Question

2) In the circuit below, R, = 3612, R2 = 4032, R2 =8032, R4 = 24092, and E = 120 V m R1 R4 R2 R3 w w a) Find the equivalent resistance in the circuit (5 pts) b) Find the current through the 8012 resistor (5 pts) c) Find the power delivered to all elements and show that power is conserved (4 pts)

Answer 1

The circuit can be redrawn as below: R₃ w Ra www E (6) R2, Rz and Rq are parallel. 1 1 1 1 - + + + Rp R3 Rq 40 80 240 6+ 3 + 1 - 240 10 240 24 Rp Rp = 24 a R and Rp Series, are effective resistanc, R = Rit Rp = 36 + 24 = 602 V - 120 2A: b) Current in circuit R 60 2x 24 = 48 v Voltage drop across som 48 802= 0.6A Current through 80

2. Power delivered by battery p. v? (120) 240 W 60 P, = I?R, = 4 x 36 = 144 W P2 = (1.2) ² 40 = 57.6 W P₂ = (0.6)² 80 = 28.8 W ; Pq = (0.2) ²x 240 = 9.6W Pit Pa + P3 + Pq = .144+ 57.6 + 28.8+ 9.6 = 240 W = p. Hence power is conserved,