Question

A $750\text{Hz},20\text{V}$ (rms) source connected to a resistance of $100\Omega ,$ an inductance of $0.1803\text{H}$ and a capacitance of $10\mu \text{F}$ all in series. The time in which the resistance (heat capacity $2{\text{J}}^{\circ }{\text{C}}^{-1}$ ) will get heated by ${10}^{\circ }\text{C}$ is close to-

[Assume no loss of heat to the surroundings ]

• A. $418\text{s}$
• B. $245\text{s}$
• C. $365\text{s}$
• D. $348\text{s}$

Answer 1

The correct option is D $348\text{s}$


Here, $R=100\Omega X_L=L\omega =0.1803\times 750\times 2\pi \approx 850\Omega X_C=\frac{1}{\omega C}=\frac{1}{{10}^{-5}\times 2\pi \times 750}=21.23\Omega$

Impedance, $Z=\sqrt{R^2+(X_L-X_C{)}^2}$

$=\sqrt{{100}^2+(850-21.23{)}^2}=834.77\simeq 835\Omega$


$H=I_{rms}^{2}Rt={\left(\frac{V_{rms}}{|Z|}\right)}^{2}Rt=\left(ms\right)\Delta T$

$\Rightarrow \frac{20}{835}\times \frac{20}{835}\times 100t=\left(2\right)\times 10[\because V_{rms}=\text{20 V and}\Delta t={10}^{\circ }\text{C}]$

$\therefore \text{Time, t = 348.61 s}\approx 348\text{s}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.