrac{\alpha}{5} = \dfrac{3}{5} \implies \alpha = 3$. (Reversing the other cell gives the same magnitude.)

Answer: 3

Let the EMF be $E$ and internal resistance be $r$. For each load, $E = I(R + r)$.

Case 1 ($R = 5\ \Omega$, $I = 0.25\ \text{A}$):

$$E = 0.25(5 + r)$$

Case 2 ($R = 2\ \Omega$, $I = 0.5\ \text{A}$):

$$E = 0.5(2 + r)$$

Equate:

$$0.25(5 + r) = 0.5(2 + r)$$

$$1.25 + 0.25r = 1 + 0.5r$$

$$0.25 = 0.25r \implies r = 1\ \Omega$$

(Then $E = 0.25(5 + 1) = 1.5\ \text{V}$, consistent with Case 2.)

Answer: 1