A resistor R=4.0 Ω is mounted in parallel with a resistor R=8.0 Ω, and this combination is connected in series with a resistor R=12.0 Ω and a 12 Volts battery. What is the power dissipated in the R2 resistor?

Power = 2.16²/8 =0.58watts

Explanation:

Total resistance =[ (R1 * R2)/(R1 + R2) ] + R3 since it is combination of both parallel and series connection

Where R1 = 4ohms

R2 = 8 ohms

R3 = 12 ohms

So therefore Tr = [ (4 * 8)/(4 +8) ] + 12 = 14.67ohms

So therefore the current across the circuit = 12v/14.67ohms = 0.82 amps

So the potential across the parallel connection = 12v - (0.82 * 12 ohms)

= 12 - 9.84 = 2.16v

Since for parallel connection voltage is equally across the resistor

So: power in R2 = V²/ R2

Power = 2.16²/8 =0.58watts

answer:

a

explanation: