In the circuit below, the capacitor is initially uncharged when Switch A is clos
ID: 2280380 • Letter: I
Question
In the circuit below, the capacitor is initially uncharged when Switch A is closed for 1.5 s, then opened. Switch B is then closed, allowing the capacitor to discharge into the resistor (R = 45.7
In the circuit below, the capacitor is initially uncharged when Switch A is closed for 1.5 s, then opened. Switch B is then closed, allowing the capacitor to discharge into the resistor (R = 45.7 ). What is the instantaneous power dissipated in the resistor after the switch has been closed for 1.57 s? (Hint: the capacitor is not fully charged to the same voltage as the battery.) Write your answer to three significant figures.Explanation / Answer
capacitor is initially uncharged
for switch A during charging till 1.5 sec
Vc(t) = 12 (1- e^(-t/T))
t= time in seconds
T= time period = RC= 40*50*10^-3 = 2 secs
Vc(1.5) = 12 (1-e^(-1.5/2)) = 6.332 volts
after switch A opened and B is closed
new time constant for discharge = T1 = R1*C = 45.7*50*10^-3 = 2.285 sec
Vc(t) = 6.332*e^(-t/T1)
Vc(1.57) = 6.332*e^(-1.57/2.285) = 3.185 volts
Vc+Vr1=0
=> Vr2= -Vc (since discharge)
instantaneous power dissipated = V^2/R = 3.185^2/45.7 = 0.222 watt