Question
For the circuit shown here, calculate the potential difference between points B and D.
✓
Answer
According to Kirchhoff’s first law the distribution of currents is shown in fig.
Applying Kirchhoff’s second law to mesh BADB, $–2(i – i_1) +2 – 1 – 1. (i – i_1) + 2i_1 = 0$
$\Rightarrow 3i – 5i_1 = 1 …(i)$
Applying Kirchhoff’s law to mesh DCBD, $-3i + 3 - 1 - 1\times i - 2i_1 = 0$
$\Rightarrow 4i + 2i_1 = 2$ Or $2i + i_1 = 1 …(ii)$
Multiplying equation (ii) with 5, we get
$10i + 5i_1 = 5 …(iii)$
Adding (i) and (iii), we get $13\text{i}=6\Rightarrow\text{i}=\frac{6}{13}\text{A}$
From (ii), $\text{i}_1=1-2\text{i}=1-\frac{12}{13}=\frac{1}{13}\text{A}$
Potential difference between B and D is,
$\text{V}_\text{n}-\text{v}_\text{n}=\text{i}-1\times2=\frac{2}{13}\text{V}$
Applying Kirchhoff’s second law to mesh BADB, $–2(i – i_1) +2 – 1 – 1. (i – i_1) + 2i_1 = 0$
$\Rightarrow 3i – 5i_1 = 1 …(i)$
Applying Kirchhoff’s law to mesh DCBD, $-3i + 3 - 1 - 1\times i - 2i_1 = 0$
$\Rightarrow 4i + 2i_1 = 2$ Or $2i + i_1 = 1 …(ii)$
Multiplying equation (ii) with 5, we get
$10i + 5i_1 = 5 …(iii)$
Adding (i) and (iii), we get $13\text{i}=6\Rightarrow\text{i}=\frac{6}{13}\text{A}$
From (ii), $\text{i}_1=1-2\text{i}=1-\frac{12}{13}=\frac{1}{13}\text{A}$
Potential difference between B and D is,
$\text{V}_\text{n}-\text{v}_\text{n}=\text{i}-1\times2=\frac{2}{13}\text{V}$
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