Alternating Current JEE Advanced previous year questions with solutions

4 solved JEE Advanced questions on Alternating Current, free to read — no sign-in needed. The full chapter has 25 questions; sign in to attempt the remaining 21 in the exam simulator.

1. Q1JEE Advanced Adv 2023 (Paper 1)
   A series LCR circuit is connected to a $45\sin (\omega t)$ Volt source. The resonant angular frequency of the circuit is ${10}^{5}rad{s}^{-1}$ and current amplitude at resonance is ${I}_{0}$. When the angular frequency of the source is $\omega =8\times {10}^{4}rad{s}^{-1}$, the current amplitude in the circuit is $0.05{I}_{0}$. If $L=50mH$, match each entry in List-$I$ with an appropriate value from List-$II$ and choose the correct option. List-$I$ List-$II$ $P$ ${I}_{0}$ in $mA$ $1$ $44.4$ $Q$ The quality factor of the circuit $2$ $18$ $R$ The bandwidth of the circuit in $rad{s}^{-1}$ $3$ $400$ $S$ The peak power dissipated at resonance in Watt $4$ $2250$ $5$ $500$
   1. A.$P\rightarrow 2,Q\rightarrow 3,R\rightarrow 5,S\rightarrow 1$
   2. B.$P\rightarrow 3,Q\rightarrow 1,R\rightarrow 4,S\rightarrow 2$
   3. C.$P\rightarrow 4,Q\rightarrow 5,R\rightarrow 3,S\rightarrow 1$
   4. D.$P\rightarrow 4,Q\rightarrow 2,R\rightarrow 1,S\rightarrow 5$
   Show answer & solution

   Answer: (B)

   Resonant angular frequency is given by, $\dfrac{1}{\sqrt{LC}}={10}^{5}$ $\dfrac{1}{\sqrt{\left(50\times {10}^{-3}\right)C}}={10}^{5} \Rightarrow C=2\times {10}^{-9}F$ Given: $V=45\sin \left(\omega t\right)$. Therefore, ${V}_{0}=45$. Now, ${I}_{0}=\dfrac{{V}_{0}}{R}=\dfrac{45}{R}...\left(ii\right)$ Inductive reactance, ${X}_{L}=\omega L=\left(8\times {10}^{4}\right)\times \left(50\times {10}^{-3}\right)=4000\Omega$. and capacitive reactance, ${X}_{C}=\dfrac{1}{\omega C}=\dfrac{1}{\left(8\times {10}^{4}\right)\times \left(2\times {10}^{-9}\right)}=6250\Omega$. For new current amplitude, we can write $0.05{I}_{0}=\dfrac{45}{\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}} \Rightarrow 0.05{I}_{0}=\dfrac{45}{\sqrt{{R}^{2}+{\left(6250-4000\right)}^{2}}} \Rightarrow 0.05\times \dfrac{45}{R}=\dfrac{45}{\sqrt{{R}^{2}+{\left(6250-4000\right)}^{2}}} \Rightarrow {R}^{2}+{\left(6250-4000\right)}^{2}=\dfrac{{R}^{2}}{{\left(0.05\right)}^{2}} \Rightarrow {R}^{2}+{\left(2250\right)}^{2}=400{R}^{2} \Rightarrow R=\dfrac{2250}{\sqrt{399}}=112.67\Omega$ Where ${X}_{{L}_{0}}={X}_{{C}_{0}}$ are at resonant frequencies On solving, $\Rightarrow {I}_{0}=\dfrac{45}{R}≃400mA$ Quality factor $Q=\dfrac{{X}_{L}}{R}≃44.44$ Now, $Q=\dfrac{{\omega }_{0}}{∆\omega }\Rightarrow ∆\omega ≃2250rad{s}^{-1}$. Peak power $=45\times \dfrac{400}{1000}W=18W$ Therefore, $P\rightarrow 3,Q\rightarrow 1,R\rightarrow 4,S\rightarrow 2$.
2. Q2JEE Advanced Adv 2011 (Paper 2)
   A series $R-C$ combination is connected to an $\mathrm{AC}$ voltage of angular frequency $\omega=500 \mathrm{rad} / \mathrm{s}$. If the impedance of the $R-C$ circuit is $R \sqrt{1.25}$, the time constant (in millisecond) of the circuit is
   Show answer & solution

   Answer: 4

   $Z=\sqrt{R^2+X_C^2}=R \sqrt{1.25}$ $$\therefore \quad R^2+X_C^2=1.25 R^2$$ or $\quad X_C=\frac{R}{2}$ or $\quad \frac{1}{\omega C}=\frac{R}{2}$ $\begin{aligned} \therefore \text { Time constant } & =C R=\frac{2}{\omega} \\ & =\frac{2}{500} \mathrm{~s}=4 \mathrm{~ms}\end{aligned}$ $\therefore$ Answer is 4 . Analysis of Question (i) Question is very simple. (ii) I think this is one of the simplest formula based question of this paper.
3. Q3JEE Advanced Adv 2011 (Paper 2)
   A series $R$ - $C$ circuit is connected to $A C$ voltage source. Consider two cases; $(A)$ when $C$ is without a dielectric medium and $(B)$ when $C$ is filled with dielectric of constant 4. The current $I_R$ through the resistor and voltage $V_C$ across the capacitor are compared in the two cases. Which of the following is/are true?
   1. A.$I_R^A\gt I_R^B$
   2. B.$I_R^A \lt I_R^B$
   3. C.$V_C^A\gt V_C^B$
   4. D.$V_C^A \lt V_C^B$
   Show answer & solution

   Answer: B,C

   $Z=\sqrt{R^2+X_C^2}=\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}$ In case (b) capacitance $C$ will be more. Therefore, impedance $Z$ will be less. Hence, current will be more. $\therefore$ Option (b) is correct. Further, $\quad \begin{aligned} V_C & =\sqrt{V^2-V_R^2} \\ & =\sqrt{V^2-(I R)^2}\end{aligned}$ In case (b), since current $I$ is more. Therefore, $V_C$ will be less. $\therefore$ Option (c) is correct. $\therefore$ Correct options are (b) and (c). Analysis of Question (i) Question is moderately difficult. (ii) In my opinion problems of alternating currents are not very difficult. (iii) Topic of $A C$ is small. One can feel comfortable in this topic by putting less efforts.
4. Q4JEE Advanced Adv 2010 (Paper 1)
   An AC voltage source of variable angular frequency $\omega$ and fixed amplitude $V_0$ is connected in series with a capacitance $C$ and an electric bulb of resistance $R$ (inductance zero). When $\omega$ is increased
   1. A.the bulb glows dimmer
   2. B.the bulb glows brighter
   3. C.total impedance of the circuit is unchanged
   4. D.total impedance of the circuit increases
   Show answer & solution

   Answer: (B)

   $Z=\sqrt{R^2+X_C^2}: I_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{Z}: P=I_{\mathrm{rms}}^2 R$ where, $\quad X_C=\frac{1}{\omega C}$ As $\omega$ is increased, $X_C$ will decrease or $Z$ will decrease. Hence $I_{\mathrm{mms}}$ or $P$ will increase. Therefore, bulb glows brighter. Hence the correct option is (b).