Question

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.

Answer 1

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