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23 Nov 2019

Capacitors won't hold a charge indefinitely; as time goes on,charge gradually migrates from the positive to the negative plate.We can model this as a discharge of the capacitor through aninternal "leakage resistance." A 0.48 capacitor charged to 2.9 willinitially discharge with a leakage current of 0.40 .What is theleakage resistance? How long will it take for the capacitor voltageto drop to 1.0 ?

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Nelly Stracke
Nelly StrackeLv2
26 Oct 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

A heart dysfunction that can cause death is ventricular fibrillation. This is an uncoordinated quivering of the heart as opposed to regular beating. An electric shock to the chest can cause momentary paralysis of the heart muscle, after which the heart will sometimes start organized beating again. A defibrillator is a device that applies a strong electric shock to the chest over a time of a few milliseconds. The device contains a capacitor of several microfarads, charged to several thousand volts. Electrodes called paddles, about 8 cm across and coated with conducting paste, are held against the chest on both sides of the heart, their handles are insulated to prevent injury to the operator, who calls clear and pushes a button on one paddle to discharge the capacitor through the patient's chest.
Assume that the capacitor in the defibrillator is 10 uF and is charged to 5,000 volts.
a. how much charge is stored in the capacitor before it is discharged?
b. how much energy is released when the capacitor is discharged?
c. if the capacitor completely discharges in 2.5 ms, what is the average current delivered by the defibrillator?
d. what is the average power delivered?

This set of problems is designed to help you prepare for the lab oncapacitors. Here are some things you should know.
The charge on a capacitor is proportional to the voltage. Theconstant of proportionality is the capacitance:
q=VC.
The simplest type of capacitor is a parallel plate capacitorconsisting of twoparallel conducting plates...DUH. If the plateseach have area A and
are separated by a distance d, the capacitance will be
C=Ke0A/d,
where e0 is the permittivity of free space (you measured that lastweek) and K is the dielectric constant of what ever material isbetween the plates. If the space between the plates is filled withvacuum, the dielectric constant is exactly 1. Air is pretty darnclose to 1.
When a capacitor is charged in a series circuit with a resistor,the charge builds up with time. Consider the circuit below.



When the switch, S, is thrown to position A, the battery, withvoltage V0, will supply a current through the resistor, R, thatcharges the capacitor C. The voltage across the capacitor at anygiven time is given by
V=V0(1-e-t/RC).
The product RC is known as the time constant. After waiting a timeequal to 1 time constant, the voltage on the capacitor is(1-e-1)=0.632 of V0. After 5RC the capacitor is nearly fullycharged and has a voltage of 0.993V0. Eventually, the voltagebuilds up to V0.
When the switch is thrown to position B, the capacitor willdischarge through the resistor. The voltage as a fuction of timewill be given by
V=V0(e-t/RC).
After a long enough time, the capacitor will fully discharge theresistor.

1. [1pt]
A parallel plate capacitor has plates with length and width of 5.7cm and are separated by 7.0 mm. The capacitance is measured to be1.479×10-11 F. What is the dielectric constant of the materialbetween the plates?

Correct, computer gets: 3.60

2. [1pt]
Consider a series circuit like the one shown above with a resistorof R=80. kilo-ohms and a capacitor of C=8.0 uF. What is the timeconstant of the circuit?

Correct, computer gets: 0.64 s


Hint: Don't forget to convert to Farads and ohms from uF andkilo-ohms.
3. [1pt]
If the battery has a voltage of 12 V, what is the voltage acrossthe capacitor after the switch has been in position A for 2 timeconstants?


Correct, computer gets: 10.38 V
4. [1pt]
What is the final charge on the capacitor after a long time?


Correct, computer gets: 9.60E-05 C
5. [1pt]
Now let's say that the capacitor has been fully charged and wethrow the switch to position B. How long will it take for thevoltage on the capacitor to be 0.891 V?


6. [1pt] What is the charge on the capacitor at this time?

Answer: ____________

This set of problems is designed to help you prepare for the lab oncapacitors. Here are some things you should know.
The charge on a capacitor is proportional to the voltage. Theconstant of proportionality is the capacitance:
q=VC.
The simplest type of capacitor is a parallel plate capacitorconsisting of twoparallel conducting plates...DUH. If the plateseach have area A and
are separated by a distance d, the capacitance will be
C=Ke0A/d,
where e0 is the permittivity of free space (you measured that lastweek) and K is the dielectric constant of what ever material isbetween the plates. If the space between the plates is filled withvacuum, the dielectric constant is exactly 1. Air is pretty darnclose to 1.
When a capacitor is charged in a series circuit with a resistor,the charge builds up with time. Consider the circuit below.



When the switch, S, is thrown to position A, the battery, withvoltage V0, will supply a current through the resistor, R, thatcharges the capacitor C. The voltage across the capacitor at anygiven time is given by
V=V0(1-e-t/RC).
The product RC is known as the time constant. After waiting a timeequal to 1 time constant, the voltage on the capacitor is(1-e-1)=0.632 of V0. After 5RC the capacitor is nearly fullycharged and has a voltage of 0.993V0. Eventually, the voltagebuilds up to V0.
When the switch is thrown to position B, the capacitor willdischarge through the resistor. The voltage as a fuction of timewill be given by
V=V0(e-t/RC).
After a long enough time, the capacitor will fully discharge theresistor.

--------------------------------------------------------------------------------


For multiple choice questions, enter the letter corresponding tothe correct answer.

--------------------------------------------------------------------------------

1. [1pt]
A parallel plate capacitor has plates with length and width of 5.7cm and are separated by 7.0 mm. The capacitance is measured to be1.479×10-11 F. What is the dielectric constant of the materialbetween the plates?

Correct, computer gets: 3.60


Hint: Don't forget to convert distances to meters.



--------------------------------------------------------------------------------

2. [1pt]
Consider a series circuit like the one shown above with a resistorof R=80. kilo-ohms and a capacitor of C=8.0 uF. What is the timeconstant of the circuit?

Correct, computer gets: 0.64 s


Hint: Don't forget to convert to Farads and ohms from uF andkilo-ohms.
3. [1pt]
If the battery has a voltage of 12 V, what is the voltage acrossthe capacitor after the switch has been in position A for 2 timeconstants?


Correct, computer gets: 10.38 V
4. [1pt]
What is the final charge on the capacitor after a long time?


Correct, computer gets: 9.60E-05 C
5. [1pt]
Now let's say that the capacitor has been fully charged and wethrow the switch to position B. How long will it take for thevoltage on the capacitor to be 0.891 V?


Answer:

6. [1pt]
What is the charge on the capacitor at this time?


Answer:

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