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  • A 750 Hz, 20 V source is connected to a resistance of 100  , an inductance of 0.1803 H and a capacitance of 1

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 \mathrm{F} all in series. Calculate the time in which the resistance (thermal capacity 2 \mathrm{~J}^{\circ} \mathrm{C}^{-1} ) will get heated by 10°C:

Option: 1

348 s


Option: 2

353 s


Option: 3

365 s


Option: 4

370 s


Answers (1)

best_answer

As, \mathrm{X_L=\omega L=2 \pi \mathrm{vl}=2 \pi \times 750 \times 0.1803=849.2 \Omega }
\mathrm{and \mathrm{X}_{\mathrm{C}}=\frac{1}{\omega \mathrm{C}}=\frac{1}{2 \pi \mathrm{vC}}=\frac{1}{2 \pi \times 750 \times 10^{-5}}=21.2 \Omega }
\mathrm{So, \mathrm{X}=\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}=849.2-21.2=828 \Omega }
Hence, \mathrm{\mathrm{Z}=\sqrt{\mathrm{R}^2+\mathrm{X}^2}=\sqrt{(100)^2+(828)^2}=834 \Omega }
But as in case of AC, \mathrm{\quad P_{\mathrm{av}}=\mathrm{V}_{\mathrm{mms}} \mathrm{I}_{\mathrm{mms}} \cos \phi=\mathrm{V}_{\mathrm{mms}} \times \frac{\mathrm{V}_{\mathrm{mms}}}{\mathrm{Z}} \times \frac{\mathrm{R}}{\mathrm{Z}} }
i.e., \mathrm{P_{a v}=\left(\frac{V_{m a s}}{Z}\right)^2 \times R=\left(\frac{20}{834}\right)^2 \times 100=0.0575 \mathrm{~W} }
And \mathrm{as, \mathrm{U}=\mathrm{P} \times \mathrm{t}=\mathrm{mc} \Delta \theta \therefore \quad \mathrm{t}=\frac{\mathrm{mc} \times \Delta \theta}{\mathrm{P}}=\frac{2 \times 10}{0.0575}=348 \mathrm{~s} }

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vishal kumar

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