Chapter 5, Problem 26P

In the RL circuit shown in Fig. P5.26, the impedances of R and L are $Z_R=R\Omega \text{and}{Z}_C=-j{X}_C\Omega$, where $j=\sqrt{-1}$. Suppose $R=100\Omega$ and $X_L=50\Omega$.

Figure P5.26 An RL circuit for problem P5-26.

1. Express the impedances $Z_R$ and $Z_L$, as complex number in both rectangular and polar forms.
2. Find the total impedance $Z=Z_R+Z_L$
3. as a complex number in both rectangular and polar forms.
4. If $V=100\angle 0°V$, find the current $I=\frac{V}{Z}$
5. as a complex number in both rectangular and polar forms.
6. Knowing the current in part ©, find the voltage phasor $V_R=I{Z}_R$ and $V_L=I{Z}_L$, in both rectangular and polar forms.
7. Show that the KVL is satisfied for the circuit shown in Hg. P5.26; in other words, show $V=V_R+V_L$

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