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

Exam code: H556

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

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The Time Constant

  • The time constant of a capacitor discharging through a resistor is a measure of how long it takes for the capacitor to discharge

  • The definition of the time constant is:

The time taken for the charge, current or voltage of a discharging capacitor to decrease to 37% of its original value

  • Alternatively, for a charging capacitor:

The time taken for the charge or voltage of a charging capacitor to rise to 63% of its maximum value

  • 37% is 0.37 or 1e (where e is the exponential function) multiplied by the original value (I0, Q0 or V0)

    • This is represented by the Greek letter tau, τ, and measured in units of seconds (s)

  • The time constant provides an easy way to compare the rate of change of similar quantities e.g. charge, current and p.d.

  • It is defined by the equation:

τ = RC

  • Where:

    • τ = time constant (s)

    • R = resistance of the resistor (Ω)

    • C = capacitance of the capacitor (F)

  • The time to half, t1/2 (half-life) for a discharging capacitor is:

The time taken for the charge, current or voltage of a discharging capacitor to reach half of its initial value

  • This can also be written in terms of the time constant, τ:

t1/2 = ln(2) τ ≈ 0.69 τ = 0.69RC

Worked Example

A capacitor of 7 nF is discharged through a resistor of resistance, R. The time constant of the discharge is 5.6 × 10−3 s.

Calculate the value of R.

Answer:

Step 1: Write out the known quantities

  • Capacitance, C = 7 nF = 7 × 10-9 F

  • Time constant, τ = 5.6 × 10-3 s

Step 2: Write down the time constant equation

τ = RC

Step 3: Rearrange for resistance R

R = τC

Step 4: Substitute in values and calculate R

R = 5.6 × 10−37 × 10−9 = 8 × 105 Ω = 800 kΩ

Examiner Tips and Tricks

Note that the time constant is not the same as half-life. Half-life is how long it takes for the current, charge or voltage to halve whilst the time constant is to 37% of its original value (not 50%).

Although the time constant is given on the datasheet, you will be expected to remember the half-life equation t1/2 = 0.69RC

Charging & Discharging Equations

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

  • These equations can be used to determine:

    • The amount of current, charge or p.d. gained after a certain amount of time for a charging capacitor

    • The amount of current, charge or p.d. remaining after a certain amount of time for a discharging capacitor

Capacitor Discharge Equations

  • 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=I0e−tRC

  • 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

  • The time taken to fall to a given fraction of I0 depends only on RC, not on the size of I0

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

A circuit of a capacitor C connected across a resistor R, next to a graph of current I (A) against time t (s) for the discharging capacitor. The current falls exponentially from I₀, following I = I₀e^(−t/RC); dashed lines show that the current is I at time t.

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=Q0e−tRC

  • 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=V0e−tRC

  • Where:

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

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

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 = I0e(−tRC) 

Step 3: Calculate the time constant

τ = RC

τ = 450 × (620 × 10-6) = 0.279 s

Step 4: Substitute into the current equation

0.4 = 0.6 × e−t0.279

Step 5: Rearrange for the time t

0.40.6 = e−t0.279

  • The exponential can be removed by taking the natural log of both sides:

ln(0.40.6) = −t0.279

t = −0.279 × ln(0.40.6) = 0.1131 = 0.1 s

Capacitor Charge Equations

  • When a capacitor is charging, the way the charge Q and potential difference V increase still shows exponential decay

    • 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−e−tRC)

  • 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−e−tRC)

  • 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 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 − e−tRC)

Step 3: Rearrange for t

VV0 = 1 − e−tRC

  • Rearrange so the exponential is on its own

e−tRC = 1 − VV0

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

ln (e−tRC)= 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 = 0.90 s

Examiner Tips and Tricks

Knowledge of the exponential constant, a number which is approximately equal to e = 2.718..., is crucial for mastering charge and discharge equations, so make sure you know how to use it:

  • 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, so make sure you know that the true definition is:

Time constant = The time taken for the charge of a capacitor to decrease to 1eof its original value

Where 1e= 0.3678

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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.