A capacitor of 100 µF, a resistor of resistance $50 \Omega$ and an inductor of inductance 0.5 H are connected in series with a 110 V, 50 Hz AC source.

The impedance of the circuit is:

$134.6 \Omega$

The correct answer is Option (4) → $134.6 \Omega$

Impedance, $Z = \sqrt{R^2 + \left(X_L - X_C\right)^2}$

where:

$X_L$, Inductive resistance = $ωL$

$X_C$, Capacitive resistance = $\frac{1}{ωC}$

R (Resistor) = 50 Ω

Now,

$C=100μF=100×10^{-6}F$

$L=0.5H$

$f=50Hz$

$V=110V$

Angular frequency, $ω=2\pi f$

$=2\pi ×50$

$=100\pi\, rad/s$

$∴X_L=ωL$

$=100\pi ×0.5=50\pi Ω$

$≃157.1Ω$

$X_C=\frac{1}{ωC}=\frac{1}{100\pi ×100×10^{-6}}$

$=\frac{1}{0.01\pi}≃31.8Ω$

$∴Z=\sqrt{50^2+(157.8-31.8)^2}$

$=\sqrt{18208.1}$

$≃134.6Ω$