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Wheatstone Bridge

In the circui...

Question

In the circuit shown in Figure above, $E_{1}$ and $E_{2}$ are two cells having emfs 2V and 3V respectively, and negligible internal resistances. Applying Kirchhoff's laws of electrical networks, find the values of currents $I_{1}$ and $I_{2}$ .

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Solution

Consider the direction of currents as shown in the figure above.
By Kirchhoff's second law (law of conservation of energy) :
in mesh (1)
$R_{3} (I_{1} + I_{2}) + R_{1} I_{1} - E_{1} = 0$
$2 (I_{1} + I_{2}) + 1 \times I_{1} = 2 \Rightarrow 3 I_{1} + 2 I_{2} = 2$ ...(i)
in mesh (2)
$R_{3} (I_{1} + I_{2}) + R_{2} I_{2} - E_{2} = 0$
$2 (I_{1} + I_{2}) + 6 I_{2} = 3 \Rightarrow 2 I_{1} + 8 I_{2} = 3$ ...(ii)
By equations (i) and (ii)
$I_{1} = \frac{1}{2} A$ and $I_{2} = \frac{1}{4} A$

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