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# A resistor of 200 ohm and a capacitor of 15micro farad are connected in series to a 220V ,50hz ac source.calculate the current in the circuit and the RMS voltage across the resistor and the capacitor.is the algebraic sum of these voltages more than the source voltage?if yes resolve the paradox

## 3 Answers

2 years ago

Given that

R = 200Ω

C = 15 μF = 15 × 10–6 F

ξ = ξ sinωt

ξ = 220 V

f = 50 Hz

W = 2π= (100π) rad/s

ξ = 220 sin (100π)

Reactance of capacitor

Xc = 1/ We

= 1 / 100 * pi * 15 * 10^-6

= 10^6 / 314 * 15

= 212.31 ohm

So, impedance

Z = sqrt (R^2 + X^2)

Z = sqrt (200^2 + 212.31^2)

= 291.67 ohm

I =  E / Z = 220 sin (100* pi * t) / 291.67

= 0.75 sin (100 * pi *t)

Irms = 0.75A

Voltage across capacitor VR = Irms × XC

= 0.75 × 200

= 150 volt

Voltage across capacitor VC = Irms × XC

= 0.75 × 212.31

= 159.23 volt

2 years ago
Dear student

Given that

R = 200Ω

C = 15 μF = 15 × 10–6 F

ξ = ξ sinωt

ξ = 220 V

f = 50 Hz

W = 2π= (100π) rad/s

ξ = 220 sin (100π)

Reactance of capacitor

Xc = 1/ We

= 1 / 100 * pi * 15 * 10^-6

= 10^6 / 314 * 15

= 212.31 ohm

So, impedance

Z = sqrt (R^2 + X^2)

Z = sqrt (200^2 + 212.31^2)

= 291.67 ohm

I =  E / Z = 220 sin (100* pi * t) / 291.67

= 0.75 sin (100 * pi *t)

Irms = 0.75A

Voltage across capacitor VR = Irms × XC

= 0.75 × 200

= 150 volt

Voltage across capacitor VC = Irms × XC

= 0.75 × 212.31

= 159.23 volt

Regards

Arun (askIITians forum expert)

11 months ago
Dear Student,
Please find below the solution to your problem.

Given that
R = 200Ω
C = 15 μF
= 15 × 10–6F
ξ = ξ sinωt
ξ = 220 V
f= 50 Hz
W = 2πf
= (100π) rad/s
ξ = 220 sin (100π)
Reactance of capacitor
Xc = 1/ We
= 1 / 100 * pi * 15 * 10^-6
= 10^6 / 314 * 15
= 212.31 ohm
So, impedance
Z = sqrt (R^2 + X^2)
Z = sqrt (200^2 + 212.31^2)
= 291.67 ohm
I = E / Z
= 220 sin (100* pi * t) / 291.67
= 0.75 sin (100 * pi *t)
Irms= 0.75A
Voltage across capacitor VR= Irms× XC
= 0.75 × 200
= 150 volt
Voltage across capacitor VC= Irms× XC
= 0.75 × 212.31
= 159.23 volt

Thanks and Regards

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