Question 296: AC fundamentals

Question

An a.c. source is applied to a series LR circuit with $\mathrm{X}_{\mathrm{L}}=3 \mathrm{R}$ and power factor is $\mathrm{X}_1$. Now a capacitor with $\mathrm{X}_{\mathrm{c}}=\mathrm{R}$ is added in series to LR circuit and the power factor is $\mathrm{X}_2$. The ratio $\mathrm{X}_1$ to $\mathrm{X}_2$ is

Accepted Answer

Given:
In the LR circuit, $X_L = 3R$.
Power factor (without capacitor) is $X_1$.
When a capacitor is added, $X_C = R$, new power factor is $X_2$.
Asked: Find ratio $\dfrac{X_1}{X_2}$.
Step 1: Power factor for LR circuit ($X_1$)
Power factor, $X_1 = \cos \phi_1 = \dfrac{R}{Z_1}$
Where,
$$ Z_1 = \sqrt{R^2 + (X_L)^2} $$
Given $X_L = 3R$, so
$$ Z_1 = \sqrt{R^2 + (3R)^2} = \sqrt{R^2 + 9R^2} = \sqrt{10}R $$
Therefore,
$$ X_1 = \dfrac{R}{\sqrt{10}R} = \dfrac{1}{\sqrt{10}} $$
Step 2: Power factor for LCR circuit ($X_2$)
Now, circuit is LCR in series.
$X_L = 3R$, $X_C = R$
Net reactance:
$$ X = X_L - X_C = 3R - R = 2R $$
Impedance:
$$ Z_2 = \sqrt{R^2 + (X)^2} = \sqrt{R^2 + (2R)^2} = \sqrt{1R^2 + 4R^2} = \sqrt{5}R $$
Power factor,
$$ X_2 = \cos \phi_2 = \dfrac{R}{Z_2} = \dfrac{R}{\sqrt{5}R} = \dfrac{1}{\sqrt{5}} $$
Step 3: Find the ratio $\dfrac{X_1}{X_2}$
$$ \dfrac{X_1}{X_2} = \dfrac{\dfrac{1}{\sqrt{10}}}{\dfrac{1}{\sqrt{5}}} = \dfrac{\sqrt{5}}{\sqrt{10}} $$
Simplify:
$$ \dfrac{\sqrt{5}}{\sqrt{10}} = \dfrac{\sqrt{5}}{\sqrt{5 	imes 2}} = \dfrac{\sqrt{5}}{\sqrt{5}\sqrt{2}} = \dfrac{1}{\sqrt{2}} $$
So,
$$ \dfrac{X_1}{X_2} = \dfrac{1}{\sqrt{2}} $$
Or,
$$ X_1 : X_2 = 1 : \sqrt{2} $$
Correct Answer:
Option C: $1 : \sqrt{2}$

Answer

$1: 2$

Answer

$2: 1$

Answer

$1: \sqrt{2}$

Answer

$\sqrt{2}: 1$

AC fundamentals

An a.c. source is applied to a series LR circuit with \mathrm{X}_{\mathrm{L}}=3 \mathrm{R} and power factor is \mathrm{X}_1. Now a capacitor with \mathrm{X}_{\mathrm{c}}=\mathrm{R} is added in series to LR circuit and the power factor is \mathrm{X}_2. The ratio \mathrm{X}_1 to \mathrm{X}_2 is

Choose the correct answer: