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Using source conversion, reduce the circuit shown in figure into single current source in parallel with single resistance.
written 9.4 years ago by gravatar for teamques10 teamques10 ★ 70k •   modified 5.6 years ago

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basic electrical engineering
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written 9.4 years ago by gravatar for teamques10 teamques10 ★ 70k

enter image description here

V1=IR= (3)(3)=9V

3Ω resistors are in series and 6Ω resistors are in parallel.

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$I_1=\dfrac{V}{R}=\dfrac96=1.5A (\uparrow)$

As 1.5A & 2A currents flow in upward direction. We add them 6Ω & 3Ω resistors are in parallel.

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