AC Power Accounting (1-17-2021)

Electricity and Magnetism Level pending

Two AC voltage sources excite a circuit with three impedances as shown in the diagram. Source V S V_S supplies what fraction of the active power consumed by Z F Z_F ?

Details and Assumptions:
1) j = − 1 j = \sqrt{-1}
2) V S = 0 + j 10 V_S = 0 + j 10
3) V R = 10 + j 0 V_R = 10 + j 0
4) Z S = Z R = 0 + j 5 Z_S = Z_R = 0 + j 5
5) Z F = 1 + j 0 Z_F = 1 + j 0


The answer is 0.7.

Steven Chase by Steven Chase , 37, USA

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1 solution

Karan Chatrath
Jan 17, 2021

Circuit equations in matrix form:

[ Z S + Z F Z F Z F Z R + Z F ] [ I S I R ] = [ V S V R ] ⟹ A I = b \left[\begin{matrix} Z_S+Z_F&Z_F\\Z_F&Z_R+Z_F\end{matrix}\right]\left[\begin{matrix} I_S\\I_R\end{matrix}\right]= \left[\begin{matrix} V_S\\V_R\end{matrix}\right]\implies AI = b ⟹ I = A − 1 b \implies I = A^{-1}b I F = I S + I R = [ 1 1 ] I I_F = I_S+I_R = \left[\begin{matrix} 1&1\end{matrix}\right]I I S = [ 1 0 ] I I_S = \left[\begin{matrix} 1&0\end{matrix}\right]I I R = [ 0 1 ] I I_R = \left[\begin{matrix} 0&1\end{matrix}\right]I

The matrix notation eases the number crunching while using the calculation tool of my choice. Active power supplied by V S V_S :

P S = r e a l ( V S I S ∗ ) P_S = \mathrm{real} \left(V_S I_S^{*}\right) Active power consumed by Z F Z_F :

P F = r e a l ( ( I F Z F ) I F ∗ ) P_F = \mathrm{real} \left((I_FZ_F) I_F^{*}\right)

P S P F = 0.7 \boxed{\frac{P_S}{P_F} = 0.7}

The ' ∗ * ' superscript denotes complex conjugate.

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