JEE Main 2019 — Alternating Current Question with Solution

(Power factor), $P_f=\cos \text{ϕ}=\frac{R}{Z}$

Where $R, Z$ are the resistance and impedance respectively.
(Energy dissipated) $E=\frac{V_{rms}^2}{Z}\times \cos \text{ϕ}\times t$
$\Rightarrow E=\frac{V_{rms}^2}{Z}\times \frac{R}{Z}\times t$

$\Rightarrow E=\frac{V_{rms}^2}{Z^2}\times R\times t$

$V_{rms}, t$ are the RMS voltage and time respectively.

We know that the Impedance of a LCR circuit is given by the formula,
$$\begin{gathered} Z^2=R^2+{\left(\text{ω}L-\frac{1}{\text{ω}C}\right)}^2 \\ \Rightarrow Z^2={60}^2+{\left(2\text{π}\times 50\times 20\times {10}^{-3}-\left(\frac{1}{2\text{π}\times 50\times 120\times {10}^{-6}}\right)\right)}^2 \end{gathered}$$

Where $\text{ω}, L, C$ are the angular frequency, inductance and capacitance in the circuit respectively.
$\Rightarrow Z^2={60}^2+{\left(2\text{π}-\frac{1000}{12\text{π}}\right)}^2$

$\Rightarrow Z^2=4009.7$

$\therefore E=\frac{{24}^2}{4009.7}\times 60\times 60 \mathrm{J}$ 

$\mathrm{=\; 517}.14\mathrm{J}$

$=5.17\times {10}^2 \mathrm{J}$