Current Electricity JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Current Electricity, free to read — no sign-in needed. The full chapter has 30 questions; sign in to attempt the remaining 25 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   A metal wire of cross-sectional area $0.5$ mm$^2$ and length $100$ m is connected across a battery of e.m.f. $2$ V and internal resistance $1\ \Omega$. The density, atomic mass and electrical conductivity of the metal are $6.35 \times 10^3$ kg m$^{-3}$, $63.5$ gm/mole and $2 \times 10^8$ mho m$^{-1}$, respectively. Assuming one conduction electron per atom of the metal, the drift velocity (in mm s$^{-1}$) of the electrons in the wire is: [Take Avogadro's number as $6 \times 10^{23}$ and charge of the electron as $1.6 \times 10^{-19}$ C.]
   1. A.$0.052$
   2. B.$0.104$
   3. C.$0.208$
   4. D.$0.156$
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   Answer: (C)

   The number density of conduction electrons $n$ is given by the number of atoms per unit volume, since there is one conduction electron per atom: $n = \dfrac{d \times N_A}{M}$ Substituting the given values: $n = \dfrac{6.35 \times 10^3 \times 6 \times 10^{23}}{63.5 \times 10^{-3}} = 6 \times 10^{28} \text{ m}^{-3}$ The resistance of the wire $R$ is: $R = \dfrac{L}{\sigma A}$ $R = \dfrac{100}{2 \times 10^8 \times 0.5 \times 10^{-6}} = \dfrac{100}{10^2} = 1\ \Omega$ The total resistance of the circuit is $R_{total} = R + r = 1 + 1 = 2\ \Omega$. The current in the circuit is: $I = \dfrac{E}{R_{total}} = \dfrac{2}{2} = 1 \text{ A}$ The drift velocity $v_d$ is given by the relation $I = n e A v_d$: $v_d = \dfrac{I}{n e A}$ $v_d = \dfrac{1}{6 \times 10^{28} \times 1.6 \times 10^{-19} \times 0.5 \times 10^{-6}}$ $v_d = \dfrac{1}{4.8 \times 10^3} \text{ m s}^{-1}$ Converting to mm s$^{-1}$: $v_d = \dfrac{1000}{4800} \text{ mm s}^{-1} = \dfrac{5}{24} \text{ mm s}^{-1} \approx 0.208 \text{ mm s}^{-1}$ Answer: $0.208$
2. Q2JEE Advanced Adv 2014 (Paper 1)
   Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40K?
   1. A.4 if wires are in parallel
   2. B.2 if wires are in series
   3. C.1 if wires are in series
   4. D.0.5 if wires are in parallel.
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   Answer: B,D

   Joules law, $\mathrm{H}=\mathrm{I}^{2} \mathrm{Rt}=\frac{\mathrm{V}^{2} \mathrm{t}}{\mathrm{R}}$ Also, $\mathrm{R}=\rho \frac{1}{\mathrm{~A}} \Rightarrow \mathrm{R} \propto \frac{1}{\mathrm{~A}} \Rightarrow \mathrm{R} \propto \frac{1}{\mathrm{r}^{2}}$ $\therefore R^{\prime}=\frac{R}{4}$ parallel $\Rightarrow \mathrm{R}_{\mathrm{p}}=\left(\frac{1}{\mathrm{R} / 4}+\frac{1}{\mathrm{R} / 4}\right)^{-1}=\frac{\mathrm{R}}{8}$ $\text { series } \Rightarrow \mathrm{R}_{\mathrm{s}}=\frac{\mathrm{R}}{4}+\frac{\mathrm{R}}{4}=\frac{\mathrm{R}}{2}$ $\therefore \frac{\mathrm{V}^{2} 4}{\mathrm{R}}=\frac{\mathrm{V}^{2} \mathrm{t}_{\mathrm{p}}}{\mathrm{R}_{\mathrm{p}}}=\frac{\mathrm{V}^{2} \mathrm{t}_{\mathrm{s}}}{\mathrm{R}_\mathrm{s}}$ $\Rightarrow \mathrm{t}_{\mathrm{p}}=0.5 \mathrm{~min}, \mathrm{t}_{\mathrm{s}}=2 \mathrm{~min}$
3. Q3JEE Advanced Adv 2010 (Paper 1)
   When two identical batteries of internal resistance $1 \Omega$ each are connected in series across a resistor $R$, the rate of heat produced in $R$ is $J_1$. When the same batteries are connected in parallel across $R$, the rate is $J_2$. If $J_1=2.25 J_2$ then the value of $R$ in $\Omega$ is
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   Answer: 4

   In series, $i=\frac{2 E}{2+R}$ $$\therefore J_1=i^2 R=\left(\frac{2 E}{2+R}\right)^2 \cdot R$$ In parallel, $i=\frac{E}{0.5+R}$ $$\begin{aligned} \therefore \quad J_2 & =i^2 R=\left(\frac{E}{0.5+R}\right)^2 \cdot R \\ \frac{J_1}{J_2} & =2.25=\frac{4(0.5+R)^2}{(2+R)^2} \end{aligned}$$ or $1.5=\frac{2(0.5+R)}{(2+R)}$ Solving we get, $R=4 \Omega$ $\therefore$ The answer is 4 .
4. Q4JEE Advanced Adv 2010 (Paper 1)
   Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, $100 \mathrm{~W}, 60 \mathrm{~W}$ and $40 \mathrm{~W}$ bulbs have filament resistances $R_{100}, R_{60}$ and $R_{40}$, respectively, the relation between these resistances is
   1. A.$\frac{1}{R_{100}}=\frac{1}{R_{40}}+\frac{1}{R_{60}}$
   2. B.$R_{100}=R_{40}+R_{60}$
   3. C.$R_{100}\gt R_{60}\gt R_{40}$
   4. D.$\frac{1}{R_{100}}\gt \frac{1}{R_{60}}\gt \frac{1}{R_{40}}$
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   Answer: (D)

   $R=\frac{V^2}{P}$ or $R \propto \frac{1}{P}$ $\therefore \quad \frac{1}{R_{100}}\gt \frac{1}{R_{60}}\gt \frac{1}{R_{40}}$ Hence, the correct option is (d).
5. Q5JEE Advanced Adv 2008 (Paper 1)
   Statement $1 \mathrm{In}$ a Meter Bridge experiment, null point for an unknown resistance is measured. Now, the unknown resistance is put inside an enclosure maintained at a higher temperature. The null point can be obtained at the same point as before by decreasing the value of the standard resistance. and Statement 2 Resistance of a metal increase with increase in temperature.
   1. A.Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.
   2. B.Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.
   3. C.Statement 1 is true, Statement 2 is false.
   4. D.Statement 1 is false, Statement 2 is true
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   Answer: (D)

   With increase in temperature, the value of unknown resistance will increase. In balanced Wheatstone bridge condition, $\frac{R}{X}=\frac{l_1}{l_2}$ Here, $R=$ value of standard resistance, $X=$ value of unknown resistance . To take null point at same point or $\frac{l_1}{l_2}$ to remain unchanged, $\frac{R}{X}$ should also remain unchanged. Therefore, if $X$ is increasing $R$ should also increase. $\therefore$ correct option is (d).