The power factor of an AC circuit having resistance R and inductance L (connected in series) and an

Q: Why is the power factor defined as $\cos \phi$ and not $\sin \phi$?

Power factor represents the ratio of true power (which does actual work) to apparent power. True power is calculated using the component of current that is in phase with the voltage, which is given by $I \cos \phi$.

Q: Can we just add resistance $R$ and reactance $X_L$ algebraically to get total impedance?

No! Resistance and reactance are out of phase by $90^\circ$. They must be added as vectors (phasors), which is why we use the Pythagorean theorem $Z = \sqrt{R^2 + X_L^2}$.

Q: What happens to the power factor if we increase the frequency of the AC source?

As frequency ($\omega$) increases, the inductive reactance ($X_L = \omega L$) increases. This makes the impedance triangle taller, increasing the phase angle $\phi$ and thereby decreasing the power factor $\cos \phi$.

Q: Is it possible for the power factor to be negative?

In standard passive circuits (containing only R, L, and C), the phase angle $\phi$ is always between $-90^\circ$ and $+90^\circ$. Since the cosine of any angle in this range is positive, the power factor is always between 0 and 1.

Q: What would the power factor be if the circuit only had a resistor?

If there is no inductor ($L=0$), the reactance is zero. The impedance $Z$ simply equals $R$, making $\cos \phi = \frac{R}{R} = 1$. The voltage and current are perfectly in phase!

Question

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

Options:

(A) $\frac{R}{\omega L}$

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

(C) $\frac{\omega L}{R}$

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

Sigma Percentile · Accepted Answer

### Understanding Power Factor in AC Circuits

When dealing with Alternating Current (AC) circuits, the concept of **power factor** is absolutely crucial. It tells us how effectively the electrical power is being converted into useful work output. In a purely resistive circuit, all the power is useful, but when we introduce inductors or capacitors, things get a bit more interesting!

Imagine you are analyzing a series $R-L$ circuit, which consists of a resistor with resistance $R$ and an inductor with inductance $L$. The AC source provides an angular frequency $\omega$. Because of the inductor, the current and voltage are no longer perfectly in sync; they are out of phase by an angle $\phi$.

### The Impedance Triangle

To visualize this phase difference and the overall opposition to current flow, we use a brilliant geometric tool called the **impedance triangle**.

Think of it as a right-angled triangle where the horizontal base represents the resistance $R$. The vertical perpendicular side represents the inductive reactance, denoted as $X_L$. We know that inductive reactance is given by the formula $X_L = \omega L$.

Now, what about the hypotenuse? According to Pythagoras' theorem, the hypotenuse represents the total impedance $Z$ of the circuit.

$$Z = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + \omega^2 L^2}$$

### Calculating the Power Factor

The power factor is mathematically defined as the cosine of the phase angle $\phi$. In our impedance triangle, $\phi$ is the angle between the base (resistance $R$) and the hypotenuse (total impedance $Z$).

Using basic trigonometry from our right-angled triangle:

$$\cos \phi = \frac{	ext{Base}}{	ext{Hypotenuse}}$$

Substituting the values we established:

$$\cos \phi = \frac{R}{Z}$$

$$\cos \phi = \frac{R}{\sqrt{R^2 + \omega^2 L^2}}$$

And there we have it! The power factor for a series $R-L$ circuit is elegantly expressed as $\frac{R}{\sqrt{R^2 + \omega^2 L^2}}$. This perfectly matches option (b).

This geometric approach not only gives us the formula but also provides a deep physical intuition about how resistance and reactance compete to determine the efficiency of an AC circuit.

Animated Solution for Physics - Electromagnetic Induction: The power factor of an AC circuit having resistance $R$ and inductance $L$ (connected in series) and an angular velocity $ω$ is

Select Answer:

Visualized Solution

Impedance Triangle

Power Factor

Calculating Impedance

Evaluating $cos ϕ$

Final Answer

The Way Forward

The Sigma Insight: Alternating Current (AC) and Voltage

Understanding Power Factor in AC Circuits

The Impedance Triangle

Calculating the Power Factor

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