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
According to Kirchoff‟s Junction law at $B$
$\text{i}_{3} = \text{i}_{1} + \text{i}_{2} \therefore\text{i}_{3} = \text{i}_{1}$
$($As $I_2=0 ($given$)$
Applying second law to loop $\text{AFEB}$
$\text{i}_{3}\times2 + \text{i}_{3} \times3 + \text{i}_{2}\text{R}_{1} = 1+ 3 + 6 $
$\therefore\text{i}_{3} = \text{i}_{1} = 2 \text{A}$
From $A$ to $D$ along $\text{AFD} \therefore\text{V}_{\text{AD}} = 2\text{i}_{3} -1 +3 \times\text{i}_{3}$
$ = ( 4- 1 + 6 ) \text{V}$
$= 9 V.$
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