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The average power in an AC circuit expressed in terms of the rms voltage and current as \begin{align} P_\text{avg}=V_\text{rms}I_\text{rms}\cos\phi \end{align} where $\phi$ is the phase angle between the voltage and the current. The term $\cos\phi$ is called the power factor. It is given by \begin{align} \cos\phi=\frac{R}{Z}. \end{align} For a purely resistive circuit $Z=R$ and the power factor is 1.

Power factor is a measure of the efficiency of an AC power system. A power factor of 1 (or 100%) means that all the power supplied to the circuit is being used by the load and that there is no wasted energy in the form of reactive power. On the other hand, a power factor less than 1 indicates that some of the energy supplied to the circuit is not being used by the load, but is instead being stored in the circuit components and then returned back to the source. This results in inefficient use of energy and can lead to higher energy costs.

Improving the power factor of an AC system can be done by adding reactive components, such as capacitors or inductors, to the circuit to counteract the effect of the reactive power and bring the power factor closer to 1.

**Problem (JEE Mains 2005):**
A circuit has a resistance of 12 ohm and an impedance of 15 ohm. The power factor of the circuit will be

- 0.8
- 0.4
- 1.25
- 0.125

**Solution:**
The power factor is given by
\begin{align}
\cos\phi &=\frac{R}{Z} \\
&=\frac{12}{15} \\
&=0.8\nonumber
\end{align}

**Problem (JEE Mains 2002):**
The power factor of an AC circuit having resistance ($R$) and inductance ($L$) connected in series and an angular velocity $\omega$ is

- $\frac{R}{\omega L}$
- $\frac{R}{(R^2+\omega^2 L^2)^{\! 1/2}}$
- $\frac{\omega L}{R}$
- $\frac{R}{(R^2 - \omega^2 L^2)^{\! 1/2}}$

**Solution:**
The power factor of an LR circuit is given by,
\begin{align}
\cos\phi &=\frac{R}{Z} \\
&=\frac{R}{\sqrt{R^2+\omega^2 L^2}}.\nonumber
\end{align}

**Problem (JEE Mains 2007):**
In an AC circuit the applied voltage is
\begin{align}
V=V_0 \sin\omega t.
\end{align}
The resulting current in the circuit is
\begin{align}
I=I_0\sin\left(\omega t-\frac{\pi}{2}\right).
\end{align}
The power consumption in the circuit is given by

- $\sqrt{2}V_0 I_0$
- $V_0 I_0/\sqrt{2}$
- $0$
- $V_0 I_0/2$

**Solution:**
The voltage $V$ **leads** the current $I$ by a phase angle $\phi=\pi/2$. It is a purely inductive circuit. The power consumption in the circuit is given by,
\begin{align}
P_\text{avg}=V_\text{rms}I_\text{rms} \cos\phi=0.\nonumber
\end{align}

**Problem (JEE Mains 2022):**
In a series LR circuit $X_L=R$ and power factor of the circuit is $P_1$. When capacitor with capacitance $C$ such that $X_L=X_C$ is put in series, the power factor becomes $P_2$. The ratio $P_1/P_2$ is

- 1/2
- $1/\sqrt{2}$
- $\sqrt{3}/\sqrt{2}$
- 2/1

**Problem (JEE Mains 2021):**
In a series LCR circuit, the inductive reactance $X_L$ is $10\Omega$ and the capacitive reactance $X_C$ is $4\Omega$. The resistance $R$ in the circuit is $6\Omega$. The power factor of the circuit is

- $1/\sqrt{2}$
- $\sqrt{3}/2$
- 1/2
- $1/2\sqrt{2}$

**Problem (JEE Mains 2020):**
In a series LR circuit, power of 400 W is dissipated from a source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In order to bring the power factor to unity, a capacitor of value $C$ is added in series to the $L$ and $R$. Taking the value of $C$ as $\frac{n}{3\pi}\;\mu$F, then value of $n$ is________