An alternating emf $$E = 440\sin 100\pi t$$ is applied to a circuit containing an inductance of $$\frac{\sqrt{2}}{\pi}$$ H. If an a.c. ammeter is connected in the circuit, its reading will be:

We have an alternating emf $$E = 440\sin(100\pi t)$$ applied to a pure inductor of inductance $$L = \frac{\sqrt{2}}{\pi}$$ H.

From the emf expression, the peak voltage is $$E_0 = 440$$ V and the angular frequency is $$\omega = 100\pi$$ rad/s.

The inductive reactance is $$X_L = \omega L = 100\pi \times \frac{\sqrt{2}}{\pi} = 100\sqrt{2}$$ $$\Omega$$.

The peak current through the inductor is $$I_0 = \frac{E_0}{X_L} = \frac{440}{100\sqrt{2}} = \frac{4.4}{\sqrt{2}}$$ A.

An a.c. ammeter reads the rms value of current, which is $$I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{4.4}{\sqrt{2} \times \sqrt{2}} = \frac{4.4}{2} = 2.2$$ A.

Hence, the correct answer is Option C.