Q.An AC circuit consists of only an inductor of inductance 2 H. If the current is represented by a sine wave of amplitude 0.25 A and frequency 60 Hz, calculate the effective potential difference across the inductor. (Use $\pi = 3.142$)

[!TLDR] $X_L=2\pi fL=2\pi(60)(2)\approx754.1\,\Omega$; $i_{rms}=i_0/\sqrt2=0.25/\sqrt2\approx0.1768$ A; $V_{rms}=i_{rms}X_L\approx133.4$ V. [!ANSWER] Ef …

Given L = 2 H, current amplitude $i_0=0.25$ A, f = 60 Hz, $\pi=3.142$. Inductive reactance: $X_L=2\pi fL=2\times3.142\times60\times2=754.08\,\Omega$.\n\nThe rms (effective) current is $i_{rms}=\dfrac{i_0}{\sqrt2}=\dfrac{0.25}{1.414}\approx0.1768$ A. Since a pure inductor obeys $V=iX_L$ for both peak and rms values (Ohm's-law-like relation, using reactance in place of resistance), the effective potential d …

Compute $X_L=2\pi fL$, convert peak current to rms ($i_{rms}=i_0/\sqrt2$), …

Multiplying the reactance by the PEAK current $i_0$ instead of the rms current $i_{rms}$ when the question specificall …