l1-61 An unloaded 47 uF capacitor is connected to a 12 kl resistor in series and

Question

l1-61 An unloaded 47 uF capacitor is connected to a 12 kl resistor in series and a 15 V battery is connected to this circuit at time t-0. 1. Find the charge of the capacitor at t-5 s. 4. At what time t, the capacitor is quarter full? a. 100 HO b. 300 uC c. 500 HC d. 700 uC e. 900 uC a. 0.141 s b. 0.162 s c. 0.282s d. 0.391 s e. 0.782s Find the charge of the capacitor at time t-0.7 s. 2. What current runs on the circuit at time t-0? 5. a. 100 uC b. 300 uC c. 50011C d. 700 uC 900 C a. zerO b. 0.75 mA c. 1.00 mA d. 1.25 mA e. 1.50 mA 3. At what time t, the capacitor is half full. What current runs on the circuit at time t-5 s? 6. a. 0.162 s b. 0.282 s c. 0.391 s d. 0.782 s e. 0.814 s a. zero b. 0.75 mA c. 1.00 mA d. 1.25 mA e. 1.50 mA 7. If you want to discharge a 29 mF capacitor in one minute, what should be the resistance of the resistor that you should be connected to it? Assume that the capacitor is practically empty at three percent of the original charge. a. Zero b. 26 c. 590 e. 200 k

Explanation / Answer

1. Q = Q0 [1 - e^(-t/T)]
  
Q0 = C V0 = (47 uF) (15) = 705 uC

T = RC = (47 x 10^-6 F)(12 x 10^3) = 0.564s

Q = (705)[1 - e^(-5/0.564)] = 704 uC

Ans(d) 700 uC

2. t = 0.7s

Q = (705)[1 - e^(-0.7/0.564)] = 501 uC

Ans(c) 500 uC

3. 1 - e^(-t/0.564) = 0.5

- t / 0.564 = ln(2)

t = 0.391 s Ans(C)

4. 1 - e^(-t/0.564) = 0.25

- t / 0.564 = ln(0.75)

t = 0.162 s Ans(B)

5. at t = 0 , I = 15/12 = 1.25 mA

Ans(d)

6. at t = 5s, I= 0

Ans(A)

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