Two resistances of 6 Ω and 12 Ω are connected in parallel. Their net resistance is ______?

Answer: 4 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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