Question

For the circuit shown here, calculate the potential difference between points B and D.

Answer 1

According to Kirchhoff’s first law the distribution of currents is shown in fig.

Applying Kirchhoff’s second law to mesh BADB,

–2(i – i₁) + 2 – 1 – 1. (i – i₁) + 2i₁ = 0

⇒ 3i – 5i1 = 1 …(i)

Applying Kirchhoff’s law to mesh DCBD,

–3i + 3 – 1 – 1× i – 2i₁ = 0

⇒ 4i + 2i₁ = 2

Or 2i + i₁ = 1 …(ii)

Multiplying equation (ii) with 5, we get

10i + 5i1 = 5 …(iii)

Adding (i) and (iii), we get

\(13i=6\Rightarrow i=\frac{6}{13}A\)

From (ii), \(i_1=1-2i=1-\frac{12}{13}=\frac{1}{13}A\)

Potential difference between B and D is 

\(V_B-V_D=i_1\times2=\frac{2}{13}V\)