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gruba
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Homework Statement
[itex][/itex]
Three receivers, with complex impedance [itex]\underline{Z_1}=(125+j375)\Omega,\underline{Z_2}=(700+j100)\Omega,\underline{Z_3}=(500-j500)\Omega[/itex], and two sinusoidal current generators of effective values [itex]I_{g2}=40mA[/itex] and unknown [itex]I_{g1}[/itex] are connected. When the switch is open, apparent power of receiver [itex]\underline{Z_2}[/itex] is [itex]S_2=\frac{\sqrt 2}{2}VA[/itex]. After the switch is closed, active power of all three receivers is reduced two times.
Find phase angle difference between [itex]\underline{I_{g1}}[/itex] and [itex]\underline{I_{g2}}[/itex]
Homework Equations
Apparent power: [itex]S=UI[VA][/itex]
Active power: [itex]P=S\cos\phi [W][/itex] where [itex]\phi[/itex] is the phase angle between current and voltage.
Reactive power: [itex]Q=S\sin\phi[var][/itex]
[itex]P=S\cos\phi=\mathfrak{R}(\underline{S})=\mathfrak{R}(\underline{U}\cdot\underline{I^{*}})[/itex] where [itex]\underline{S}[/itex] is the complex apparent power.
Current phase angle is given by [itex]\psi=\theta-\phi[/itex] where [itex]\theta[/itex] is voltage phase angle.
The Attempt at a Solution
From the impedances, we can find the phase angles [itex]\phi_1,\phi_2,\phi_3[/itex].
[itex]\phi_1=\tan^{-1}\frac{375}{125}=\tan^{-1}(3)\approx 71.565^{o}[/itex]
[itex]\phi_2=\tan^{-1}\frac{100}{700}=\tan^{-1}(1/7)\approx 8.138^{o}[/itex]
[itex]\phi_3=\tan^{-1}\frac{-500}{500}=\tan^{-1}(-1)=-45^{o}[/itex]
When the switch is opened, we know [itex]S_2[/itex], and the active power of [itex]\underline{Z_2}[/itex] is then [itex]P_2=S_2\cos\phi_2\approx 0.699W[/itex].
Reactive power is [itex]Q_2=S_2\sin\phi_2\approx 0.099 var[/itex]
Complex apparent power of [itex]\underline{Z_2}[/itex] is now [itex]\underline{S_2}=(0.699+j0.099)VA[/itex].
But, because [itex]I_{g1}[/itex] is unknown, from [itex]S_2=U_{23}I_2[/itex] we can't find [itex]U_{23}[/itex] and [itex]I_2[/itex].
Question: How to find [itex]\underline{I_{g1}},\underline{U_{23}},\underline{I_2}[/itex] when the switch is opened?
When the switch is closed, we have [itex]{P_2}^{c}=\frac{{P_2}^{o}}{2}\approx 0.349W[/itex].
But, from the first case we need [itex]\underline{I_{g1}}[/itex].
When we know [itex]\underline{I_{g1}},\underline{I_{g2}}[/itex], then we will know the phase angles [itex]\psi_1,\psi_2[/itex], and the phase difference is [itex]|\psi_1-\psi_2|[/itex].
Question: How to find [itex]\underline{I_{g1}}[/itex] and [itex]\underline{I_{g2}}[/itex]?