Question

Use Kirchhoff’s rules to determine the potential difference between the points A and D when no current flows in the arm BE of the electric network shown in the figure.

Answer 1

According to Kirchhoff’s junction rule at E or B

I₃ = I₂ + I₁

Since I₂ = 0 in the arm BE as given in the question

⇒ I₃ = I₁

Using loop rule in the loop AFEBA

+6V – 2I₃ + 1V – 3I₃ – I₂ R₁ + 3V = 0

⇒ 2I₃ + 3I₃ + I₂R₁ = 10V

Since I₂ = 0, so

5I₃ = 10V

⇒ I₃ = 2 A

∴ I₃ = I₂ = 2 A

The potential difference between A and D, along the branch AFED of the closed circuit.

VA – 2I₃ + 1V – 3 I₃ – VD = 0

⇒ V_A – V_D = 2I₃ –1 + 3I₃

= 2 × 2 – 1 + 3 × 2 

= 9 V