Question

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

• A. 348 s
• B. 353 s
• C. 365 s
• D. 370 s

Answer 1

Answer: (A)

As,
$X_L=\omega L=2\pi \upsilon L=2\pi \times 750\times 0.1803=849.2\Omega$
and $X_C=\frac{1}{\omega C}=\frac{1}{2\pi \upsilon C}=\frac{1}{2\pi \times 750\times 10^{-5}}=21.2\Omega$
so, $X=X_L-X_C=849.2-21.2=828\Omega$
Hence, $Z=\sqrt{R^2+X^2}=\sqrt{{\left(100\right)}^2+{\left(828\right)}^2}=834\Omega$
$P_{a\upsilon }=V_{rms}{I}_{rms}cos\varphi =V_{rms}\times \frac{V_{rms}}{Z}\times \frac{R}{Z}$
i.e., $P_{a\upsilon }={\left(\frac{V_{rms}}{Z}\right)}^2\times R={\left(\frac{20}{834}\right)}^2\times 100=0.0575W$
And as, $U=P\times t=mc\Delta \theta$
$\therefore t=\frac{mc\times \Delta \theta }{P}=$$\frac{2\times 10}{0.0575}$= 348 s