As shown in the figure an inductor of inductance $$200$$ mH is connected to an AC source of emf $$220$$ V and frequency $$50$$ Hz. The instantaneous voltage of the source is $$0$$ V when the peak value of current is $$\frac{\sqrt{a}}{\pi}$$ A. The value of $$a$$ is ______.

We need to find the value of $$a$$ for a purely inductive AC circuit where an inductor is connected across an alternating electromotive force (emf) source.

1. Identify the Given Data

• Inductance ($$L$$) = $$200\text{ mH} = 200 \times 10^{-3}\text{ H} = 0.2\text{ H}$$
• RMS Voltage ($$V_{rms}$$) = $$220\text{ V}$$
• Frequency ($$f$$) = $$50\text{ Hz}$$
• Peak Current ($$I_0$$) = $$\frac{\sqrt{a}}{\pi}\text{ A}$$

Note: In a purely inductive circuit, the current lags the voltage by $$90^\circ$$ ($$\frac{\pi}{2}$$ radians). Consequently, when the instantaneous voltage of the source drops to $$0\text{ V}$$, the current reaches its maximum magnitude (peak value $$I_0$$).

2. Calculate the Inductive Reactance ($$X_L$$)

The opposition offered by the inductor to the AC flow is given by:

$$X_L = \omega L = 2\pi f L$$

Substituting the given values:

$$X_L = 2 \times \pi \times 50 \times 0.2 = 20\pi\ \Omega$$

3. Calculate the Peak Voltage ($$V_0$$)

The relationship between the alternating source's peak voltage and its RMS value is:

$$V_0 = V_{rms} \times \sqrt{2} = 220\sqrt{2}\text{ V}$$

4. Calculate the Peak Current ($$I_0$$)

Using Ohm's law for an alternating inductive circuit network:

$$I_0 = \frac{V_0}{X_L}$$

Substituting our expressions for $$V_0$$ and $$X_L$$ yields:

$$I_0 = \frac{220\sqrt{2}}{20\pi} = \frac{11\sqrt{2}}{\pi}\text{ A}$$

5. Solve for $$a$$

We equate our derived value of peak current to the given mathematical form:

$$\frac{\sqrt{a}}{\pi} = \frac{11\sqrt{2}}{\pi}$$

Canceling out the common denominator $$\pi$$ from both sides gives:

$$\sqrt{a} = 11\sqrt{2}$$

Squaring both sides of the equation to isolate $$a$$:

$$a = (11\sqrt{2})^2 = 121 \times 2 = 242$$

Therefore, the value of $$a$$ is 242.