Question

Chapter 20, Problem 062 A 83.9-12 resistor is connected in parallel with a 145.1-2 resistor. This parallel group is connected in series with a 27.4-12 resistor. The total combination is connected across a 12.3-V battery. Find (a) the current and (b) the power dissipated in the 145.1-12 resistor. R R3 R2 V (a) Number Units (b) Number Units

Answer 1

- I = The resistors R, and R, are connected in parallel. So the equivalent resistance is RR (83.9) (145.1) = 53.16 0 R+R (83.9)+(145.1) The equivalent resistance of the circuitis Rq = R +R = (27.4 22) +53.16 Q = 80.56 22 The current in the circuit is 12.3 V 0.15 A Rea 80.562 The voltage across the resistors R or R, is V12 = IR, = (0.15 A) (53.16 0) = 8.1164 V The current passing through the resistor 145.12 is 8.1164 V 214.10 -= 0.0559 = 0.056 A 145.12 145.12 b) The power dissipated in the resistor 145.1Qis P= 145.1nº (145.10) = (0.0559) (145.1) = 0.454 W 12