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$$
\begin{array}{l}{\text { The sinusoidal current source in the circuit shown }} \\ {\text { produces the current } i_{5}=8 \cos (200,0001) \mathrm{A}} \\ {\text { a) Construct the frequency-domain equivalent circuit. }} \\ {\text { b) Find the steady-state expressions for } v, i_{1}, i_{2}, \text { andi_{3} }}\end{array}
$$

الخطوات

(1) حساب قيمة Z لكل عنصر

(2) تحويل المصادر لل phasor

(3) إيجاد تيارات الأفرع

$$
R=6 , L=40 \mathrm{\mu H}
$$

$$
Z=R+J \omega L=6+J200000*40*10^{-6}=6+J8 \ (\Omega)
$$

$$
R=0, C=1 \mu F
$$

$$
Z=R+J\left(\omega L-\frac{1}{\omega c}\right)=\frac{-J}{\omega c}=\frac{-J}{200000*1*10^{-6}}=-J5 \ (\Omega)
$$

$$
I (t)=8 cos [200000t]=I_m \angle \phi=8 \angle 0
$$

$$
Z_{eq}=\frac{Z_{1} Z_{2}}{Z_{1}+Z_{2}}=\frac{(-5J)(6+J8)}{-5J+6 + 8J}
$$

$$
Z_{eq}=3.33-J6.66
$$

$$
Z_ { eq }=\frac{Z_1 Z_2}{Z_1+Z_2}=\frac{10[3.33-J6.66]}{10+3.33-J6.66}
$$

$$
Z_{eq}=4-3J
$$

$$
\overline{V}=\overline{I} \ \overline{Z}
$$

$$
\overline{V}=8 \angle0(4-3J)
$$

$$
\overline{V}=40 \angle -36.87^{\circ}
 \quad (V)$$

$$
\omega=200000 \mathrm{rad} / \mathrm{sec}
$$

$$
V(t)=40 \cos \left[200000t-36.87^{\circ}\right]
$$

$$
\overline I_{1}=\frac{\overline{V}}{\overline Z_{1}}=\frac{40 \angle-36.87}{10}=4 \angle{36.87} \quad \mathrm{(A)}=4 \cos [200000 t-36.87] \ (A)
$$

$$
\overline I_{2}=\frac{\overline{V}_{1}}{\overline Z_{1}}=\frac{40\angle-36 .87}{6+J8}=4 \angle{-90} \quad \mathrm{(A)}=4 \cos [200000 t-90] \ (A)
$$

$$
\overline{I}_{3}=\frac{\overline{V}}{\overline{Z}_{3}}=\frac{40 \angle-56.87}{-5J}=8 \angle53.13 \quad \mathrm{(A)}=8 \cos [2000000 t+53.13] \ (A)
$$

KCL

$$
\sum I_{node}=0
$$

$$
\sum I=\overline I_5-\overline I_{1}-\overline I_{2}-\overline I_{3}=8 \angle0-4 \angle-36.8-4 \angle-90-8 \angle53.13
$$

$$
\sum I=0
$$            KVL

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