Current Electricity JEE Main previous year questions with solutions

4 solved JEE Main questions on Current Electricity, free to read — no sign-in needed. The full chapter has 380 questions; sign in to attempt the remaining 376 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 (31 Jan, Shift 1)Conduction
   Two conductors have the same resistances at $0^{\circ}C$ but their temperature coefficients of resistance are ${\alpha }_{1}$ and ${\alpha }_{2}$. The respective temperature coefficients for their series and parallel combinations are :
   1. A.${\alpha }_{1}+{\alpha }_{2},\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}$
   2. B.$\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2},\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}$
   3. C.${\alpha }_{1}+{\alpha }_{2},\dfrac{{\alpha }_{1}{\alpha }_{2}}{{\alpha }_{1}+{\alpha }_{2}}$
   4. D.$\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2},{\alpha }_{1}+{\alpha }_{2}$
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   Answer: (B)

   Change in resistance due to temperature is given by, ${R}^{'}=R\left(1+\alpha ∆T\right)$ For Series Combination: ${R}_{eq}={R}_{1}+{R}_{2}$ $\Rightarrow 2R\left(1+{\alpha }_{eq}∆T\right)=R\left(1+{\alpha }_{1}∆T\right)+R\left(1+{\alpha }_{2}∆T\right)$ $\Rightarrow 2R\left(1+{\alpha }_{eq}∆T\right)=2R+\left({\alpha }_{1}+{\alpha }_{2}\right)R∆T$ $\Rightarrow {\alpha }_{eq}=\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}$ For Parallel Combination: $\dfrac{1}{{R}_{eq}}=\dfrac{1}{{R}_{1}}+\dfrac{1}{{R}_{2}}$ $\Rightarrow \dfrac{1}{\dfrac{R}{2}\left(1+{\alpha }_{eq}∆T\right)}=\dfrac{1}{R\left(1+{\alpha }_{1}∆T\right)}+\dfrac{1}{R\left(1+{\alpha }_{2}∆T\right)}$ $\Rightarrow \dfrac{2}{1+{\alpha }_{eq}∆T}=\dfrac{1}{1+{\alpha }_{1}∆T}+\dfrac{1}{1+{\alpha }_{2}∆T}$ $\Rightarrow \dfrac{2}{1+{\alpha }_{eq}∆T}=\dfrac{1+{\alpha }_{2}∆T+1+{\alpha }_{1}∆T}{\left(1+{\alpha }_{1}∆T\right)\left(1+{\alpha }_{2}∆T\right)}$ $\Rightarrow 2\left[\left(1+{\alpha }_{1}∆T\right)\left(1+{\alpha }_{2}∆T\right)\right]=\left[2+\left({\alpha }_{1}+{\alpha }_{2}\right)∆T\right]\left[1+{\alpha }_{eq}∆T\right]$ $\Rightarrow 2\left[1+{\alpha }_{1}∆T+{\alpha }_{2}∆T+{\alpha }_{1}{\alpha }_{2}∆T\right]=2+2{\alpha }_{eq}∆T+\left({\alpha }_{1}+{\alpha }_{2}\right)∆T+{\alpha }_{eq}\left({\alpha }_{1}+{\alpha }_{2}\right)∆{T}^{2}$ Neglecting small terms $2+2\left({\alpha }_{1}+{\alpha }_{2}\right)∆T=2+2{\alpha }_{eq}∆T+\left({\alpha }_{1}+{\alpha }_{2}\right)∆T$ $\Rightarrow \left({\alpha }_{1}+{\alpha }_{2}\right)∆T=2{\alpha }_{eq}∆T$ $\Rightarrow {\alpha }_{eq}=\dfrac{{\alpha }_{1}+{\alpha }_{2}}{2}$
4. Q4JEE Main 2023 (12 Apr, Shift 1)Electric Current and Drift of Electrons
   <p>The current flowing through a conductor connected across a source is $2A$ and 1.2A at ${0}^{o}$C and ${100}^{o}C$ respectively. The current flowing through the conductor at ${50}^{o}C$ will be _____ $\times {10}^{2}mA$.</p>
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   Answer: 15

   <p>By Ohm's law, $V = IR$. Let the current and resistance at $0^\circ\text{C}$ be i_0 and $R_0$ respectively. Let the current and resistance at $100^\circ\text{C}$ be $i_{100}$ and $R_{100}$ respectively.</p> <p>Given the condition $i_0 R_0 = i_{100} R_{100}$:</p> <p>$2R_0 = 1.2R_0(1 + 100\alpha)$</p> <p>$1 + 100\alpha = \frac{2}{1.2} = \frac{20}{12} = \frac{5}{3}$</p> <p>$100\alpha = \frac{5}{3} - 1 = \frac{2}{3}$</p> <p>$50\alpha = \frac{1}{3}$</p> <p>Since the voltage remains constant $( i_{50}R_{50} = i_0R_0 )$, we calculate the current at $50^\circ\text{C}:$</p> <p>$i_{50} = \frac{i_0 R_0}{R_{50}} = \frac{2 \times R_0}{R_0(1 + 50\alpha)}$</p> <p>Substituting<strong> $50\alpha = \frac{1}{3}:$</strong></p> <p><strong>$i_{50} = \frac{2}{1 + \frac{1}{3}} = \frac{2}{\frac{4}{3}} = 1.5\text{ A}$</strong></p> <p>Converting to milliamperes (mA):</p> <p>$i_{50} = 1.5 \times 10^3\text{ mA} = 15 \times 10^2\text{ mA}$</p>