At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I0 cos ( \omega t), with I0 = 1 A and \omega = 500 rad s-1 starts flowing in it with the initial direction shown in the figure. At t = 7\pi \over 6\omega , the key is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 \mu F, R = 10 \Omega and the battery is ideal with emf of 50 V, identify the correct statement(s).

The current in the left part of the circuit just before t = 7\pi \over 6\omega is clockwise, Immediately after A is connected to D, the current in R is 10 A, Q = 2 \times 10 - 3 C

Magnitude of the maximum charge on the capacitor before t = 7\pi \over 6\omega is 1 \times 10 - 3 C

The current in the left part of the circuit just before t = 7\pi \over 6\omega is clockwise

Immediately after A is connected to D, the current in R is 10 A

Alternating Current PYQ — JEE Advanced 2014 · Shift 1 · Q48 with Solution

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Correct answer: B, C, D

Current in the left loop before switching

Given $I(t)=I_0\cos(\omega t),\qquad I_0=1\text{ A},\quad \omega=500\text{ rad s}^{-1}$

The switching time is $t_1=\frac{7\pi}{6\omega}$

So, $I(t_1)=\cos\left(\omega t_1\right)=\cos\left(\frac{7\pi}{6}\right)=-\frac{\sqrt3}{2}\text{ A}$

Since the initial direction shown corresponds to positive current, a negative value means that just before switching the current is opposite to the shown direction.

Hence, if the shown direction was anticlockwise initially, then just before switching it is clockwise.

So Option B is correct.

Charge on the capacitor before switching

Before switching, the capacitor is in the AC branch. For a capacitor, $i = \frac{dq}{dt}$

Thus $q(t)=\int i\,dt=\int I_0\cos(\omega t)dt=\frac{I_0}{\omega}\sin(\omega t)+\text{constant}$

Since at $t=0$ , current starts and capacitor charge may be taken zero, $q(t)=\frac{I_0}{\omega}\sin(\omega t)$

Maximum magnitude of charge is $q_{\max}=\frac{I_0}{\omega}=\frac{1}{500}=2\times 10^{-3}\text{ C}$

So Option A says $1\times 10^{-3}$ C, which is incorrect.

Hence Option A is false.

Charge on capacitor at the instant of switching

At $t=t_1=\frac{7\pi}{6\omega}$

we have

=\frac{1}{500}\left(-\frac12\right) =-\frac{1}{1000}\text{ C} =-1\times10^{-3}\text{ C}$$ So the capacitor has charge of magnitude $10^{-3}$ C, with polarity opposite to the positive reference. Its voltage at that instant is $$V_C(t_1)=\frac{q}{C}= rac{-10^{-3}}{20\times10^{-6}}=-50\text{ V}$$ Thus just before switching, the capacitor has **50 V** across it, opposite in polarity to the battery's charging polarity. --- 4. **Immediately after connecting A to D** After switching, the capacitor is connected to the ideal battery of emf 50 V through resistor $R=10\,\Omega$. Initial capacitor voltage is $-50$ V and final voltage is $+50$ V. Therefore initial potential difference across the resistor is $$\Delta V = 50-(-50)=100\text{ V}$$ Hence initial current through the resistor is $$I_R(0^+) = \frac{100}{10}=10\text{ A}$$ So **Option C is correct**. --- 5. **Total charge supplied by battery to fully charge capacitor** Initially, $$q_i=-CV=-20\times10^{-6}\times 50=-10^{-3}\text{ C}$$ Finally, when fully charged by the battery, $$q_f=+CV=+20\times10^{-6}\times 50=+10^{-3}\text{ C}$$ Total charge flowing from the battery is the increase in capacitor charge: $$Q=q_f-q_i=10^{-3}-(-10^{-3})=2\times10^{-3}\text{ C}$$ So **Option D is correct**. --- 6. **Final evaluation of options** - **A:** False - **B:** True - **C:** True - **D:** True Therefore, the correct options are $$\boxed{B,\ C,\ D}$$ --- 7. **Comparison with stored answer** Stored correct answer: **C, D** My derived answer: **B, C, D** I disagree with the stored answer because the current at switching time is $$I\left(\frac{7\pi}{6\omega}\right)=\cos\left(\frac{7\pi}{6}\right)<0,$$ so the current direction is opposite to the initially shown direction; hence statement **B** should also be correct.

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Correct answer: B, C, D

Current in the left loop before switching

Given $I(t)=I_0\cos(\omega t),\qquad I_0=1\text{ A},\quad \omega=500\text{ rad s}^{-1}$

The switching time is $t_1=\frac{7\pi}{6\omega}$

So, $I(t_1)=\cos\left(\omega t_1\right)=\cos\left(\frac{7\pi}{6}\right)=-\frac{\sqrt3}{2}\text{ A}$

Since the initial direction shown corresponds to positive current, a negative value means that just before switching the current is opposite to the shown direction.

Hence, if the shown direction was anticlockwise initially, then just before switching it is clockwise.

So Option B is correct.

Charge on the capacitor before switching

Before switching, the capacitor is in the AC branch. For a capacitor, $i = \frac{dq}{dt}$

Thus $q(t)=\int i\,dt=\int I_0\cos(\omega t)dt=\frac{I_0}{\omega}\sin(\omega t)+\text{constant}$

Since at $t=0$ , current starts and capacitor charge may be taken zero, $q(t)=\frac{I_0}{\omega}\sin(\omega t)$

Maximum magnitude of charge is $q_{\max}=\frac{I_0}{\omega}=\frac{1}{500}=2\times 10^{-3}\text{ C}$

So Option A says $1\times 10^{-3}$ C, which is incorrect.

Hence Option A is false.

Charge on capacitor at the instant of switching

At $t=t_1=\frac{7\pi}{6\omega}$

we have

=\frac{1}{500}\left(-\frac12\right) =-\frac{1}{1000}\text{ C} =-1\times10^{-3}\text{ C}$$ So the capacitor has charge of magnitude $10^{-3}$ C, with polarity opposite to the positive reference. Its voltage at that instant is $$V_C(t_1)=\frac{q}{C}= rac{-10^{-3}}{20\times10^{-6}}=-50\text{ V}$$ Thus just before switching, the capacitor has **50 V** across it, opposite in polarity to the battery's charging polarity. --- 4. **Immediately after connecting A to D** After switching, the capacitor is connected to the ideal battery of emf 50 V through resistor $R=10\,\Omega$. Initial capacitor voltage is $-50$ V and final voltage is $+50$ V. Therefore initial potential difference across the resistor is $$\Delta V = 50-(-50)=100\text{ V}$$ Hence initial current through the resistor is $$I_R(0^+) = \frac{100}{10}=10\text{ A}$$ So **Option C is correct**. --- 5. **Total charge supplied by battery to fully charge capacitor** Initially, $$q_i=-CV=-20\times10^{-6}\times 50=-10^{-3}\text{ C}$$ Finally, when fully charged by the battery, $$q_f=+CV=+20\times10^{-6}\times 50=+10^{-3}\text{ C}$$ Total charge flowing from the battery is the increase in capacitor charge: $$Q=q_f-q_i=10^{-3}-(-10^{-3})=2\times10^{-3}\text{ C}$$ So **Option D is correct**. --- 6. **Final evaluation of options** - **A:** False - **B:** True - **C:** True - **D:** True Therefore, the correct options are $$\boxed{B,\ C,\ D}$$ --- 7. **Comparison with stored answer** Stored correct answer: **C, D** My derived answer: **B, C, D** I disagree with the stored answer because the current at switching time is $$I\left(\frac{7\pi}{6\omega}\right)=\cos\left(\frac{7\pi}{6}\right)<0,$$ so the current direction is opposite to the initially shown direction; hence statement **B** should also be correct.

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