For the circuit shown below, find the potential difference between a and d:
Fig. 6(a)
Applying KVL in 1st mesh, we have
4I1-3=0
Or I1=3/4 Amp.
Vab=2I1=2×3/4=3/2 V
Applying KVL at mesh 2nd, we have</p
10I2-5=0
I2=1/2 Amp.
Vcd=5I2=5×1/2=5/2 V
Now Vad=Vab+Vbc+Vcd=3/2-2+5/2=2 V
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