Question 126 of 190

A $750Hz,20V\left(rms\right)$ source is connected to a resistance of $100\Omega ,$ an inductance of $0.1803H$ and a capacitance of $10\mu F$ all in series. The time in which the resistance (heat capacity $2J/^{\circ}C$ ) will get heated by $10^{\circ}C.$ (assume no loss of heat to the surroundings) is close to :

A$418s$B$245s$C$365s$D$348s$

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Answer: (D) $348s$

The AC circuit is shown below.

Here,$R=100,{X}_{L}=L\omega =0.1803\times 750\times 2\pi =850\Omega$
The capacitive reactance ${X}_{c}=\dfrac{1}{C\omega }=\dfrac{1}{{10}^{-5}\times 2\pi \times 750}=21.23\Omega$
Total impedance $Z=\sqrt{{R}^{2}+{\left({X}_{L}-{X}_{C}\right)}^{2}}$
$=\sqrt{{100}^{2}+{\left(850-21.23\right)}^{2}}=834.77\approx 835$
The heat loss $H={i}_{rms}^{2}Rt=\left(ms\right)∆t$
$\dfrac{20}{835}\times \dfrac{20}{835}\times 100t=\left(2\right)\times 10$
$t=348.61\sec$