A capacitor (C) and resistor (R) are connected in series with an ac source of voltage of frequency 50\; Hz. The potential difference across C and R are respectively 120\; V,\; 90\; V. and the current in the circuit is 3A. Calculate (i) the impedance of the circuit (ii) the value of the inductance, which when connected in series with C and R will make the power factor of the circuit unity.

 

 
 
 
 
 

Answers (1)

Calculating the impedance of the circuit.

Z=\sqrt{R^{2}+X^{2}_{C}} ____(1)

R=\frac{V_{R}}{I_{R}}=30\; \Omega

That's,

X_{C}=\frac{V_{C}}{I_{C}}=\frac{120}{30} Where, V_{C} is the potential difference accross C.

X_{C}=40\; \Omega

Now putting the values in eqn (1), we get inpedance

Z=\sqrt{(30)^{2}+(40)^{2}}=50\; \Omega

Such that, C and R are connected in series so, 

As the power factor =1

X_{C}=X_{L}

2\pi f L=40

 100\pi L=40

L=\frac{2}{5\pi }    Henry

L=\frac{2}{5\pi } is the value of inductance.

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