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Question

In Column I, the currents through different branches and voltage between different points of the circuit shown are specified as A, B, C and D. In Column II, values are given, one or more of which may correspond to those in Column I.

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Solution

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+I1I2+2(I1I2I3)=2
4I13I22I3=2 .. (2)
(iii) In loop debcd
2(I2+I3)2(I1I23)+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=45x0.2=3 or I3=0.3A
Hence, I1=20.60.6=0.8A
Now, the potential difference between points a and c is
Va2+0.8=Vc or VaVc=1.2V
and the potential difference between points b and c is
Vb2(0.5)=Vc or VbVc=1.0V
And,
I1I2I3=0.80.20.3=0.3A

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