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

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

## Answer

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.

The relationship between voltage, charge, and capacitor is.

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

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

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:

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

Substitute

to C and

for V in expression

.

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

.

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

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

.

Substitute HTMLmedia_tag_24$ to HTMLmedia_tag_25$ and HTMLmedia_tag_26$.

Substitute

to replace

or

to replace

within expression

.

Or

(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:

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.

to *U* with expression #media_tag_41$.

Natural log both ends of the equation

.

For time *t*, rewrite the equation

.

Substitute

to

, and

to

within equation

.

Or

Ans: Part A

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

Part b

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**.