Question

Two cells of emf 1.5 Volt and 2 Volt having respective internal resistances 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

Given: E₁ = 1.5 V, E₂ = 2V, r₁ = 1 Ω, r₂ = 2 Ω, R = 5 Ω, I = ?

Let I₁ and I₂,  be the currents flowing through the two branches as shown in the following figure. The current through the 5 Ω resistor will be I₁ + I₂ [Kirchhoff's current law]. 

Applying Kirchhoff's voltage law to loop ABCDEFA, we get,

- (I₁ + I₂) R - I₁r₁ + E₁ = 0

- (I₁ + I₂)5 - I₁ + 1.5 = 0

- 5 I₁ - 5 I₂ - I₁ = 1.5

- 6 I₁ - 5 I₂ = - 1.5

∴ 6 I₁ + 5 I₂ = 1.5       ....(1)

Applying Kirchhoff's voltage law to loop BCDEB, we get,

- (I₁ + I₂)R - I₂r₂ + E₂ = 0

- (I₁ + I₂)5 - 2I₂ + 2 = 0

- 5 I₁ - 5 I₂ - 2 I₂ = - 2

- 5 I₁ - 7 I₂ = - 2

∴ 5I₁ + 7I₂ = 2       ....(2)

∴ Multiplying Eq. (1) by 5 and Eq. (2) by 6, we get,

6 I₁ + 5 I₂ = 1.5   ...(× 5)

5I₁ + 7I₂ = 2       ....(× 6)

30 I₁ + 25 I₂ = 7.5       ....(3)
30 I₁ + 42 I₂ = 12       ....(4)
-         -            -    
  17 I₂ = 4.5

∴ I₂ = ${\displaystyle \frac{4.5}{17}}$ A

Substituting this value of I₂ in Eq. (1), we get,

6 I₁ + 5 I₂ = 1.5

${\displaystyle 6{\text{I}}_1+5\left(\frac{4.5}{17}\right)=1.5}$

∴ ${\displaystyle 6{\text{I}}_1+\frac{22.5}{17}=1.5}$

∴ ${\displaystyle 6{\text{I}}_1=1.5-\frac{22.5}{17}}$

∴ ${\displaystyle 6{\text{I}}_1=\frac{25.5-22.5}{17}}$

∴ ${\displaystyle 6{\text{I}}_1=\frac{3}{17}}$

∴ ${\displaystyle {\text{I}}_1=\frac{3}{17\times 6}}$

∴ ${\displaystyle {\text{I}}_1=\frac{0.5}{17}}$ A

Current through the 5 Ω resistance (external resistance) = ${\displaystyle {\text{I}}_1+{\text{I}}_2=\frac{0.5}{17}+\frac{4.5}{17}=\frac{5}{17}}$A.