A voltage [latex]\nu(t)=V_0 \cos (\omega t)[/latex] with [latex]V_0=25 V[/latex] and [latex]f=1 kHz[/latex] is applied to a load consisting of a resistor having resistance [latex] R=1.5 k \Omega[/latex] in parallel with an inductor having inductance L=50 mH . Find the average power dissipated in the load and the average power delivered by the source.

The impedance of the load is [latex] \begin{aligned} Z &=\frac{j \omega L R}{R+j \omega L}=\frac{j(2 \pi)(1 kHz )(50 mH )(1.5 k \Omega)}{1.5 k \Omega+j(2 \pi)(1 kHz )(50 mH )} \\\\ & \cong 63.0+j 301 \Omega \cong 307 \angle 1.36 \Omega \end{aligned} [/latex] The load current is [latex] \tilde{I}=\frac{\tilde{V}}{Z} \cong \frac{25 V \angle 0}{307 \Omega \angle 1.36} \cong 81.0 \angle-1.36 mA [/latex] The complex power is [latex] \begin{aligned} S &=\frac{1}{2} \tilde{V} \tilde{I}^* \cong \frac{1}{2}(25 V \angle 0)(81.0 mA \angle 1.36) \\\\& \cong 1.01 \angle 1.36 VA . \end{aligned} [/latex]

Question 13.6: A voltage ν(t)=V0cos(ωt) with V0=25 V and f=1 kHz is applie...

Introduction to Circuit Analysis and Design

A voltage $\nu(t)=V_0 \cos (\omega t)$ with $V_0=25 V$ and $f=1 kHz$ is applied to a load consisting of a resistor having resistance $R=1.5 k \Omega$ in parallel with an inductor having inductance L=50 mH . Find the average power dissipated in the load and the average power delivered by the source.

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