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 same direction through an external resistance of 5 Ω. Find the current through the external resistance.

Answer 1

Let I₁ and I₂ be the currents through the two branches as shown in below figure. The current through the 5 Q resistor will be I₁ + I₂ [Kirchhoff’s current law].

Applying Kirchhoff’s voltage law to loop w ABCDEFA, we get,

– 5(I₁ + I₂ ) – I₁ + 1.5 = 0 

∴ 6I₁ + 5I₂ = 1.5 ……………. (1)

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

– 5(I₁ + I₂ ) – 2I₂ + 2 = 0 

∴ 5I₁ + 7I₂ = 2 ……………(2)

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

30I₁ + 25I₂ = 7.5 …………… (3) 

and 30I₁ + 42I₂ = 12 …………. (4)

Subtracting Eq. (3) from Eq. (4), we get,

17I₂ = 4.5

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

Current through the 5 Q resistance (external resistance)