First slide

combination of resistance

Question

In the circuit shown in figure, current through the conductor BC is

Moderate

A

2A
B

4A
C

3A
D

6A

Solution

The resistance 2Ω is short circuited. therefore current through 3Ω, $i_{3} = \frac{12}{3} A$ = 4A and current through 6Ω, $i_{6} = \frac{12}{6} A$ = 2A.
Current through 2Ω is zero and current through BC is (4 + 2) A i.e. 6 A.

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