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Question: A 0.2\(k\Omega\) resistor and \(15\mu F\) capacitor are connected in series to a 220 V, 50 Hz ac sou...

A 0.2kΩk\Omega resistor and 15μF15\mu F capacitor are connected in series to a 220 V, 50 Hz ac source. The impedance of the circuit is

A

250 Ω\text{250 }\Omega

B

268 Ω\text{268 }\Omega

C

29.15 Ω\text{29.15 }\Omega

D

291.5 Ω\text{291.5 }\Omega

Answer

291.5 Ω\text{291.5 }\Omega

Explanation

Solution

: Here, ,R=0.2KΩ=200ΩR = 0.2K\Omega = 200\Omega

C=15μF=15×106FC = 15\mu F = 15 \times 10^{- 6}F

Vrms=220V,υ=50HzV_{rms} = 220V,\upsilon = 50Hz

Capacitive reactance, XC=12πυCX_{C} = \frac{1}{2\pi\upsilon C}

=12×3.14×50×15×106= \frac{1}{2 \times 3.14 \times 50 \times 15 \times 10^{- 6}}

=212Ω= 212\Omega

The impedance of the RC circuit is,

Z=R2+XC2=(200)2+(212)2=291.5ΩZ = \sqrt{R^{2} + X_{C}^{2}} = \sqrt{(200)^{2} + (212)^{2}} = 291.5\Omega