If a 30-Ω R and a 40-Ω XC are in series with 100 V applied, find the following: Z T , I, VR, VC, and θZ. What is the phase angle between VC and VR with respect to I? Prove that the sum of the series voltage drops equals the applied voltage VT.

Question

If a 30-[latex]\Omega[/latex] R and a 40-[latex]\Omega X_C [/latex] are in series with 100 V applied, find the following: [latex]Z_T , I, V_R, V_C, and θ_Z[/latex]. What is the phase angle between[latex]V_C and V_R[/latex] with respect to I? Prove that the sum of the series voltage drops equals the applied voltage [latex]V_T.[/latex]

Accepted Answer

[latex]Z_T=\sqrt{R^2+X^2_C}=\sqrt{30^2+40^2}[/latex]

[latex]=\sqrt{900+1600}[/latex]

[latex]=\sqrt{2500}[/latex]

[latex]=50 \Omega[/latex]

[latex]I=\frac{V_T}{Z_T}=\frac{100 V}{50\Omega}=2A[/latex]

[latex]V_R = IR = 2 A imes 30 \Omega = 60 V[/latex]

[latex]V_C = IX_C= 2 A imes 40 \Omega = 80 V[/latex]

[latex] an heta _Z=-\frac{X_C}{R}=-\frac{40}{30}=-1.333[/latex]

[latex] heta _Z[/latex]=-53.1°

Therefore, [latex]V_T[/latex] lags I by 53.1°. Furthermore, I and [latex]V_R [/latex]are in phase, and [latex]V_C[/latex] lags I by 90°. Finally,

[latex]V_T=\sqrt{V^2_R+V^2_C}=\sqrt{60^2+80^2}=\sqrt{3600+6400}[/latex]

[latex]=\sqrt{10,000}[/latex]

=100 V

Note that the phasor sum of the voltage drops equals the applied voltage [latex]V_T[/latex].

Question 17.1: If a 30-Ω R and a 40-Ω XC are in series with 100 V applied, ...

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