(a) Calculate the phase shift of the circuit in Fig. 9.82.

(b) State whether the phase shift is leading or lagging (output with respect to input).

(c) Determine the magnitude of the output when the input is 120 V.

Question

(a) Calculate the phase shift of the circuit in Fig. 9.82.

(b) State whether the phase shift is leading or lagging (output with respect to input).

(c) Determine the magnitude of the output when the input is 120 V.

Accepted Answer

(a) Consider the circuit as shown.

[latex]\begin{aligned}&{Z}_{1}=j 30 \|(30+j 60)=\frac{(j 30)(30+j 60)}{30+j 90}=3+j 21\\&{Z}_{2}=\mathrm{j} 10 \|\left(40+\mathbf{Z}_{1} ight)=\frac{(\mathrm{j} 10)(43+\mathrm{j} 21)}{43+\mathrm{j} 31}=1.535+\mathrm{j} 8.896=9.028 \angle 80.21^{\circ}\\& ext { Let } {V}_{\mathrm{i}}=1 \angle 0^{\circ}\\&{V}_{2}=\frac{\mathbf{Z}_{2}}{{Z}_{2}+20} {V}_{\mathrm{i}}=\frac{\left(9.028 \angle 80.21^{\circ} ight)\left(1 \angle 0^{\circ} ight)}{21.535+\mathrm{j} 8.896}\\&{V}_{2}=0.3875 \angle 57.77^{\circ}\\&{V}_{1}=\frac{{Z}_{1}}{{Z}_{1}+40} {V}_{2}=\frac{3+\mathrm{j} 21}{43+\mathrm{j} 21} {V}_{2}=\frac{\left(21.213 \angle 81.87^{\circ} ight)\left(0.3875 \angle 57.77^{\circ} ight)}{47.85 \angle 26.03^{\circ}}\\&{V}_{1}=0.1718 \angle 113.61^{\circ}\\&{V}_{\mathrm{o}}=\frac{\mathrm{j} 60}{30+\mathrm{j} 60} {V}_{1}=\frac{\mathrm{j} 2}{1+\mathrm{j} 2} {V}_{1}=\frac{2}{5}(2+\mathrm{j}) {V}_{1}\\&{V}_{\mathrm{o}}=\left(0.8944 \angle 26.56^{\circ} ight)\left(0.1718 \angle 113.6^{\circ} ight)\\&{V}_{\mathrm{o}}=0.1536 \angle 140.2^{\circ}\end{aligned}[/latex]

Therefore, the phase shift is 140.2°

(b) The phase shift is leading.

(c) If [latex]V_{i}[/latex] = 120 V , then

[latex]V_{0}[/latex] = (120)(0.1536∠140.2°) = 18.43∠140.2° V

and the magnitude is 18.43 V.

Question 9.79: (a) Calculate the phase shift of the circuit in Fig. 9.82. (...

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