SSC JE Electrical Engineering 2013 Paper

Question : 157 of 200

Marks: +1, -0

Two 2000 Ω, 2 watt resistors are connected in parallel. Their combined resistance value and wattage rating are

Solution:

Req.

= R_1 || R_2

={2000×2000}/{2000+2000}

Req. = 1000Ω
P =

P_1 + P_2

= 2 + 2 = 4W

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