Problem 3.60 Fundamental of Electric Circuits (Alexander/Sadiku) 5th Edition

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Calculate the power dissipated in each resistor in the circuit of Fig. 3.104
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you explain it so well that at the end we feel like experts and can catch little mistakes you made... thank you for the videos!

johnhughes
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I swear you are the best professor in the world, wish you all the best

Mohammeds
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Thanks for solve this problem, but the P on (4 ohms resistor) is P= (-4-8)^2 x 4 = 576
because
it's influenced by i1 and i3

luiscolares
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I1 and i0 are flowing opposite side then why loop no 1 eq is not
(i1-i0)

ssifatkhan
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Hello, Thanks for this video. I used Nodal Analysis and get V1=8v, V2=56v, V3=17.6v and also get power dissipated through (1ohm.... p=64w), ( 8ohm.... p= 184.32w ), ( 2ohm... p=154.88w ) but (4ohm... p=576w). Can you check again ? I used for finding ( i1= V1-56/4) where V1=8v and V2=56v so ( i1=8-56/4)= -12 A . p1=(-12 * -12) * 4 = 576w

shuvoahmed
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Hello sir can you upload me problems on two port network practice problem 18.2. Z parameter solution on alexander sadiku 3rd edition

suba
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Thank you .But you made a mistake in finding the power in 4 ohm resistor.

shawonjoy
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Are you on insta or telegram??..i need help with one more question

shatakshiverma
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power dissiopated at 4ohm resistor is wrong

msmamun