Question

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 _________.

Hint:

The power factor measures how effectively electrical power is being used. A value of 1 (unity power factor) is ideal. Notice that adding the capacitor brought the net reactance down from 3R to R, moving the circuit closer to resonance and improving the power factor (from \(1/\sqrt{10} \approx 0.316\) to \(1/\sqrt{2} \approx 0.707\)).

Answer 1

Correct Answer: 1

Step 1: Understanding the Question:
We have an AC circuit that initially is an LR circuit and then becomes an LCR circuit. We need to find the ratio of the power factors in the two cases and use it to determine the value of x.
Step 2: Key Formula or Approach:
The power factor (PF) of an AC circuit is given by \( \cos\phi = \frac{R}{Z} \), where R is the resistance and Z is the impedance.
- For an LR circuit, \( Z = \sqrt{R^2 + X_L^2} \).
- For an LCR circuit, \( Z = \sqrt{R^2 + (X_L - X_C)^2} \).
Step 3: Detailed Explanation:
Case 1: Old Circuit (LR circuit)
- Resistance = R
- Inductive reactance, \( X_L = 3R \).
The impedance of the old circuit is:
\[ Z_{\text{old}} = \sqrt{R^2 + X_L^2} = \sqrt{R^2 + (3R)^2} = \sqrt{R^2 + 9R^2} = \sqrt{10R^2} = R\sqrt{10} \] The old power factor is:
\[ PF_{\text{old}} = \frac{R}{Z_{\text{old}}} = \frac{R}{R\sqrt{10}} = \frac{1}{\sqrt{10}} \] Case 2: New Circuit (LCR circuit)
A capacitor is added in series.
- Resistance = R
- Inductive reactance, \( X_L = 3R \).
- Capacitive reactance, \( X_C = 2R \).
The net reactance is \( X = X_L - X_C = 3R - 2R = R \).
The impedance of the new circuit is:
\[ Z_{\text{new}} = \sqrt{R^2 + X^2} = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2} \] The new power factor is:
\[ PF_{\text{new}} = \frac{R}{Z_{\text{new}}} = \frac{R}{R\sqrt{2}} = \frac{1}{\sqrt{2}} \] Step 4: Finding the value of x
We are given the ratio:
\[ \frac{PF_{\text{new}}}{PF_{\text{old}}} = \frac{\sqrt{5}}{x} \] Let's calculate the ratio from our results:
\[ \frac{PF_{\text{new}}}{PF_{\text{old}}} = \frac{1/\sqrt{2}}{1/\sqrt{10}} = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{\frac{10}{2}} = \sqrt{5} \] Now, compare this with the given ratio:
\[ \sqrt{5} = \frac{\sqrt{5}}{x} \] This implies \( x = 1 \).
Step 5: Final Answer:
The value of x is 1.