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Question
In the circuit shown in Figure below, E1 and E2 are batteries having emfs of 25V and 26V. They have an internal resistance of 1 Ω and 5 Ω respectively. Applying Kirchhoff’s laws of electrical networks, calculate the currents I1 and I2.
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Solution
Applying Kirchhoff's voltage law to loop ABCDEF, we get
2(I1 + I2) + 3I2 - 26 + 5I2 = 0
or 2I1 + 2I2 + 3I2 + 5I2 = 26
or I1 + 5I2 = 13 ...(i)
Now, applying Kirchhoff’s voltage law to loop HJBCDEG.
2(I1 + I2) + 4I1 + I1 - 25 = 0
or 2I1 + 2I2 + 5I1 = 25
or 7I1 + 2I2 = 25 ...(ii)
Equation (i) × 2 - equation (ii) × 5
2I1 + 10I2 = 26
35I1 + 10I2 = 125
- - -
- 33I1 = - 99
I1 = 3
From equation (i)
3 + 5I2 = 13
or I2 = 2
Hence I1 = 3 A
I2 = 2 A.
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