If a 30-Ω R and a 40-Ω XL are in series with 100 V applied, find the following: ZT, I, VR, VL, and θZ. What is the phase angle between VL 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[/latex] [latex]X_L[/latex] are in series with 100 V applied, find the following: [latex]Z_T, I, V_R, V_L, and heta _Z[/latex]. What is the phase angle between [latex]V_L[/latex] and [latex]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_L^2}=\sqrt{900+1600}[/latex]

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

[latex]=50 \Omega [/latex]

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

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

[latex]V_L = IX_L = 2 imes 40 = 80 V[/latex]

[latex] an heta _Z=\frac{X_L}{R}=\frac{40}{30}=\frac{4}{3}=1.33[/latex]

[latex] heta _Z[/latex]=53.1°

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

[latex]V_T=\sqrt{V^{2}_R+V^{2}_L}=\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.

Question 20.1: If a 30-Ω R and a 40-Ω XL are in series with 100 V applied, ...

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