What is the current at the instant a capacitor starts to charge from a supply of p.d. V through a resistor R?

V ÷ R. The empty capacitor has no p.d. across it, so the resistor has all of it.

What happens to the current as a capacitor charges?

It falls. The p.d. across the resistor falls as the capacitor fills.

What is the current when a capacitor is fully charged?

zero. The p.d. across the capacitor equals the supply p.d., so nothing drives a current.

A capacitor is charged through a bigger resistor than before. How does the charging change?

It takes longer. The time constant RC is bigger.

Charging a capacitor · Edexcel IAL Physics WPH revision

In plain words

Filling a capacitor through a resistor is like filling a bucket from a tap that slows down as the bucket fills. At first charge rushes on. As the capacitor's voltage climbs towards the supply's there is less and less push, and the flow dies away.

5 things to know

At the start the capacitor is empty, so the whole supply p.d. is across the resistor and the current is at its greatest: I₀ = V ÷ R.
As charge builds up, the p.d. across the capacitor rises and the p.d. across the resistor falls, so the current falls. It falls exponentially: I = I₀e^(−t/RC).
The charge and the p.d. across the capacitor rise quickly at first and then level off, at Q₀ = CV and at the supply p.d. After one time constant, RC, they have reached about 63% of their final values.
At every moment, the p.d. across the capacitor plus the p.d. across the resistor equals the supply p.d.
The curves can be shown by connecting an oscilloscope or a data logger across the capacitor.

Worked example

A 220 µF capacitor is charged from a 9.0 V supply through a 10 kΩ resistor. Find the current at the start, the time constant and the final charge.

Current at the start = V ÷ R = 9.0 ÷ 10 × 10³ = 9.0 × 10⁻⁴ A (0.90 mA).
Time constant = RC = 10 × 10³ × 220 × 10⁻⁶ = 2.2 s.
Final charge = CV = 220 × 10⁻⁶ × 9.0 = 2.0 × 10⁻³ C.

Tips and tricks

While charging, the current falls and the charge rises, both with the same time constant.
After one time constant a charging capacitor has reached 63% of its final p.d., not 37%.

5 questions, about 2 minutes.

It lands in your notebook with its questions as flashcards.

Charging a capacitor: 5 questions and answers

These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.

What is the current at the instant a capacitor starts to charge from a supply of p.d. V through a resistor R?
- V ÷ R (the answer)
- zero
- VR
- V ÷ C
The empty capacitor has no p.d. across it, so the resistor has all of it.
What happens to the current as a capacitor charges?
- It falls. (the answer)
- It rises.
- It stays the same.
- It reverses.
The p.d. across the resistor falls as the capacitor fills.
After one time constant, what fraction of the supply p.d. is across a charging capacitor?
- 37%
- 50%
- 63% (the answer)
- 100%
1 − 1/e is 0.63.
What is the current when a capacitor is fully charged?
- zero (the answer)
- V ÷ R
- its greatest value
- half its starting value
The p.d. across the capacitor equals the supply p.d., so nothing drives a current.
A capacitor is charged through a bigger resistor than before. How does the charging change?
- It takes longer. (the answer)
- It is quicker.
- It takes the same time.
- It never finishes charging at all.
The time constant RC is bigger.

5 questions

Tap an answer and you’ll see straight away whether it’s right, and why.

Worksheet

4 questions, 10 marks. Write your answers on paper, then check them.

Charging a capacitor

Edexcel International A Level Physics WPH · 10 marks · papermunch.org

Name ______________________________ Date ______________

A capacitor is charged from a 12 V supply through a resistor. Find the p.d. across it after one time constant.[2]
Show answerHide answer
7.6 V. 0.63 × 12.
I got this one
Describe how the p.d. across a capacitor changes with time while it charges through a resistor.[2]
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It rises quickly at first, then more and more slowly, levelling off at the supply p.d.
I got this one
A capacitor is being charged from a 6.0 V supply through a 3.0 kΩ resistor. At one moment the p.d. across the capacitor is 4.5 V. Find the current at that moment.[3]
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0.50 mA. The p.d. across the resistor is 6.0 − 4.5 = 1.5 V, and 1.5 ÷ 3000 = 5.0 × 10⁻⁴ A.
I got this one
Describe how to display and analyse the charging of a capacitor.[3]
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Connect the capacitor and a resistor in series with a d.c. supply and a switch, with a data logger (or an oscilloscope) across the capacitor. Close the switch and record the p.d. against time. Read off the time for the p.d. to reach 63% of the supply p.d.: that is the time constant, which should equal RC.
I got this one

Answers: Charging a capacitor

1. 7.6 V. 0.63 × 12.
2. It rises quickly at first, then more and more slowly, levelling off at the supply p.d.
3. 0.50 mA. The p.d. across the resistor is 6.0 − 4.5 = 1.5 V, and 1.5 ÷ 3000 = 5.0 × 10⁻⁴ A.
4. Connect the capacitor and a resistor in series with a d.c. supply and a switch, with a data logger (or an oscilloscope) across the capacitor. Close the switch and record the p.d. against time. Read off the time for the p.d. to reach 63% of the supply p.d.: that is the time constant, which should equal RC.