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प्रश्न
Two cells of emf 4 V and 2 V having a respective internal resistance of 1 Ω and 2 Ω are connected in parallel, so as to send current in the same direction through an external resistance of 5 Ω. Find the current through the external resistance.
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उत्तर
Applying Kirchhoff's voltage law to loop ABCDEFA. We get,
-5 (l1 + l2) - l1 + 4 = 0
∴ - 5 l1 - 5 l2 - l1 = - 4
6 l1 + 5 l2 = 4 ...(i)
Applying Kirchhoff's voltage law to loop BCDEB, We get
-5 (l1 + l2) - 2 l2 + 2 = 0
∴ - 5 l1 - 5 l2 - 2 l2 = - 2
- 5 l1 - 7 l2 = 2 ...(ii)
Multiplying equation (i) by 5 and equation (ii) by 6 we get
30 l1 + 25 l2 = 20 ...(iii)
And 30 l1 + 42 l2 = 12 ...(iv)
Subtracting equation (iii) from equation (iv) we get,
17 l2 = - 8
I2 = `(- 8)/17` A
Substituting this value of l2 in equation (i), we get
`6"I"_1 + 5 xx (- 8)/17` = 4
`6"I"_1 - 40/17` = 4
`6"I"_1 = 4 + 40/17 = 108/17`
∴ `"I"_1 = 18/17`A
∴ The current through external resistance
`"I"_"S" = "I"_1 + "I"_2`
`= 18/17 - 8/17`
`= 10/17`A
∴ The current through external resistance is `"l"_"s" 10/17`A
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