Alternating Current — JEE Main practice

1. When a charged capacitor is connected to an inductor, energy oscillates between the capacitor's electric field and the inductor's magnetic field, with charge on the capacitor varying as

$$q(t) = Q_0\cos(\omega t), \qquad \omega = \frac{1}{\sqrt{LC}}$$

2. Initially, all the energy resides in the capacitor. [IMAGE: LC circuit diagram showing charge oscillating between capacitor and inductor over time]

$$U_0 = \frac{Q_0^2}{2C}$$

3. The energy stored in the capacitor at time $t$ follows from the charge expression.

$$U_C(t) = \frac{q(t)^2}{2C} = U_0\cos^2(\omega t)$$

4. By conservation of energy, whatever leaves the capacitor goes into the inductor as magnetic energy.

$$U_L(t) = U_0-U_C(t) = U_0\sin^2(\omega t)$$

5. Given that 25% of the initial energy is transferred to the inductor, set up the equation.

$$U_0\sin^2(\omega t) = \frac{1}{4}U_0 \;\Rightarrow\; \sin^2(\omega t) = \frac{1}{4} \;\Rightarrow\; \sin(\omega t) = \frac{1}{2}$$

6. Solve for $\omega t$ using the smallest positive angle satisfying this.

$$\omega t = \frac{\pi}{6}$$

7. Substitute $\omega=\dfrac{1}{\sqrt{LC}}$ and solve for $t$.

$$\frac{t}{\sqrt{LC}} = \frac{\pi}{6} \;\Rightarrow\; t = \frac{\pi\sqrt{LC}}{6}$$

Hence, the value of $t$ is $\dfrac{\pi\sqrt{LC}}{6}$ s, so the answer is option B.

1. The power factor of an AC circuit is the ratio of true resistance to total impedance.

$$\cos\phi = \frac{R}{Z}$$

2. In a series LCR circuit, inductive and capacitive reactances oppose each other, so the net reactance is their difference.

$$X = |X_L-X_C|$$

3. Substituting the given reactance values from the figure ($X_L=70\Omega$, $X_C=150\Omega$).

$$X = |70-150| = 80\Omega$$

4. The total impedance combines resistance and net reactance. [IMAGE: Impedance diagram showing R=60Ω and net reactance X=80Ω forming the impedance triangle]

$$Z = \sqrt{R^2+X^2} = \sqrt{(60)^2+(80)^2} = \sqrt{10000} = 100\Omega$$

5. Compute the power factor.

$$\cos\phi = \frac{R}{Z} = \frac{60}{100} = \frac{6}{10}$$

6. Compare with the given form $\dfrac{\alpha}{10}$ to solve for $\alpha$.

$$\frac{\alpha}{10} = \frac{6}{10} \;\Rightarrow\; \alpha = 6$$

Hence, the value of $\alpha$ is 6, so the answer is option A.

1. The r.m.s value of a time-varying current over a period $T$ is defined as the square root of the average of its square.

$$i_{rms} = \sqrt{\frac{1}{T}\int_0^T i^2\,dt}$$

2. Given $i=i_o\left(\dfrac{t}{T}\right)$, substitute this into the formula.

$$i_{rms}^2 = \frac{1}{T}\int_0^T i_o^2\frac{t^2}{T^2}\,dt$$

3. Since $i_o$ and $T$ are constants, factor them out of the integral and evaluate.

$$i_{rms}^2 = \frac{i_o^2}{T^3}\left[\frac{t^3}{3}\right]_0^T = \frac{i_o^2}{T^3}\times\frac{T^3}{3} = \frac{i_o^2}{3}$$

4. Take the square root to find $i_{rms}$.

$$i_{rms} = \frac{i_o}{\sqrt3}$$

Hence, the r.m.s current is $\dfrac{i_o}{\sqrt3}$, so the answer is option C.