SSC JE Electrical Engineering 22 Jan 2018 Evening Paper

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Question : 109 of 200

Marks: +1, -0

Determine the power (in W) delivered by the current source to the given electrical circuit.

92.5
107.5
104.2
93.7

Solution:

∴\R_\eq={3×10}/{3+10}=30/13\Ω

Now

30/13

Ω and 2 Ω are in series

∴\R_\eq=30/13+2=56/13\Ω

Now

P=I^2R=[5^2×56/13]\watt

= 107.69 watt.

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