Question

A resistor of 200 omega & a capacitor 15.0 microfarad are are connected in series to a 220V, 50 Hz a.c. source. Calculate the current in the circuit and the r.m.s voltage across the resistor and capacitor.

Answer 1

(a)

The rms current in the circuit is

I_rms = V_rms / Z (here V_rms = V_max/√2)

here Z = √[R²+(X_C )²]

so,

I_rms = V_rms/ √[R²+(X_C )²] ( here capacitive reactance X_C = 1/ωC = 1/2Ï€fC)
= V_rms / √[R² +(1/(2Ï€fC))²]
given:

V_max = 220 V; V_rms = 220/√2 = 155.57 V

R = 200 ohms

f = 50 Hz

C = 15 uF = 15 x 10^-6 F

so,

I_rms = 155.57 / √[200² +(1/(2Ï€ x 50 x 5 x 10^-6 ))²]

which can be calculated

.
b)

Now, the rms voltage drop across the resistor is
ΔVrms = Irms x R

.
c)

The rms voltage drop across the capacitor is
ΔV_rms = I_rms*X_C
= I_rms*1/ ωC
= I_rms*1/(2Ï€fC)