Q 12-07-133JEE MainJEE Main 2021 (27 Aug, Shift 2)Easy
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 the new power factor with the old power factor of the circuit is $\sqrt{5} : x$. The value of $x$ is
Numerical value type. Enter your answer.
Answer: 1
Old: $Z_1 = \sqrt{R^2 + 9R^2} = \sqrt{10}R$, $\cos\phi_1 = \dfrac{1}{\sqrt{10}}$.
New: $Z_2 = \sqrt{R^2 + (3R - 2R)^2} = \sqrt{2}R$, $\cos\phi_2 = \dfrac{1}{\sqrt{2}}$.
$$\frac{\cos\phi_2}{\cos\phi_1} = \frac{\sqrt{10}}{\sqrt{2}} = \sqrt{5} = \sqrt{5} : 1 \Rightarrow x = 1$$
Solution by Sreeraj P, M.Sc Physics