First slide

Kirchoff's Law

Question

In the given circuit diagram a wire is joining points B and D. The current in this wire is

Moderate

A

0
B

1 A
C

2 A
D

4 A

Solution

$\begin{array}{l} 1 Ω and 4 Ω are in parallel \frac{1 (4)}{1 + 4} = \frac{4}{5} Ω \\ 2 Ω and 3 Ω are in parallel \frac{2 (3)}{2 + 3} = \frac{6}{5} Ω \\ \frac{4}{5} Ω and are in series \\ R_{eff} = \frac{4}{5} + \frac{6}{5} = \frac{10}{5} = 2 Ω \\ i = \frac{V}{R_{eff}} = \frac{10}{2} = 5 A \end{array}$
Apply junction rule at point B
$L e t R_{1} = 4 Ω, R_{2} = 1 Ω, R_{3} = 3 Ω a d R_{4} = 2 Ω c u r r e n t t h r o u g h R_{1} = 4 Ω i s i_{1} c u r r e n t t h r o u g h R_{2} = 1 Ω i s i_{2} c u r r e n t t h r o u g h R_{3} = 3 Ω i s i_{3} c u r r e n t t h r o u g h R_{4} = 2 Ω i s i_{4} {Current across 4Ω,I}_{1} = \frac{R_{2} i}{R_{1} + R_{2}} = = \frac{1 (5)}{1 + 4} = 1 A {Current across 3Ω,I}_{3} = \frac{R_{4} i}{R_{3} + R_{4}} = = \frac{2 (5)}{2 + 3} = 2 A c u r r e n t t h r o u g h B C = i = I_{3} - I_{1} = 2 - 1 = 1 A$

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