An inductor of self inductance 1 H connected in series with a resistor of $$100\pi$$ ohm and an ac supply of $$100\pi$$ volt, 50 Hz. Maximum current flowing in the circuit is ______ A.

Given
Self-inductance $$L = 1 \text{ H}$$, Resistance $$R = 100\pi \, \Omega$$, Frequency $$f = 50 \text{ Hz}$$, RMS supply voltage $$V_{rms}=100\pi \text{ V}$$.

First find the angular frequency.
Formula: $$\omega = 2\pi f$$ $$-(1)$$
Using $$f = 50$$ Hz,
$$\omega = 2\pi \times 50 = 100\pi \text{ rad s}^{-1}$$.

Inductive reactance $$X_L$$ is
$$X_L = \omega L$$ $$-(2)$$
Substituting $$\omega = 100\pi$$ and $$L = 1$$ H,
$$X_L = 100\pi \, \Omega$$.

The impedance of an $$R-L$$ series circuit is
$$Z = \sqrt{R^2 + X_L^2}$$ $$-(3)$$.
Here $$R = 100\pi$$ and $$X_L = 100\pi$$, so
$$Z = \sqrt{(100\pi)^2 + (100\pi)^2} = 100\pi\sqrt{2}\, \Omega$$.

RMS current:
$$I_{rms} = \frac{V_{rms}}{Z}$$ $$-(4)$$.
Putting $$V_{rms}=100\pi$$ V and $$Z = 100\pi\sqrt{2}$$ Ω,
$$I_{rms} = \frac{100\pi}{100\pi\sqrt{2}} = \frac{1}{\sqrt{2}} \text{ A}$$.

Maximum (peak) current relates to RMS current by
$$I_0 = \sqrt{2}\, I_{rms}$$ $$-(5)$$.
Hence
$$I_0 = \sqrt{2} \times \frac{1}{\sqrt{2}} = 1 \text{ A}$$.

Therefore, the maximum current flowing in the circuit is 1 A.