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Question
In the circuit shown below:
The potential difference across the 3 Ω resistor is:
Options
`1/9`V
`1/2` V
1 V
2 V
Solution
1V
The resistors of 1 Ω, 2 Ω and 3 Ω are connected in series. Therefore, the net resistance,
R = R1 +R2 +R3
R = 1 Ω + 2 Ω + 3 Ω = 6 Ω
Current in the circuit will be,
I = V/R
or I = 2/6 = 1/3 A
Current = `1/3`A
Therefore, the voltage across the 3 Î resistor,
V = IR
or V= `1/3xx3` = 1 V
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