Problem 23.60 How much power does bulb A dissipate when the switch is open? If the battery in (Figure 1) were ideal, lightbulb A would not dim when the switch is closed. However, real batteries have a small internal resistance, which we can model as the 0.3 2 resistor shown inside the battery. In this case, the brightness of bulb A changes when the switch is closed. Assume 2.0 V Express your answer with the appropriate units IA ropen- value Units Figure 〈 1of1 Submit Request Answer ▼ Part B How much power does bulb A dissipate when the switch is closed? 0.3 2 A(6.0 6.0 Ω Express your answer with the appropriate units Pclosed = Value Units
Answer
[A] When switch is open then total resistance is R = 6 + 0.3 R = 6.3 Ohm So current flowing through bulb A is I = E/R, E = 2.0 V So I = 2/6.3 = 0.317 A So power dissipate at bulb A is P_open = I^2 R = (0.317)^2 times 6 = 0.6029 Watt P_open = 0.603 Watt Part [B] When switch is class then total resistance is R = 6 times 6/6 + 6 + 0.3 (R' = R_1 R_2/R_1 + R_2) in parallel combination R = 3 + 0.3 R = 3.3 Ohm So current flowing through bulb A is I' = I/2 = E/2R = 2/2 times 3.3 = 0.303 A So power dissipate at bulb 'A' is P_close = (I')^2 R = (0.303)^2 times 6 = 0.5508 Watt P_close = 0.551 Watt