Question

A $750 Hz , 20 V ( rms )$ 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. The time in which the resistance (heat capacity $2 J /{ }^{\circ} C$ ) will get heated by $10^{\circ} C$. (assume no loss of heat to the surroundings) is close to :

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

Answer 1

Answer: (C)

$f =750 Hz , V _{ rms }=20 V$
$ R =100 \Omega, L =0.1803 H$
$C =10 \mu F , S =2 J /{ }^{\circ} C$
$Z =\sqrt{ R ^{2}+\left( X _{ L }- X _{ C }\right)^{2}}$
$=\sqrt{ R ^{2}+(\omega L -1 / \omega C )^{2}}$
$=\sqrt{ R ^{2}+\left(2 \pi fL -\frac{1}{2 \pi fC }\right)^{2}}$
Putting values
$|Z|=834 \Omega$
In AC power $P = V _{ rms } i _{ rms } \cos \phi$
Cos $\phi=\frac{ R }{| Z |} \quad i _{ rms }=\frac{ V _{ rms }}{| Z |}$
$=\frac{ V _{ rms }^{2} R }{(| Z |)^{2}}$
$=\left(\frac{20}{834}\right)^{2} \times 100=0.0575 J / s$
$H = Pt = S \Delta \theta$
$t =\frac{2(10)}{0.0575}=348$ sec