Find the power factor of the given A.C circuit

The correct answer is 1.

To find the power factor of the given A.C. circuit, we analyze the circuit diagram which consists of an inductor, a capacitor, and a resistor connected in series (an RLC series circuit) to an AC source.

From the given image, we can identify the following values for the components:
Inductance, $L=\frac{1}{10\pi }\text{ H}$
Capacitance, $C=\frac{1}{1000\pi }\text{ F}$
Resistance, $R=10\Omega$
AC Source frequency, $f=50\text{ Hz}$

Step 1: Calculate the angular frequency ($\omega$)
The angular frequency of the AC source is given by:
$\omega =2\pi f$
Substituting the given value of frequency:
$\omega =2\pi \left(50\right)=100\pi \text{ rad/s}$

Step 2: Calculate the inductive reactance ($X_L$)
The inductive reactance is given by:
$X_L=\omega L$
Substituting the values of $\omega$ and $L$:
$X_L=100\pi \times \frac{1}{10\pi }=10\Omega$

Step 3: Calculate the capacitive reactance ($X_C$)
The capacitive reactance is given by:
$X_C=\frac{1}{\omega C}$
Substituting the values of $\omega$ and $C$:
$X_C=\frac{1}{100\pi \times \frac{1}{1000\pi }}=\frac{1}{1/10}=10\Omega$

Step 4: Find the total impedance ($Z$)
The impedance of a series RLC circuit is calculated as:
$Z=\sqrt{R^2+{(X_L-X_C)}^2}$
Since $X_L=X_C=10\Omega$, the circuit is in a state of electrical resonance:
$Z=\sqrt{{10}^2+{(10-10)}^2}=\sqrt{{10}^2}=10\Omega$

Step 5: Determine the power factor ($cos\varphi$)
The power factor ($cos\varphi$) of the AC circuit is the ratio of resistance to impedance:
$cos\varphi =\frac{R}{Z}$
Substituting the values:
$cos\varphi =\frac{10}{10}=1$

Therefore, the power factor of the given AC circuit is 1.