# 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.

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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