Dear Student, impeadance Z=5+15-3j=5+12jZ=25+144=13power=Vrms*IrmscosϕZ=26×2613cosϕϕ is the phase difference between the current and the voltagecosϕ=513power=26×2613513=20 WOption 4Regards

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Solve this: Find average power loss¯in the following RLC circuit 150 (1) 13 w (3) 52 w 26f2 sin(mt) (2) 26 w (4) 20 w

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i m p e a d a n c e Z = 5 + (15 - 3) j = 5 + 12 j |Z| = \sqrt{25 + 144} = 13 p o w e r = \frac{V_{r m s} * I_{r m s} \cos ϕ}{Z} = \frac{26 \times 26}{13} \cos ϕ ϕ i s t h e p h a s e d i f f e r e n c e b e t w e e n t h e c u r r e n t a n d t h e v o l t a g e \cos ϕ = \frac{5}{13} p o w e r = \frac{26 \times 26}{13} (\frac{5}{13}) = 20 W O p t i o n 4 R e g a r d s

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Solve this: Solve this: Find average power loss¯in the following RLC circuit 150 (1) 13 w (3) 52 w 26f2 sin(mt) (2) 26 w (4) 20 w

Solve this:

Solve this: Find average power loss¯in the following RLC circuit 150 (1) 13 w (3) 52 w 26f2 sin(mt) (2) 26 w (4) 20 w