Question

In the circuit shown , 1 $K\Omega$ resistor is connected in series to a reactive impedance of $X\Omega$ for ac source of 200 V, 50 Hz. The power factor of the circuit is found to be 0.7 leading, Then:

• A. the power absorbed by the circuit is 20 W
• B. the maximum current through the source must be 200 mA.
• C. the average rate of doing work by the source in half-cycle should be 10 W.
• D. the power absorbed is maximum when $X=0$,

Answer 1

Answer: (A), (B), (D)

the power absorbed by the circuit is 20 W
the maximum current through the source must be 200 mA.
the power absorbed is maximum when $X=0$,

$V_o=200\times \sqrt{2}=200\times 1.414$

= $282.800\simeq 283V$

$P_f=0.7=cos\varphi \Rightarrow \varphi =\frac{\pi }{4}$ leading

$\therefore$ current leads the voltage by ${45}^{\circ }$.

$Z=\sqrt{R^2+X^2,}{P}_f=\frac{R}{Z}=0.7=\frac{1}{\sqrt{2}}$

$\therefore Z=\sqrt{2}R=1414\Omega$

$I_o=\frac{200\sqrt{2}}{\sqrt{2}R}=\frac{200}{1000}=0.2A=200mA$

Power absorbed = $V_m{I}_m\cos \varphi =\frac{200\times 200}{1000\sqrt{2}}.\frac{1}{\sqrt{2}}=20W$

$e=V_o\sin \omega t,i=I_o\sin (\omega t+\frac{\pi }{4}),\omega =100\pi$

$P=V_o{I}_o\sin \omega t(\sin \omega t\cdot \cos \frac{\pi }{4}+\cos \omega t\cdot \sin \frac{\pi }{4})$

$\frac{V_o{I}_o}{\sqrt{2}}[{\sin }^2\omega t+\sin \omega t\cos \omega t]$

$\frac{V_o{I}_o}{\sqrt{2}}[{\sin }^2\omega t+\frac{1}{2}\sin 2\omega t]$

$\therefore P_{half}=\frac{{\int }_o^{\frac{\pi }{\omega }}Pdt}{{\int }_o^{\frac{\pi }{\omega }}dt}=\frac{\omega }{\pi }\cdot \frac{V_o{I}_o}{\sqrt{2}}[{\int }_o^{\frac{\pi }{\omega }}\frac{1-\cos 2\omega t}{2}dt+\frac{1}{2}{\int }_o^{\frac{\pi }{\omega }}\sin 2\omega tdt]$

$=\frac{\omega }{\pi }\cdot \frac{V_o{I}_o}{\sqrt{2}}[\frac{\pi }{2\omega }-\frac{1}{4\omega }{\left[\sin 2\omega t\right]}_o^{\frac{\pi }{\omega }}-\frac{1}{4\omega }{\left[\cos 2\omega t\right]}_o^{\frac{\pi }{w}}]$

$=\frac{V_o{I}_o}{2}\cdot \frac{1}{\sqrt{2}}=V_m{I}_m\cos \varphi =20W$