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Wheatstone Bridge

The current t...

Question

The current through the 4Ω resistor shown in the figure is.

A

0 A

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B

1 A

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C

3 A

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D

2 A

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Solution

The correct option is A 0 A
Let's assume the currents to be like this

Applying KVL on ADCA
$+ 3 - 3 i_{1} + 4 (i_{2} - i_{1}) = 0$
$7 i_{1} - 4 i_{2} = 3$ ....(1)
Applying KVL on ACBA
$- 4 (i_{2} - i_{1}) - 6 i_{2} + 6 = 0$
$10 i_{2} - 4 i_{1} = 6$ ....(2)
Solving (1) and (2) we get
$i_{1} = 1 A, i_{2} = 1 A$
$∴ i_{2} - i_{1} = 0 A .$
Current through 4Ω resistor is 0 A.

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