A 6V cell with 0.5Ω internal resistance, a 10V cell with 1Ω internal resistance and a 12Ω external resistance are connected in parallel. The current (in amperes) through the 10V cell is
A
0.6
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B
2.27
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C
2.87
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D
5.14
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Solution
The correct option is B 2.87 The potential difference between AB,CD and EF are same. Applying Kirchhoff's law, 6−(0.5)I1=10−I2=12(I1+I2) solving above equations, we get I1=2.87A.