A sinusoidal voltage of peak value 283 , V an | vedclass

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Question

(B) Given: Peak voltage $V_0 = 283 \, V$,angular frequency $\omega = 320 \, rad/s$,resistance $R = 5 \, \Omega$,inductance $L = 25 \, mH = 25 	imes 1

Vedclass · Accepted Answer

$7 \, \Omega$ and $45^{\circ}$

Vedclass · Answer

(B) Given: Peak voltage $V_0 = 283 \, V$,angular frequency $\omega = 320 \, rad/s$,resistance $R = 5 \, \Omega$,inductance $L = 25 \, mH = 25 	imes 10^{-3} \, H$,and capacitance $C = 1000 \, \mu F = 10^{-3} \, F$.First,calculate the inductive reactance $X_L$:$X_L = \omega L = 320 	imes 25 	imes 10^{-3} = 8 \, \Omega$.Next,calculate the capacitive reactance $X_C$:$X_C = \frac{1}{\omega C} = \frac{1}{320 	imes 10^{-3}} = \frac{1}{0.32} = 3.125 \, \Omega$.Total impedance $Z$ of the circuit is given by:$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{5^2 + (8 - 3.125)^2} = \sqrt{25 + (4.875)^2} \approx \sqrt{25 + 23.76} \approx \sqrt{48.76} \approx 6.98 \, \Omega \approx 7 \, \Omega$.The phase difference $\phi$ is given by:$	an \phi = \frac{X_L - X_C}{R} = \frac{8 - 3.125}{5} = \frac{4.875}{5} \approx 0.975 \approx 1$.Since $0.975$ is close to $1$,$	an \phi \approx 1$,so $\phi \approx 45^{\circ}$.However,checking the options,$X_L - X_C = 4.875 \approx 5$. Thus,$	an \phi = \frac{5}{5} = 1$,which is $45^{\circ}$.Given the options provided,the closest match for impedance is $7 \, \Omega$ and the phase angle calculation $\frac{X_L - X_C}{R} = \frac{8 - 3.125}{5} = 0.975$. The option $D$ is the intended answer based on standard approximation.

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