Current Electricity JEE Main previous year questions with solutions

5 solved JEE Main questions on Current Electricity, free to read — no sign-in needed. The full chapter has 377 questions; sign in to attempt the remaining 372 in the exam simulator.

1. Q1JEE Main 2026 (02 Apr, Shift 2)Combination of Resistances
   Two resistors of $200 \, \Omega$ and $400 \, \Omega$ are connected in series with a battery of $100$ V. A bulb rated at $200$ V, $100$ W is connected across the $400 \, \Omega$ resistance. The potential drop across the bulb is _______ V.
   1. A.$25$
   2. B.$50$
   3. C.$66.6$
   4. D.$100$
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   Answer: (B)

   The resistance of the bulb is calculated using its power rating: $R_b = \dfrac{V^2}{P} = \dfrac{200^2}{100} = 400 \, \Omega$ The bulb is connected in parallel with the $400 \, \Omega$ resistor. The equivalent resistance of this parallel combination is: $R_p = \dfrac{400 \times 400}{400 + 400} = 200 \, \Omega$ This parallel combination is in series with the $200 \, \Omega$ resistor. The total equivalent resistance of the circuit is: $R_{eq} = 200 + R_p = 200 + 200 = 400 \, \Omega$ The current drawn from the battery is: $I = \dfrac{V}{R_{eq}} = \dfrac{100}{400} = 0.25 \text{ A}$ The potential drop across the bulb is the same as the potential drop across the parallel combination: $V_p = I \times R_p = 0.25 \times 200 = 50 \text{ V}$ Answer: $50$
2. Q2JEE Main 2025 (22 Jan, Shift 1)Electric Cell or Battery
   <p>Given below are two statements : Statement-I : The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs. Statement-II : The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries. In the light of the above statements, choose the correct answer from the options given below.</p>
   1. A.Both Statement-I and Statement-II are false
   2. B.<p>Statement-I is false but Statement-II is true</p>
   3. C.Both Statement-I and Statement-II are true
   4. D.<p>Statement-I is true but Statement-II is false</p>
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   Answer: (B)

   <p>.In parallel connections $\begin{aligned} & \frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} \end{aligned}$ &nbsp;$\begin{aligned}\frac{E_{eq}}{r_{eq}} = \frac{E_1}{r_1} + \frac{E_2}{r_2} \end{aligned}$</p> <p>If&nbsp;$E_1=E_2$ and $r_1=r_2, E_{\text {eq }}=E_1=E_2$ $\therefore$ Statement 1 is false. $r_{\text {eq }}$ is less then both $r_1$ and r_2 $\therefore$ Statement II is true</p>
3. Q3JEE Main 2024 (01 Feb, Shift 1)Electric Current and Drift of Electrons
   The current in a conductor is expressed as $I=3{t}^{2}+4{t}^{3}$, where $I$ is in Ampere and $t$ is in second. The amount of electric charge that flows through a section of the conductor during $t=1s$ to $t=2s$ is ____________ $C.$
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   Answer: 22

   For variable current, integrate to get the total charge flown in a given interval.$q={\int }_{1}^{2}idt={\int }_{1}^{2}\left(3{t}^{2}+4{t}^{3}\right)dt$ ${\Rightarrow q=\left({t}^{3}+{t}^{4}\right)}_{1}^{2}$ $\Rightarrow q=22C$
4. Q4JEE Main 2023 (13 Apr, Shift 1)Electric Instruments
   When a resistance of $5\Omega$ is shunted with a moving coil galvanometer, it shows a full scale deflection for a current of $250mA,$ however when $1050\Omega$ resistance is connected with it in series, it gives full scale deflection for $25$ volt. The resistance of galvanometer is$\underline{\phantom{0000}}\Omega .$
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   Answer: 50

   The formula to calculate the maximum current through the galvanometer, when a shunt resistance ${r}_{1}$ is connected is given by${\left({i}_{G}\right)}_{\max }=\dfrac{{r}_{1}}{{r}_{1}+{R}_{G}}{i}_{\max }...\left(1\right)$ When a series resistance ${r}_{2}$ is connected, the potential difference across the galvanometer is given by $V={\left({i}_{G}\right)}_{\max }\left[{R}_{G}+{r}_{2}\right]...\left(2\right)$ From equations (1) and (2), it can be written that $\begin{matrix}V & = & \dfrac{{r}_{1}}{{r}_{1}+{R}_{G}}{i}_{\max }\left({R}_{G}+{r}_{2}\right) \\ & \Rightarrow & V{r}_{1}+V{R}_{G}={i}_{\max }{r}_{1}{R}_{G}+{i}_{\max }{r}_{1}{r}_{2} \\ & \Rightarrow & \left(V-{i}_{\max }{r}_{1}\right){R}_{G}={i}_{\max }{r}_{1}{r}_{2}-V{r}_{1} \\ & \Rightarrow & {R}_{G}=\dfrac{{i}_{\max }{r}_{1}{r}_{2}-V{r}_{1}}{V-{i}_{\max }{r}_{1}}...\left(3\right)\end{matrix}$ Substitute the values of the known parameters into equation (3) to calculate the required galvanometer resistance. $\begin{matrix}{R}_{G} & = & \dfrac{0.250A\times 5\Omega \times 1050\Omega -25V\times 5\Omega }{25V-0.250A\times 5\Omega } \\ & = & 50\Omega \end{matrix}$
5. Q5JEE Main 2022 (29 Jun, Shift 1)Electric Power and Heating Effect of Current
   Two coils require $20$ minutes and $60$ minutes respectively to produce same amount of heat energy when connected separately to the same source. If they are connected in parallel arrangement to the same source; the time required to produce same amount of heat by the combination of coils, will be _____ $\min$.
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   Answer: 15

   If $H$ is the heat required to heat the water, then ${P}_{1}\times 60=H\Rightarrow {P}_{1}=\dfrac{H}{60}$ and ${P}_{2}\times 20=H\Rightarrow {P}_{2}=\dfrac{H}{20}$ When the coils are connected in parallel their powers will get added. If the required time now is $t$, then $\left({P}_{1}+{P}_{2}\right)t=H \Rightarrow t=\dfrac{H}{\dfrac{H}{60}+\dfrac{H}{20}}=15\min$