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Mixed Combination of Cells

In Column I, ...

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

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