Question
If $\mathbf{V}_{o}=8 \angle 30^{\circ} \mathrm{V}$ in the circuit of Fig. 9.55 find $\mathbf{I}_{s}$.
Step 1
From the circuit, it is clear that the 5 ohm resistor is in parallel with the -j5 ohm capacitor. So, the equivalent impedance can be calculated as follows: \[Z_{eq} = \frac{5 \cdot -j5}{5 + -j5} = 2.5 - j2.5 \, \Omega\] Show more…
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