Let, I1, I2 and I3 be the currents through the branches of circuit as shown in figure. Va, Vb and Vc be the potentials at points a, b and c respectively.
For current through different branches we can write,
(i) In loop abca,
I2+2(I2+I3)+I1=2
i.e., I1+3I2+2I3=2 .. (1)
(ii) In loop cadc
I1+I1−I2+2(I1−I2−I3)=2
4I1−3I2−2I3=2 .. (2)
(iii) In loop debcd
2(I2+I3)−2(I1−I2−3)+2I3=1
i.e., −2I1+4I2+6I3=1 ....(3)
Now performing, equation (2) + equation (3) x 2 we get
5I2+10I3=4 ... (4)
and performing, equation (1) x 4 - equation(2) we get
15I2+10I3=6 ....... (5)
∴10I2=2 or I2=0.2A
And, 10I3=4−5x0.2=3 or I3=0.3A
Hence, I1=2−0.6−0.6=0.8A
Now, the potential difference between points a and c is
Va−2+0.8=Vc or Va−Vc=1.2V
and the potential difference between points b and c is
Vb−2(0.5)=Vc or Vb−Vc=1.0V
And,
I1−I2−I3=0.8−0.2−0.3=0.3A