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 $10\mu \text{F}$ in series. Calculate the time in which the resistance which has thermal capacity $2\text{J}/{}^{\circ }\text{C}$ will get heated by ${10}^{\circ }\text{C}$.

• A. $10\text{s}$
• B. $134\text{s}$
• C. $225\text{s}$
• D. $348\text{s}$

Answer 1

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

Given:

$f=750\text{Hz},R=100\Omega$
$L=0.1803\text{H},C=10\mu \text{F}$

The inductive and capacitive reactance are given as:

$X_L=\omega L=\left(2\pi f\right)L=2\pi \times 750\times 0.1803=850\Omega$

$X_C=\frac{1}{\omega C}=\frac{1}{2\pi fC}=\frac{1}{(2\pi \times 750)\times {10}^{-5}}=21.2\Omega$

Impedance of series $\text{RLC}$ circuit is,

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

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

Power dissipated in the circuit is:

$P=V_{rms}{I}_{rms}\cos \varphi$

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

$P=\frac{{20}^2\times 100}{{835}^2}=0.0574\text{W}$

Heat produced in resistance to raise its temperature by, $10{}^{\circ }\text{C}$ is:

$H=2\text{J}{/}^{\circ }\text{C}\times 10{}^{\circ }\text{C}=20\text{J}$

Therefore, the time in which this heat is produced is given by:

$t=\frac{H}{P}=\frac{20}{0.0574}\approx 348\text{s}$

Hence, $\left(D\right)$ is the correct answer.