# Time it takes for a capacitor to lose half of its energy

• sliperyfrog
In summary, in the given circuit, the switch is opened and the capacitor loses half of its initial stored energy. The time elapsed before this happens can be calculated using the equation t = RCln(2)/2, where R is the equivalent resistance of the circuit and C is the capacitance. The parallel resistors in the circuit have an equivalent resistance of (1/90 + 1/90)-1 = 45Ω. Therefore, the time elapsed is approximately .0031 seconds.
sliperyfrog

## Homework Statement

In the circuit below (drew it the best I can) the switch is opened, how much time elapses before the capacitor loses half of its initial stored energy.

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q(t) = Qe-t/RC

U = q2/2C

## The Attempt at a Solution

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So first thing I did was I set (1/2)Q2/2C = q2/2C then i replaced q2 for Q2e-2t/RC so I get (1/2)Q2/2C = Q2e-2t/RC/2C I cancel out the like terms and I am left with 1/2 = e-2t/RC I than get rid of the e so I get ln(1/2) =-2t/RC then i get -ln(2) = -2t/RC getting me t = RCln(2)/2 and now this is were i got confused I know C = 200E-6 F but i don't know what R is suppose to be, I think it is suppose to be 90Ω since the 30Ω and the 60Ω resistors connect together at where the capacitor is.

#### Attachments

• Circuit.jpg
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Draw a separate circuit diagram for discharging without the 90V battery. In which fashion are the resistors connected?

Okay so i redrew it and i found out it was parallel so I did (1/60 + 1/30)-1 = 20Ω but now i am confused on whether the R in the equation is 20Ω or 40Ω since the current goes through the first 20Ω then the capacitor then another 20Ω.

#### Attachments

• Circuit 2.jpg
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sliperyfrog said:
Okay so i redrew it
That's not correct. Why have you drawn the external loop joining the two junctions? The discharging circuit will contain only the 4 resistors and the capacitor.

cnh1995 said:
Why have you drawn the external loop joining the two junctions?

I was thinking that since that loop that connects the battery was in the original circuit that the junction that connect to the battery would still be there.

sliperyfrog said:
I was thinking that since that loop that connects the battery was in the original circuit that the junction that connect to the battery would still be there.
No. The switch is "opened" and the loop is disconnected from the circuit.

So then the circuit would look like this
And now I am back to the problem i was struggling with before is the two 90Ω resistors in parallel or is the capacitor in the the middle stopping it?

sliperyfrog said:
So then the circuit would look like this View attachment 110280 And now I am back to the problem i was struggling with before is the two 90Ω resistors in parallel or is the capacitor in the the middle stopping it?
They are in parallel. In fact, the capacitor too is in parallel with them. Review the theory of series and parallel components to avoid any further confusion.

sliperyfrog
So since they are in parallel that means the R in the problem is (1/90 + 1/90)-1 = 45Ω so for the answer I would get t = (45)(200E-6)ln(2)/2 = .0031s

sliperyfrog said:
So since they are in parallel that means the R in the problem is (1/90 + 1/90)-1 = 45Ω so for the answer I would get t = (45)(200E-6)ln(2)/2 = .0031s
Looks good.

sliperyfrog

## What is a capacitor?

A capacitor is an electronic component that stores electrical energy in an electric field. It is commonly used in electronic circuits to temporarily store and release energy.

## What factors affect the time it takes for a capacitor to lose half of its energy?

The time it takes for a capacitor to lose half of its energy, also known as the half-life, is affected by several factors including the capacitance value, the resistance of the circuit, and the voltage across the capacitor.

## How do you calculate the half-life of a capacitor?

The half-life of a capacitor can be calculated using the formula t = RCln(2), where t is the time in seconds, R is the resistance in ohms, and C is the capacitance in farads.

## How does the type of capacitor affect its half-life?

The type of capacitor, such as ceramic, electrolytic, or film, can affect the half-life due to differences in material properties and construction. For example, electrolytic capacitors typically have a shorter half-life compared to ceramic capacitors.

## Can the time it takes for a capacitor to lose half of its energy be increased?

Yes, the time it takes for a capacitor to lose half of its energy can be increased by increasing the capacitance value or decreasing the resistance in the circuit. Additionally, using a capacitor with a higher voltage rating can also increase the half-life.

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