Question

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

[ $\text{Use, thermal capacity}=2\text{J}{/}^{\circ }\text{C}$]

• A. $349\text{s}$
• B. $328\text{s}$
• C. $248\text{s}$
• D. $228\text{s}$

Answer 1

The correct option is A $349\text{s}$

Given,
$\text{Voltage of source},V_{rms}=20\text{V}\text{Frequency of source}=750\text{Hz}R=100\Omega ;L=0.1803\text{H};C=10\mu \text{F}\text{Thermal capacity}=2\text{J}{/}^{\circ }\text{C}\Delta T={10}^{\circ }\text{C}$

The impedance,

$Z=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$

$X_L=\omega L=2\times 3.14\times 750\times 0.1803\approx 850\Omega$

$X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 750\times {10}^{-5}}=21.2\Omega$

$\therefore Z=\sqrt{(100{)}^2+{(850-21.2)}^2}\approx 835\Omega$

$\therefore H=I^{2}Rt=ms\Delta T$

${\left(\frac{V_{rms}}{Z}\right)}^2\times Rt=2\times 10$

${\left(\frac{20}{835}\right)}^2\times 100t=2\times 10$

$\therefore t=\frac{20}{(0.023{)}^2\times 100}\approx 349\text{s}$

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