A 1.40 F capacitor is charged through a 126 resistor and then discharged through

Question

A 1.40 F capacitor is charged through a 126 resistor and then discharged through the same resistor by short-circuiting the battery.

Part A

While the capacitor is being charged, find the time for the charge on its plates to reach 1e of its maximum value.

SubmitMy AnswersGive Up

Incorrect; Try Again; no points deducted

Part B

While the capacitor is being charged, find the current in the circuit at the time when the charge on its plates has reached 1e of its maximum value.

SubmitMy AnswersGive Up

Part C

During the discharge of the capacitor, find the time for the charge on its plates to decrease to 1e of its initial value.

SubmitMy AnswersGive Up

Part D

Find the time for the current in the circuit to decrease to 1e  of its initial value.

A 1.40 F capacitor is charged through a 126 resistor and then discharged through the same resistor by short-circuiting the battery.

Part A

While the capacitor is being charged, find the time for the charge on its plates to reach 1e of its maximum value.

t =   s  

SubmitMy AnswersGive Up

Incorrect; Try Again; no points deducted

Part B

While the capacitor is being charged, find the current in the circuit at the time when the charge on its plates has reached 1e of its maximum value.

I =   I0  

SubmitMy AnswersGive Up

Part C

During the discharge of the capacitor, find the time for the charge on its plates to decrease to 1e of its initial value.

t =   s  

SubmitMy AnswersGive Up

Part D

Find the time for the current in the circuit to decrease to 1e  of its initial value.

Explanation / Answer

a)
Capacitor charge equation .. Qt = Qo [1 - e^(-t/CR) ]
Qt/Qo = [1 - e^(-t/CR) ] = 1/e = 0.3679

1 - 0.3679 = e^(-t/CR)
Ln(0.6321) = -t/CR
t = -CR Ln(0.6321) .. - (1.40^-6F x 126 x -0.4587)

t = 8.091 x 10^-5 sec………Ans.

b)
When charge on C has reached Q/e the pd across C and R = Vo/e (VQ)
Current .. i = pd/R = (Vo/e) / R

i = Vo/(eR) = V0/2.718*126 =0.00292V0………..Ans. (battery pd(Vo) required, e= 2.718)

c)
Discharge equation .. Qt = Qo e^(-t/CR)
Qt/Qo = e^(-t/CR) = 1/e = e¹ therefore ..

(-t/CR) = -1
t = CR .. .. 1.40^-6F x 126

t = 1.764 x 10^-4 sec……..Ans.

d)
Current decay follows same decay curve as Q and V .. so the time for IoIo/e = time for QoQo/e which is calculated in part (c)

t = 1.764 x 10^-4 sec……..Ans.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.