The sinusoidal current source in the circuit shown in Fig. 9.18 produces the current $i_{s}=8\cos200,000 t A$ .

a) Construct the frequency-domain equivalent circuit.

b) Find the steady-state expressions for v, $i_{1},i_{2} ,$ and $i_{3}$ .

Question

The sinusoidal current source in the circuit shown in Fig. 9.18 produces the current [latex]i_{s}=8\cos200,000 t A [/latex].

a) Construct the frequency-domain equivalent circuit.

b) Find the steady-state expressions for v,[latex]i_{1},i_{2} ,[/latex]and [latex]i_{3}[/latex].

Accepted Answer

a) The phasor transform of the current source is [latex]8\angle 0^{°}[/latex]; the resistors transform directly to the frequency domain as 10 and 6Ω the 40μH inductor has an impedance of j 8 Ω at the given frequency of[latex]200,000 \frac{rad}{s}[/latex] ;and at this frequency the 1μF capacitor has an impedance of -j5 Ω Figure 9.19 shows the frequency-domain equivalent circuit and symbols representing the phasor transforms of the unknowns.

b) The circuit shown in Fig. 9.19 indicates that we can easily obtain the voltage across the current source once we know the equivalent impedance of the three parallel branches. Moreover, once we know V, we can calculate the three phasor currents[latex]\pmb{I}_{1},\pmb{I}_{2} ,[/latex]and [latex]\pmb{I}_{3}[/latex] by using Eq. 9.35[latex]\pmb{V} = Z\pmb{I}[/latex]. To find the equivalent impedance of the three branches, we first find the equivalent admittance simply by adding the admittances of each branch. The admittance of the first branch is

[latex]Y_{1} =\frac{1}{10} = 0.1 S[/latex],

the admittance of the second branch is

[latex]Y_{2}=\frac{1}{6 + j8} = \frac{6 - j8}{100 }= 0.06 - j0.08 S,[/latex],

and the admittance of the third branch is

[latex]Y_{3} =\frac{1}{-j5 } = = j0.2 S[/latex].

The admittance of the three branches is

[latex]Y = Y_{1}+Y_{2}+Y_{3}[/latex]

= 0.16 + j0.12

[latex]= 0.2\angle36.87^{°}S[/latex].

The impedance at the current source is

[latex]Z =\frac{1}{Y}= 5\angle -36.87^{°} \Omega[/latex].

The Voltage V is

[latex]\pmb{V} = Z\pmb{I} = 40 \angle -36.87^{°}V[/latex].

Hence

[latex]I_{1}=\frac{40 \angle -36.87^{°}}{10} = 4 \angle -36.87^{°} = 3.2 - j2.4 A,[/latex]

[latex]I_{2}=\frac{40 \angle -36.87^{°}}{6+j8} = 4 \angle -90^{°} = -j4 A,[/latex]

and

[latex]I_{3}=\frac{40 \angle -36.87^{°}}{5\angle -90^{°}} = 8\angle 53.13^{°} = 4.8 + j6.4 A.[/latex]

We check the computations at this point by verifying that

[latex] I_{1}+I_{2}+I_{3}=I [/latex]

Specifically,

3.2 - j2.4 - j4 + 4.8 + j6.4 = 8 + j0.

The corresponding steady-state time-domain expressions are

[latex]v = 40 \cos \left(200,000t - 36.87^{°}\right) [/latex]V,

[latex] i_{1} = 4 \cos \left(200,000t - 36.87^{°}\right)[/latex] A,

[latex] i_{2} = 4 \cos \left(200,000t - 90^{°}\right)[/latex] A,

[latex] i_{3 }= 8 \cos \left(200,000t +53.13^{°}\right)[/latex] A.

Question 9.7: The sinusoidal current source in the circuit shown in Fig. 9...

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