# A 750Hz. 20V (rms) source is connected to a resistance of 100 om,

Question:

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

1. $418 \mathrm{~s}$

2. $245 \mathrm{~s}$

3. $348 \mathrm{~s}$

4. $365 \mathrm{~s}$

Correct Option: , 3

Solution:

$\mathrm{f}=750 \mathrm{~Hz}, \mathrm{~V}_{\mathrm{rms}}=20 \mathrm{~V}$

$\mathrm{R}=100_{\Omega}, \mathrm{L}=0.1803 \mathrm{H}$

$\mathrm{C}=10_{\mu} \mathrm{F}, \mathrm{S}=2 \mathrm{~J} /{ }^{\circ} \mathrm{C}$

$Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}=\sqrt{R^{2}+(\omega L-1 / \omega C)^{2}}$

$=\sqrt{\mathrm{R}^{2}+\left(2 \pi \mathrm{fL}-\frac{1}{2 \pi \mathrm{f} \mathrm{C}}\right)^{2}}$

Putting values

$|Z|=834_{\Omega}$

In $A C$ power $P=V_{r m s} i_{r m s} \cos \phi$

$\operatorname{Cos} \phi=\frac{\mathrm{R}}{|\mathrm{Z}|} \quad \mathrm{i}_{\mathrm{ms}}=\frac{\mathrm{V}_{\mathrm{rms}}}{|\mathrm{Z}|}$

$=\frac{\mathrm{V}_{\mathrm{ms}}^{2} \mathrm{R}}{(|\mathrm{Z}|)^{2}}$

$=\left(\frac{20}{834}\right)^{2} \times 100=0.0575 \mathrm{~J} / \mathrm{s}$

$\mathrm{H}=\mathrm{Pt}=\mathrm{S}_{\Delta \theta}$

$\mathrm{t}=\frac{2(10)}{0.0575}=348 \mathrm{sec}$