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1. Use a series of source transformations to find the power dissipated by the 10 kOhm resistor in the circuit below. 6 ΚΩ 10
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Answer #1

Л) lok s M GOV (1) loma izkr V - 10mA 6x103 In the given circuit Gov source is in series with 6kr resists. This can be Converv=IR = (10x103) (4x103) you 1 Now the circuit becomes, uko no w ifkar bov loker you In the above circuit the current Hows in- ukr - А M DK2 2mA + uva for the given circuit to find the we have to the venin equivalent circuit bind the venir voltage vaNow 4 Va is alone acting then open Circuit source Current then, 4a = 4x4 = 16V Now circuit is, A zkr 0.C ula = 16V Y Now, the

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