An ac source of angular frequency ω is fed across a resistor R and a capacitor C in series. The current registered is I . If now the frequency of the source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of the reactance to resistance at the original frequency ω is given as √ x/5 . Find x . - Padhle.Online
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# An ac source of angular frequency ω is fed across a resistor R and a capacitor C in series. The current registered is I . If now the frequency of the source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of the reactance to resistance at the original frequency ω is given as √ x/5 . Find x .

The root-mean-square current Irms in the RC circuit is given by

Irms = Vrms/√R2 +X2C…… (1)

Here, Vrms is the root-mean-square voltage, R is the resistance and XC is the capacitive reactance.

The capacitive reactance XC is given by XC = 1/ωC …… (2)

Here, ω is the angular frequency and C is the capacitance.

We have given that the initial angular frequency of the source is ω and the current is I .

Substitute I for Irms and 1/ωC for XCin equation (1).

I = Vrms√R2+(1ω/C)2

⇒ I = Vrms/√R2+1ω2C2

⇒ I = VrmsωC/√R2ω2C2+1 …… (3)

When the angular frequency of the source is changed to ω/3 then the current becomes I/2 but the potential is the same.

Hence, the above equation becomes

⇒ I/2 = Vrmsω/3C/√R2(ω/3)2C2+1

⇒ I/2 = VrmsωC/√R2ω2C+ 9 …… (4)

Divide equation (3) by equation (4). ⇒

I/I/2 = VrmsωC/√R2ω2C+ 1/VrmsωC/√R2ω2C+ 9

⇒2 = √R2ω2C+ 9 / √R2ω2C+ 1

Take square on both sides of the above equation.

⇒ 4=R2ω2C2+9 / R2ω2C+ 1

⇒ 4R2ω2C2+4 = R2ω2C2+9

⇒ 4R2ω2C2−R2ω2C2=9−4 ⇒3R2ω2C2=5

⇒ R2ω2C= 5/3

⇒ R2=5/3 1/ω2C2

Take square root on both sides of the above equation.

⇒R =√5/3 1/ωC

Substitute XC for 1/ωC in the above equation.

XC/R = √3/5

## About the Author

Principal, Babu Daudayal SVM, Mathura

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