Animated Solution for Physics - Electromagnetic Induction: An AC circuit has an inductor and a resistor of resistance $R$ in series, such that $X_L=3R$. Now, a capacitor is added in series such that $X_C=2R$. The ratio of new power factor with the old power factor of the circuit is $\sqrt{5}:x$. The value of $x$ is.

Have you ever wondered how different components in an alternating current (AC) circuit interact with each other? It is not just simple addition like in DC circuits; it is a beautiful dance of vectors and phases! Today, we are going to dive deep into a classic JEE problem that perfectly illustrates this dance. We will explore what happens to the power factor of a circuit when we introduce a new component, specifically a capacitor, into an existing inductor-resistor setup.

Grab your mental compass and straightedge, because we are about to navigate the fascinating world of phasor diagrams!

The Initial RL Circuit

Imagine you have an AC circuit. Initially, this circuit consists of just two components connected in series: a resistor with resistance $R$, and an inductor. The problem gives us a crucial piece of information: the inductive reactance, $X_L$, is exactly equal to $3R$.

What does this mean visually? In the realm of AC circuits, we use phasor diagrams to represent these quantities. Think of the resistance $R$ as a vector pointing horizontally to the right along the x-axis. The inductor, however, causes the voltage to lead the current by $90^{\circ }$. Therefore, its reactance, $X_L$, is represented by a vector pointing straight up along the positive y-axis.

Because these two vectors are perpendicular, we cannot just add their magnitudes together to find the total opposition to the current. Instead, we must find their vector sum, which we call the impedance, denoted by $Z_1$.

Using the Pythagorean theorem, the magnitude of this impedance is:

Let's substitute the value of $X_L$ that we were given:

Now, we need to find the power factor of this initial circuit. The power factor is a measure of how effectively the circuit uses the power supplied to it. Mathematically, it is the cosine of the phase angle ${\varphi }_1$ between the total impedance $Z_1$ and the resistance $R$.

From our right-angled phasor triangle, the cosine of this angle is simply the adjacent side (Resistance) divided by the hypotenuse (Impedance):

Substituting our calculated value for $Z_1$:

We have successfully found the old power factor! Hold onto this value; we will need it soon.

Introducing the Capacitor

Now, the problem introduces a twist. A capacitor is added in series to our existing circuit. This capacitor has a capacitive reactance, $X_C$, equal to $2R$.

How does this change our phasor diagram? While an inductor causes voltage to lead, a capacitor causes voltage to lag the current by $90^{\circ }$. In our phasor diagram, this means the capacitive reactance $X_C$ is represented by a vector pointing straight down along the negative y-axis.

Notice that $X_L$ points up and $X_C$ points down. They are exactly $180^{\circ }$ out of phase! This means they directly oppose each other. To find the net reactance of the circuit, we simply subtract the smaller reactance from the larger one.

Our new net reactance is just $R$, and because $X_L$ was larger, this net reactance points upwards.

Now, let's calculate the new impedance, $Z_2$, for this RLC circuit. Our horizontal vector is still $R$, and our new vertical vector is also $R$.

With this new impedance, we can find the new power factor, $\cos {\varphi }_2$:

Finding the Ratio

We have both the new power factor and the old power factor. The question asks for the ratio of the new power factor to the old power factor, which is given as $\sqrt{5}:x$.

Let's set up our ratio:

Substitute the values we calculated:

To simplify this fraction, we multiply the numerator by the reciprocal of the denominator:

Using the properties of square roots, we can combine them:

So, our calculated ratio is $\sqrt{5}:1$.

The problem states that this ratio is equal to $\sqrt{5}:x$. By directly comparing our result with the given expression, it becomes absolutely clear:

Therefore, the value of $x$ must be 1.

This problem is a fantastic demonstration of how adding components to an AC circuit doesn't just add to the total opposition; it can actually reduce it! By adding a capacitor, we partially cancelled out the inductive reactance, which lowered the overall impedance and increased the power factor from $\frac{1}{\sqrt{10}}$ (about 0.316) to $\frac{1}{\sqrt{2}}$ (about 0.707).

This concept is widely used in real-world power systems, a process known as power factor correction, to make power delivery more efficient. Keep practicing these phasor diagrams, and soon, visualizing AC circuits will become second nature to you!