Two resistances of 6 Ω and 12 Ω are connected in parallel. Their net resistance is ______?
- 7 Ω
- 6 Ω
- 4 Ω
- 5 Ω
Explanation
When two resistances are connected in parallel, the net resistance (Rnet) is calculated using the formula:
1/Rnet = 1/R1 + 1/R2
where R1 and R2 are the individual resistances.
In this case, R1 = 6 Ω and R2 = 12 Ω, so:
1/Rnet = 1/6 + 1/12 = 1/4
Rnet = 4 Ω
Therefore, the net resistance of the two resistances connected in parallel is 4 Ω.
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