Question

Four resistors, R1 = 29.0 ?, R2 = 46.0 ?, R3 = 85.0 ? and R4 = 20.0 ?, are connected to a battery with a terminal voltage of 12.0 V, as shown in the figure below. Determine the power lost in resistor R2.

Answer 1

The resistors R1 and R2 are connected in series. Then, the equivalent resistance is, 12 75 Ω This resistance is connected in parallel with the resistance R3. Then, the equivalent resistance 1s, R2R 123 (75Ω)(85 Ω) 75 Ω +85 Ω 39.84 Ω This resistance is connected in series to the resistance R4. Therefore, the equivalent resistance of the circuit is, 123 39.84 Ω+ 20 Ω 59.8462 The total current in the circuit is, total equi 12 V 59.84 2 0.201 A

The voltage across the combination of resistors is, P,,-12 V-(0.201 A) (20 ) The current through the resistor R2 is, R12 98 V 75 Ω 0.1064 A Therefore, the power lost in resistor R2 is, (0.1064 A) (462) - 0.52 W Hence, the power lost in resistor R2 is 0.52 W