A resistor of resistance 30 Ω and a capacitor of capacitance $(250 π^{-1}) μF$ are connected in series to a 220 V, 50 Hz source. The current in the circuit is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4.4 A
Given:
Resistor: $R = 30 \, \Omega$
Capacitor: $C = \frac{250}{\pi} \, \mu F = \frac{250 \times 10^{-6}}{\pi} \, F$
AC source: $V = 220 \, V$, $f = 50 \, Hz$
Capacitive reactance: $X_C = \frac{1}{2 \pi f C}$
$X_C = \frac{1}{2 \pi (50) (250 \times 10^{-6} / \pi)}$
$X_C = \frac{1}{2 \pi \cdot 50 \cdot 250 \cdot 10^{-6} / \pi}$
$X_C = \frac{1}{2 \cdot 50 \cdot 250 \cdot 10^{-6}}$
$X_C = \frac{1}{0.025} = 40 \, \Omega$
Impedance of series RC circuit: $Z = \sqrt{R^2 + X_C^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, \Omega$
Current: $I = \frac{V}{Z} = \frac{220}{50} = 4.4 \, A$
Answer: The current in the circuit is 4.4 A.