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The average power delivered to the 6 Ω load in the circuit of figure below is ___________(a) 8 W(b) 76.68 W(c) 625 kW(d) 2.50 kWI have been asked this question in homework.I would like to ask this question from Advanced Problems on Millman’s Theorem and Maximum Power Transfer Theorem in section Useful Theorems in Circuit Analysis of Network Theory

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Right ANSWER is (b) 76.68 W

The best I can explain: I2 = \(\frac{V_2}{6}\), I1 = \(\frac{I_2}{5} = \frac{V_2}{30}\)

V1 = 5V2

50 = 400(I1 – 0.04V2) + V1

Or, V2= 21.45 V

∴ PL = \(\frac{V_2^2}{6} \)

= \(\frac{21.45^2}{6} \)

= \(\frac{460.1025}{6}\) = 76.68 W.



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