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Question
A battery of emf 4 volt and internal resistance 1 Ω is connected in parallel with another battery of emf 1 V and internal resistance 1 Ω (with their like poles connected together). The combination is used to send current through an external resistance of 2 Ω. Calculate the current through the external resistance.
Solution
Let I1 and l2 be the currents through the two branches as shown in the following figure. The current through the 2 Ω resistance will be (I1 + I2) [Kirchhoffs current law].
Applying Kirchhoff's voltage law to loop ABCDEFA, we get,
-2(I1 + I2) - 1(I1) + 4 = 0
∴ 3I1 + 2I2 = 4 .....(1)
Applying Kirchhoff's voltage law to loop BCDEB, we get,
-2(I1 + I2) - 1(I2) + 1 = 0
2I1 + 3I2 = 1 ....(2)
Multiplying Eq. (1) by 2 and Eq. (2) by 3, we get,
6I1 + 4I2 = 8 .....(3)
and 6I1 + 9I2 = 3 ....(4)
Subtracting Eq. (4) from Eq. (3), we get,
- 5I2 = 5
∴ I2 = - 1A
The minus sign shows that the direction of current I2 is opposite to that assumption. Substituting this value of I2 in Eq. (1), we get,
3I1 + 2(-1) = 4
∴ 3I1 - 4 + 2 = 6
∴ I1 = 2A
Current through the 2 Ω resistance = I1 + I2 = 2 - 1 = 1 A.
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