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.

Answer 1

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.