Alternating Current — NEET UG practice

1. Recall that in a series LCR circuit, resonance occurs when $X_L = X_C$, and the frequency at which this happens is given by the resonance frequency formula.

$$f_0 = \frac{1}{2\pi\sqrt{LC}}$$

2. Substitute the given values $L = 1\times10^{-3}\,\text{H}$ and $C = 0.1\times10^{-6}\,\text{F}$ into the formula.

$$f_0 = \frac{1}{2\pi\sqrt{(1\times10^{-3})(0.1\times10^{-6})}}$$

3. Simplify the product under the square root and evaluate.

$$f_0 = \frac{1}{2\pi\sqrt{1\times10^{-10}}} = \frac{1}{2\pi\times10^{-5}} \approx 15.9\,\text{kHz}$$

Hence, the resonance frequency is approximately 15.9 kHz, so the correct answer is option D.

1. Since the current starts from zero and grows like a sine wave, write it in the standard AC form.

$$i = i_{\text{peak}}\sin(\omega t)$$

2. Find $\omega$ using $\omega = 2\pi f$, since angular frequency relates directly to the given frequency.

$$\omega = 2\pi \times 60 = 120\pi \ \text{rad/s}$$

3. Substitute $\omega$ back into the current equation.

$$i = 5\sin(120\pi t)$$

4. The current reaches its peak value ($i = 5$ A) when the sine term equals 1, since $\sin\theta$ is maximum at $\theta = \frac{\pi}{2}$.

$$5 = 5\sin(120\pi t) \implies \sin(120\pi t) = \sin\left(\frac{\pi}{2}\right)$$

5. Solve for $t$ by equating the angles.

$$120\pi t = \frac{\pi}{2} \implies t = \frac{1}{240}\,\text{s}$$

Hence, the time taken to reach the peak value is $\frac{1}{240}$ s, so the correct answer is option D.

1. Compare the given voltage equation with the standard form $V = V_0\sin(\omega t)$ to pick out the angular frequency of the source.

$$\omega = 2\times10^3 \text{ rad/s}$$

2. Find the inductive reactance using $X_L = \omega L$, since this tells us how much the inductor opposes the changing current.

$$X_L = (2\times10^3)(10\times10^{-3}) = 20\,\Omega$$

3. Find the capacitive reactance using $X_C = \dfrac{1}{\omega C}$, since this tells us how much the capacitor opposes the changing current.

$$X_C = \frac{1}{(2\times10^3)(25\times10^{-6})} = 20\,\Omega$$

4. Notice that $X_L = X_C$, which means the circuit is at resonance. At resonance the inductive and capacitive effects cancel each other out, so the total impedance is just the resistance.

$$Z = R = 100\,\Omega$$

5. Calculate the current amplitude using $i_0 = \dfrac{V_0}{Z}$, the AC equivalent of Ohm's law for peak values.

$$i_0 = \frac{220}{100} = 2.2\,A$$

Hence, the current amplitude is 2.2 A, so the correct answer is option B.