JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Alternating Current

The source drives a series combination of $L=100\text{mH}=0.1\text{H}$, $C=100\mu \text{F}=10^{-4}\text{F}$ and $R=120\Omega$, with the supply reading $V_{rms}=20\text{V}$ at angular frequency $\omega =100{\text{rad s}}^{-1}$. The reactances are $X_L=\omega L=100\times 0.1=10\Omega$ and $X_C=\frac{1}{\omega C}=\frac{1}{100\times 10^{-4}}=100\Omega$, so the net reactance is $|X_C-X_L|=90\Omega$ and the impedance is $Z=\sqrt{R^2+(X_C-X_L{)}^2}=\sqrt{120^2+90^2}=\sqrt{22500}=150\Omega$. The rms current is therefore $i_{rms}=\frac{V_{rms}}{Z}=\frac{20}{150}=\frac{2}{15}\text{A}$. Only the resistor dissipates energy, and it does so at the average rate $P=i_{rms}^{2}R={\left(\frac{2}{15}\right)}^2\times 120=\frac{4}{225}\times 120=\frac{480}{225}\text{W}$. Raising the temperature of the resistor, whose thermal capacity is $2\text{J}{/}^{\circ }\text{C}$, by $16^{\circ }\text{C}$ requires heat $Q=2\times 16=32\text{J}$, so the time needed is $\Delta t=\frac{Q}{P}=\frac{32\times 225}{480}=15\text{s}$.