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 $10 \,μF$ all in series. Calculate the time in which the resistance (thermal capacity $2\,J^{\circ}C^{-1}$) will get heated by $10^{\circ}C$.

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

Answer 1

Answer: (A)

As,
$X_{L} = \omega L = 2\pi\upsilon L = 2\pi × 750 × 0.1803 = 849.2\, \Omega$
and $X_{C} = \frac{1}{\omega C} = \frac{1}{ 2\pi \upsilon C} = \frac{1}{2\pi × 750 ×10^{-5}} = 21.2\, \Omega $
so, $X=X_{L}-X_{C} =849.2-21.2=828\Omega$
Hence, $Z = \sqrt{R^{2}+X^{2}} = \sqrt{\left(100\right)^{2} +\left(828\right)^{2}} = 834 \,\Omega$
$P_{a\upsilon} = V_{rms}\,I_{rms} \,cos\phi = V_{rms} \times \frac{V_{rms}}{Z} \times\frac{R}{Z}$
i.e., $P_{a\upsilon} = \left(\frac{V_{rms}}{Z}\right)^{2} \times R = \left(\frac{20}{834}\right)^{2}\times 100 = 0.0575\, W$
And as, $U = P × t = mcΔ\theta$
$\therefore \quad t = \frac{mc×Δ\theta}{P} =$$\frac{2\times10}{0.0575}$= 348 s