Q.A capacitor of 100 $\mu F$, a coil of resistance 50 $\Omega$ and inductance 0.5 H are connected in series with a 110 V-50 Hz source. Calculate the rms value of current in the circuit.

[!TLDR] $X_L=2\pi fL\approx157.08\,\Omega$, $X_C=1/(2\pi fC)\approx31.83\,\Omega$; $Z=\sqrt{50^2+(157.08-31.83)^2}\approx134.86\,\Omega$; $i_{rms} …

Given C = 100 $\mu F$, R = 50 $\Omega$, L = 0.5 H, $V_{rms}=110$ V, f = 50 Hz. $X_L=2\pi fL=2\times3.1416\times50\times0.5=157.08\,\Omega$. $X_C=\dfrac{1}{2\pi fC}=\dfrac{1}{2\times3.1416\times50\times100\times10^{-6}}=\dfrac{1}{0.031416}\approx31.83\,\Omega$.\n\nImpedance: $Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{50^2+(157.08-31.83)^2}=\sqrt{2500+(125.25)^2}=\sqrt{2500+15687.6}=\sqrt{18187.6}\approx134.86\,\Omega$.\n\nRms current: $i_{rms …

Compute $X_L$ and $X_C$ at the given 50 Hz, find Z from the LCR impedance f …

Adding $X_L$ and $X_C$ instead of subtracting them before squaring -- they must be combined as $(X_L-X_C)$, no …