Law of Superposition
The Attempt at a Solution
I am reducing the circuit.
i) By law of Superposition, short circuit V2
100 ohms on the leftmost branch and 50 ohms on the rightmost branch, both are in parallel so
100*80/(100+80)=44.4 ohms seen by V1(t)
iii) short circuit V1
reduce the circuit,
60+40+80=180 ohms as all the resistors are in series.
v)Using law of super position
By shorting V2 I obtain I=0.02A
By shorting V1 I obtain I= 1/18A
I = 0.02+1/18 = 0.076A
Please correct me where I am going wrong. Your help is appreciated. Thanks!
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