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Circuit analysis using source transformation?

Use source transformation to find I in the following circuit

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  • 2 months ago
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    combine the two voltage sources and two resistors to get below.

    Then use thevenin equivalent to change current source and 2 Ω resistor to a voltage source of 4 volts and series R of 2 Ω.  (corrected)

    Now just ohms law I = 6/16 = 3/8 amp = 0.375 amps

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  • 1 month ago

    By source transformation the 2A source in parallel with the 2Ω can be replaced by it's Thevenin equivalent which is 2A*2Ω = 4V in series with a 2Ω resistance. The + side of the 4V source on the right. Then perform KVL assuming a clockwise current i:

    8 - 6*i - 2*i + 4 - 8*i - 6 = 0

    6 = 16*i and i = 3/8 A  =  375mA <<<<

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  • 1 month ago

    The voltage sources reduce to 2V.

    The series resistors reduce to 8+6=14Ω.

    The I R reduces to 4v in series with 2Ω.

    The whole circuit is now 2V+4V=6v

    in series with 2Ω+14Ω=16Ω

    I=6/16= 3/8= 375mA.

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