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In a circuit, a metal filament lamp is connected in series with a capacitor of capacitance $C \mu F$ across a $200 V , 50 Hz$ supply. The power consumed by the lamp is $500 W$ while the voltage drop across it is $100 V$. Assume that there is no inductive load in the circuit. Take $m m s$ values of the voltages. The magnitude of the phase-angle (in degrees) between the current and the supply voltage is $\varphi$.
Assume, $\pi \sqrt{3} \approx 5$. . . . .
($1$) The value of $C$ is . . . . . .
($2$) The value of $\varphi$ is
Give the answers of the questions ($1$) and ($2$)
$100,60$
$100,70$
$101,60$
$102,80$
Solution
$\sqrt{ V _{ C }^2+ V _{ R }^2}=\varepsilon_{\operatorname{mm}}$
$\Rightarrow V _{ C }^2+100^2=200^2$
$V _{ C }=100 \sqrt{3 V }$ $. . . . . . (i)$
$\tan \phi=\frac{ V _{ C }}{ V _{ R }}=\frac{100 \sqrt{3}}{100}$
$\therefore \phi=60^{\circ}$ $. . . . . . (ii)$
$P = I _{\operatorname{mmm}} \varepsilon_{\operatorname{mm}} \cos \phi$
$=\frac{\varepsilon_{\operatorname{mms}}^2}{ z } \frac{1}{2}$
$500=\frac{200}{ z } \frac{1}{2}$
$\therefore z =40 \Omega$ $. . . . . . (iii)$
$\cos \phi=\frac{ R }{ Z } \Rightarrow \frac{1}{2}=\frac{ R }{40}$
$\therefore R =20$
$\& x _{ C }=\sqrt{ z ^2- R ^2}=\sqrt{40^2-20^2}=20 \sqrt{3} \Omega$
$\Rightarrow \frac{1}{\omega C }=20 \sqrt{3} \therefore C =100$
