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Kirchoff's Junction Law

In the circui...

Question

In the circuit, given in the figure currents in different branches and value of one resisitor are shown. Then potential at point $B$ with respect to the point $A$ is

A

+ 2 V

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B

- 2 V

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C

- 1 V

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D

+ 1 V

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Solution

The correct option is D $+ 1 V$

Using Kirchoff's junction at $C$ , we get

$\Rightarrow i_{1} + i_{3} = i_{2}$
$\Rightarrow 1 A + i_{3} = 2 A \Rightarrow i_{3} = 1 A$
Now using Kirchoff's loop law along $A C D B$
$V_{A} + 1 + i_{3} (2) - 2 = V_{B}$
$\Rightarrow V_{A} + + 1 + i_{3} (2) - 2 = V_{B}$
$\Rightarrow V_{B} - V_{A} = 3 - 2 = 1 V$

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