AC Load Voltage
In the AC circuit below, there are voltage sources and complex impedances. What is the magnitude of the voltage V , shown in red?
The answer is 2.234.
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2 solutions
Nice solution. Out of curiosity, why do you prefer superposition over nodal analysis?
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Actually, I remember superposition better. Thanks for your solution, I will use node and loop analysis the next round.
The current flowing out of the top node is zero
Z 1 V − V 1 + Z 2 V − V 2 + Z L V − 0 = 0 V ( Z 1 1 + Z 2 1 + Z L 1 ) = Z 1 V 1 + Z 2 V 2 V = Z 1 1 + Z 2 1 + Z L 1 Z 1 V 1 + Z 2 V 2
By principle of superposition, the given circuit is equivalent to the addition of two circuits on the right as shown above, such that:
V = V 1 + V 2 = 2 + j 3 + ( 1 − j 2 ) ∣ ∣ ( 1 + j ) ( 1 − j 2 ) ∣ ∣ ( 1 + j ) ( 5 + j 0 ) + 1 − j 2 + ( 2 + j 3 ) ∣ ∣ ( 1 + j ) ( 2 + j 3 ) ∣ ∣ ( 1 + j ) ( 0 + j 7 ) = 3 . 4 + j 3 . 2 1 . 4 + j 0 . 2 ( 5 + j 0 ) + 1 . 6 8 − j 1 . 2 4 0 . 6 8 + j 0 . 7 6 ( 0 + j 7 ) = 1 . 2 3 8 5 3 2 1 1 − j 0 . 8 7 1 5 5 9 6 3 3 + − 3 . 4 0 3 6 6 9 7 2 5 + j 0 . 3 2 1 1 0 0 9 1 7 = − 2 . 1 6 5 1 3 7 6 1 5 − j 0 . 5 5 0 4 5 8 7 1 6 ≈ 2 . 2 3 4 ∠ 1 4 . 2 6 5 ∘