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Using Kirchhoff’s rules, calculate the potential difference between B and D in the circuit diagram as shown in the figure.

Answers (1)

Consider the loop

$ABDA$

Apply $KVL$ to $ABDA$

$1-2-2I_{2}-2(I_{1}+I_{2})-1\times I_{2}=0$

$-2I_{1}-5I_{2}=1$ _______(1)

Apply $KVL$ to $DCBD$

$3-3I_{1}-1\times I_{1}-1-2(I_{1}+I_{2})=0$

$-6I_{1}-2I_{2}=-2$

$6I_{1}+2I_{2}=2$ _______(2)

Solve (1) and (2)

multiply (1) $\times 3$

$\Rightarrow -6I_{1}-15I_{2}=3$ ______(3)

Add (3) & (2)

$-13 I_{2}=5$

$I_{2}=\frac{-5}{13}$

From (1)

$-2I_{1}-5\left ( \frac{-5}{13} \right )=1$

$-2I_{1}=1-\frac{25}{13}$

$-2I_{1}=\frac{-12}{13}$

$I_{1}=\frac{6}{13}$

Therefor current through the branch $BD$

$=I_{1}+I_{2}$

$=\frac{6}{13}-\frac{5}{13}=\frac{1}{13}A$

Magnitude voltage across BD

$=\frac{1}{13}\times 2=0.153\; volt$

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