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:
Answer & explanation
Correct answer: option 4
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, $ω=2\pi f$
$=2\pi ×50$
$=100\pi\, rad/s$
$∴X_L=ωL$
$=100\pi ×0.5=50\pi Ω$
$≃157.1Ω$
$X_C=\frac{1}{ωC}=\frac{1}{100\pi ×100×10^{-6}}$
$=\frac{1}{0.01\pi}≃31.8Ω$
$∴Z=\sqrt{50^2+(157.8-31.8)^2}$
$=\sqrt{18208.1}$
$≃134.6Ω$