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

2A
B

0.4A
C

4A
D

zero

Solution

$\begin{array}{l} R e s i s t o r s (1 | | 4) + (2 | | 3) \\ h e r e | | m e a n s p a r a l l e l a n d + m e a n s s e r i e s \\ R_{e f f} = \frac{(1 \times 4)}{(1 + 4)} + \frac{(2 \times 3)}{(2 + 3)} = 2 Ω \\ O h m' s L a w, i = \frac{E}{R} = \frac{20}{2} = 10 A \\ C u r r e n t i n B A = i_{1} \\ C u r r e n t i n A D = i - i_{1} \\ C u r r e n t i n B C = i_{2} \\ C u r r e n t i n B D = i_{1} - i_{2} \\ K i r c h h o f f L o o p i n A B C E A, \\ - (i_{1}) (1) - (i_{2}) (2) + 20 = 0 - - - - - - (1) \\ K i r c h h o f f L o o p i n A B D A, \\ - (i_{1}) (1) + (i - i_{1}) (4) = 0 \\ \Rightarrow - (i_{1}) (1) + (10 - i_{1}) (4) = 0 \\ \Rightarrow i_{1} = 8 A, a n d \\ P u t i_{1} = 8 i n e q u a t i o n (1) \\ \Rightarrow i_{2} = 6 A \\ H e n c e, C u r r e n t i n B D = i_{1} - i_{2} = 8 - 6 = 2 A \end{array}$

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