Question

A series RC circuit has R=240 and C=102uF. It is connected in parallel to an inductor of 200mH and the combination is connected across a 220V, 60Hz source. Find the current in each branch.

Answer 1

R=240 ohm

V=220V

$C=102\mu F$

f=60Hz

L=200mH

$Z=\sqrt{R^2+(X_L-X_c)^2}$

$$X_L=wL=2\pi fL=2\pi\times60\times200\times10^{-3}=75.36\Omega$$

$$X_L=\frac{1}{2\pi fC}=\frac{1}{2\pi\times60\times102\times10^{-6}}=26.01\Omega$$

$$X_L-X_C=75.36-26.01=49.35\Omega$$

Impedence

$Z=\sqrt{240^2+49.35^2}=245.02\Omega$

Current

$$I=\frac{E}{Z}=\frac{220}{245.02}=0.8978Amp$$

$Z_{RC}=\sqrt{R^2+X_C^2}$

$$Z_{RC}=\sqrt{240^2+26.01^2}=241.405\Omega$$

$Z_L=X_L$

Now current

$I_L=\frac{V}{X_L}$

$I_L=\frac{220}{75.36}=2.92Amp$

$I_{RC}=\frac{V}{Z_{RC}}=\frac{220}{241.36}=0.91Amp$