Circuit analysis using source transformation?
Use source transformation to find I in the following circuit
- billrussell42Lv 72 months agoFavourite answer
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
- oldschoolLv 71 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 <<<<
- PhilomelLv 71 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.