# Finding Current after 50s Capacitors

• Googl
In summary, the conversation is about a question that involves calculating the current after 50 seconds using the time constant formula. The first part of the question is solved correctly, but the second part is giving incorrect results. The person asking for help is trying to find the current after 50 seconds but their calculations are wrong. Some pointers are given to help them find their error, such as using quick estimates and using the number e as an exponential base. It is also pointed out that the unit for microamps is 10^-6 A, not 10^-3 A.
Googl

## Homework Statement

I have managed to calculate the first part of the question but the second part b I keep getting wrong. I was wondering whether you might give me some guidance. I am trying to find the current after 50s in part b below.

## Homework Equations

time constant = RC

## The Attempt at a Solution

a) Time constant = RC = (50 x 10-6 F) x (100, 000Ω)
Time constant = (5 x 10-5 F) x (100, 000Ω)
Time constant = 5s Correct

b) T1/2 = 5 x In 2 = 3.4657...
50 seconds corresponds to 50/3.4657 = 14.41 half-times.
Δ changing current 14.41t1/2 = 90µA/2^14.41 = 4.13 x 10^-3 µA = 4.13 x 10^-6A
Current after 50 seconds is 4.13 x 10^-6A

This is wrong. Help. Thanks.

Your setup looks okay to me, but the arithmetic in your last step is wrong. Try the last calculation again.

A couple of miscellaneous pointers:

1. A handy number to remember is that 10 half-lives corresponds to a decrease of about 1000 (actually 1024). So after more than 14 half-lives, we expect a decrease by a factor of about 10,000. Thus we can see that your answer is much too large; it's about 1/20 of the initial current, not 1/10,000! Quick estimates like this are very useful for finding calculation mistakes.

2. To cut down on calculation, you can use the number e as an exponential base instead of 2. The time-constant RC is actually the "1/e-life" (as opposed to the 1/2-life). So once you calculate RC, you don't need to translate into half-lives, you can just get the answer immediately. Numerically, e = 2.718... but your calculator may have a button labeled "e^"

In this case, RC = 5, so we have,

I (50s) = (90)e^(-50/5) = (90)e^(-10)

And you're done.

50/3.4657 is 14.427

Thanks a lot.

I cannot believe I missed that.

Googl said:
Thanks a lot.

I cannot believe I missed that.

Also, a µA is 10^(-6) A, not 10^(-3) A

## 1. What is a capacitor?

A capacitor is an electronic component that stores electrical charge. It is made of two conductive plates separated by an insulating material, also known as a dielectric. When a voltage is applied to the capacitor, it charges up and stores energy.

## 2. How do capacitors affect current flow?

Capacitors have the ability to store and release electrical energy, which can affect the flow of current in a circuit. When a capacitor is fully charged, it blocks the flow of current, but when it is discharged, it allows current to flow through. Therefore, the presence of capacitors can impact the amount and timing of current in a circuit.

## 3. What happens to capacitors after 50 seconds?

After 50 seconds, the charge on a capacitor will reach its maximum capacity. This means that the capacitor will no longer be able to store any more energy and will behave like an open circuit, blocking the flow of current. However, the capacitor will still hold its charge until it is discharged or the power source is removed.

## 4. How can I find the current after 50 seconds in a capacitor?

To find the current after 50 seconds in a capacitor, you will need to know the capacitance and the voltage of the capacitor, as well as the resistance of the circuit it is connected to. Using Ohm's Law (I = V/R) and the formula for capacitive reactance (Xc = 1/(2πfC)), you can calculate the current after 50 seconds using the values of the capacitor and the circuit. Alternatively, you can use a multimeter to directly measure the current in the circuit.

## 5. Can capacitors lose their charge over time?

Yes, capacitors can lose their charge over time due to leakage current and parasitic capacitance. Leakage current is the flow of current through the dielectric material of the capacitor, which can slowly discharge the capacitor. Parasitic capacitance refers to the unintentional capacitance that exists between the conductive plates of a capacitor and its surroundings, which can also cause the capacitor to discharge. However, high-quality capacitors have low leakage current and parasitic capacitance, so they can hold their charge for longer periods of time.

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