Given :

Two batteries of 7V and 13V of internal resistances 1Ω and 2Ω respectively are connected in parallel with a resistance of 12Ω.

To Find :

Current through each branch of the circuit and the potential difference across 12Ω resistor.

Solution :

❒ Length of question doesn't decide that question is really tough or not :)

Whole question is just based on kirchoff's second law or loop rule.

Simple concept : ∑ ε = ∑ IR

Diagram : Refer to the attached image!

Let current flow in upper branch be I₁ and that in lower branch be I₂.

Total current flow in circuit = I

• I = I₁ + I₂

A] Applying kirchoff's rule for upper branch (loop ACDEA)

B] Applying kirchoff's rule for lower branch (loop BCDFB)

Net current flow in the circuit :

On solving we get,

By substituting the given values;

★ Current flow in upper branch :

➠ IR = E₁ - I₁r₁

➠ (0.71)(12) = 7 - I₁(1)

➠ I₁ = 7 - 8.52

➠ I₁ = -1.52A

Negative sign shows opposite direction!

★ Current flow in lower branch :

➠ IR = E₂ - I₂r₂

➠ (0.71)(12) = 13 - I₂(2)

➠ 2I₂ = 13 - 8.52

➠ I₂ = 4.48/2

➠ I₂ = 2.24A

★ Potential difference across 12Ω :

➛ V = I × R

➛ V = 0.71 × 12

➛ V = 8.52 volts

Verified answer

Refer to the given attachment......