Trustsu

by


A 12.0-μF capacitor is charged to a potential of 50.0V and then discharged through a 175-Ω…

The following solution is suggested to handle the subject “A 12.0-μF capacitor is charged to a potential of 50.0V and then discharged through a 175-Ω…“. Let’s keep an eye on the content below!

Question “A 12.0-μF capacitor is charged to a potential of 50.0V and then discharged through a 175-Ω…”

A 12.0-μF capacitor is charged to a potential of 50.0V and then discharged through a 175-Ω resistor How long does it take the capacitor to lose (a) half of its charge and (b) half of its stored energy? 

Answer

Concepts and Reason

The key concepts to solving the problem are: relation between charge, capacitance and voltage, decaying charges expression and decaying energy expression.

To calculate the charge, first use the relationship between charge, capacitance and voltage. Next, calculate the final charge of the capacitor using the condition. To calculate the time, you can use the decay charge equation.

The decaying energy equation can be used to calculate how long it takes for a capacitor to lose half its stored energy.

Fundamentals

The relationship between voltage, charge, and capacitor is.


Q=CV


Q represents the charge, C the capacitance and V the voltage.

Here’s how to express the charge in a capacitor:


Q=Qe RC


Here

represents the initial charge and

the final charge. t is the time. R is resistance. C is capacitance.

Here’s how to express the energy stored in the capacitor:


U=U e RC


Here

represents the initial energy and

the final energy. t is the time. R is the resistance. C is the capacitance.

(a)

Find out how long it takes for a capacitor to lose half its charge.

The relationship between voltage, charge, and capacitor is.


Q=CV


Q represents the charge, C the capacitance and V the voltage.

Substitute

12x10F

to C and

50.0V

for V in expression

Q=CV

.


Q=(12x10ⓇF)(50.00) = 6.0x10°C


The final charge on the capacitor equals half the initial charge.



Q will be the final charge, and

will be the initial charge.

Substitute HTMLmedia_tag_14$ in expression

.


0-6.0x10*c = 3.0x10C


Here’s how to express the charge in a capacitor:


Q=Qe RC


Here

represents the initial charge and

the final charge. t is the time. R is resistance. C is capacitance.

Rearrange the equation to solve for

.

Natural log both sides of equation

Q=Qe RC

.


nek neese) KE


Substitute HTMLmedia_tag_24$ to HTMLmedia_tag_25$ and HTMLmedia_tag_26$.


RC t=RCIn(2)


Substitute

to replace

or

12x10F

to replace

within expression

t = RC In (2)

.


t = (17592)(12x10“F)In(2) = 1.46x10s


Or


t=1.46 ms


(b)

Calculate how long it takes for the capacitor to lose half its stored energy.

Here’s how to express the energy stored in the capacitor:


U=U e RC


represents the initial energy and

the final energy. t is the time. R is the resistance. C is the capacitance.

The final energy of the capacitor is equal to half the initial energy.



U will be the final charge, and

will be the initial charge.

Substitute

to U with expression #media_tag_41$.


v -lo


Natural log both ends of the equation

ere

.


(2) --(


For time t, rewrite the equation

In(2) RC

.


i ln(2)RC


Substitute

to

, and

to

within equation

i ln(2)RC

.


In(2)(17592)(12x10F) 1 = = 0.728x10s


Or


t=0.728 ms


Ans: Part A

1.46 ms

is the time it takes for the capacitor to lose half its charge.

Part b

0.728 ms

is the time it takes for a capacitor to lose half its energy.

Conclusion

Above is the solution for “A 12.0-μF capacitor is charged to a potential of 50.0V and then discharged through a 175-Ω…“. We hope that you find a good answer and gain the knowledge about this topic of science.


Related Posts:

ajax-loader