Charge & Discharge Equations (AQA A Level Physics): Revision Note

Exam code: 7408

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on

Capacitor Discharge Equation

  • The time constant is used in the exponential decay equations for the current, charge or potential difference (p.d) for a capacitor discharging through a resistor

    • These can be used to determine the amount of current, charge or p.d left after a certain amount of time for a discharging capacitor

  • This exponential decay means that no matter how much charge is initially on the plates, the amount of time it takes for that charge to halve is the same

  • The exponential decay of current on a discharging capacitor is defined by the equation:

I = I0etRC

  • Where:

    • I = current (A)

    • I0 = initial current before discharge (A)

    • e = the exponential function

    • t = time (s)

    • RC = resistance (Ω) × capacitance (F) = the time constant τ (s)

  • This equation shows that the smaller the time constant τ, the quicker the exponential decay of the current when discharging

  • Also, how big the initial current is affects the rate of discharge

    • If I0 is large, the capacitor will take longer to discharge

  • Note: during capacitor discharge, I0 is always larger than I, as the current I will always be decreasing

Capacitor Discharge Graph and Equation, downloadable AS & A Level Physics revision notes

Values of the capacitor discharge equation on a graph and circuit

  • The current at any time is directly proportional to the p.d across the capacitor and the charge across the parallel plates

  • Therefore, this equation also describes the charge on the capacitor after a certain amount of time:

Q = Q0etRC

  • Where:

    • Q = charge on the capacitor plates (C)

    • Q0 = initial charge on the capacitor plates (C)

  • As well as the p.d after a certain amount of time:

V = V0etRC

  • Where:

    • V = p.d across the capacitor (C)

    • V0 = initial p.d across the capacitor (C)

The Exponential Function e

  • The symbol e represents the exponential constant, a number which is approximately equal to e = 2.718...

  • On a calculator, it is shown by the button ex

  • The inverse function of ex is ln(y), known as the natural logarithmic function

    • This is because, if ex = y, then x = ln (y)

  • The 0.37 in the definition of the time constant arises as a result of the exponential constant, the true definition is:

The time taken for the charge of a capacitor to decrease to 1e of its original value

  • Where 1e = 0.3679

Worked Example

The initial current through a circuit with a capacitor of 620 µF is 0.6 A. The capacitor is connected across the terminals of a 450 Ω resistor.

Calculate the time taken for the current to fall to 0.4 A.

Answer:

Step 1: Write out the known quantities

  • Initial current before discharge, I0 = 0.6 A

  • Current, I = 0.4 A

  • Resistance, R = 450 Ω

  • Capacitance, C = 620 μF = 620 × 10-6 F

Step 2: Write down the equation for the exponential decay of current

I = I0etRC 

Step 3: Rearrange for t

II0 = etRC

  • The exponential is removed by taking the natural log (ln) of both sides

ln (II0) = tRC

t = RC ln (II0)

Step 4: Substitute in the values

t = 450×(620×106)×ln (0.40.6)

t = 0.1131 = 0.1 s

Examiner Tips and Tricks

The equation for Q will be given on the data sheet, however you will be expected to remember that it is similar for I and V.

Capacitor Charge Equation

  • When a capacitor is charging, the charge Q and potential difference V increase exponentially

    • Over time, they continue to increase, but at a slower rate

  • This means the equation for Q for a charging capacitor is:

Q = Q0(1  etRC)

  • Where:

    • Q = charge on the capacitor plates (C)

    • Q0 = maximum charge stored on capacitor when fully charged (C)

    • e = the exponential function

    • t = time (s)

    • RC = resistance (Ω) × capacitance (F) = the time constant τ (s)

  • Similarly, for V:

V = V0(1  etRC)

  • Where:

    • V = p.d across the capacitor (V)

    • V0 = maximum potential difference across the capacitor when fully charged (V)

  • The charging equation for the current I is the same as its discharging equation, since the current still decreases exponentially

  • The key difference with the charging equations is that Q0 and V0 are now the final (or maximum) values of Q and V that will be on the plates, rather than the initial values

Worked Example

A capacitor is to be charged to a maximum potential difference of 12 V between its plates.

Calculate how long it takes to reach a potential difference of 10 V, given that it has a time constant of 0.5 s.

Answer:

Step 1: Write down the known quantities

  • Initial p.d., V0 = 12 V

  • Final p.d., V = 10 V

  • Time constant, RC = 0.5 s

Step 2: Write down the potential difference charging equation

V = V0(1  etRC)

Step 3: Rearrange for t

VV0 = 1  etRC

  • Rearrange so the exponential is on its own

etRC = 1  VV0

  • The exponential is removed by taking the natural log (ln) of both sides

ln (etRC )= ln (1  VV0)

tRC = ln (1  VV0)

t = RC ln (1  VV0)

Step 4: Substitute in the values

t = 0.5×ln (1  1012)

t = 0.8959 = 0.9 s

Examiner Tips and Tricks

Make sure you’re confident in rearranging equations with natural logs (ln) and the exponential function (e) for both charging and discharging equations. To refresh your knowledge of this, have a look at the AS Maths revision notes on Exponentials & Logarithms.

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Katie M

Author: Katie M

Expertise: Curriculum Expert

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.