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19.3 Discharging a capacitor

Syllabus
9702–2028–2029
Topic
19.3
Level
A2

Learning objectives

19.3.1The variation with time of potential difference, charge and current• analyse graphs of the variation with time of potential difference, charge and current for a capacitor discharging through a resistor19.3.2Τ = RC for the time constant for a capacitor discharging through a resistor• recall and use τ = RC for the time constant for a capacitor discharging through a resistor19.3.3Equations of the form x = x0 e–(t / RC) where x could represent current• use equations of the form x = x0 e–(t / RC) where x could represent current, charge or potential difference for a capacitor discharging through a resistor

Voltage, charge and current magnitudes decay exponentially during capacitor discharge

Qfalls→V=Q/Cfalls→currentmagnitudeI=V/Rfalls→∣dQ/dt∣fallsSoequaltimeintervalsremoveequalfractions,notequalamounts.Q falls → V=Q/C falls → current magnitude I=V/R falls → |dQ/dt| falls So equal time intervals remove equal fractions, not equal amounts.

Quantity against time Initial value Shape and final behaviour
charge Q Q0 steepest fall initially; exponential curve asymptotic to 0
p.d. V V0=Q0/C same fractional exponential decay as Q; asymptotic to 0
current magnitude I I0=V0/R maximum initially; same fractional decay; asymptotic to 0

If current is defined positive in the charging direction, discharge current is negative and rises toward zero from -I0. If the graph shows current magnitude, it starts at +I0 and falls toward zero. State the convention.

Each decay graph has a negative gradient whose magnitude decreases with time. The initial tangent is steepest because V and I are greatest at t=0; the curve never reaches zero at a finite ideal time.

ln(Q/Q0)=ln(V/V0)=ln(I/I0)=−t/(RC):alog−ratioagainsttgraphisastraightlinethroughtheoriginwithgradient−1/(RC).ln(Q/Q0)=ln(V/V0)=ln(I/I0)=-t/(RC): a log-ratio against t graph is a straight line through the origin with gradient -1/(RC).

Do not sketch a straight-line discharge or a non-zero final plateau for an ideal RC circuit. Q and V are proportional, while I is the rate at which Q changes—not a constant.

The RC time constant τ=RC sets the timescale of capacitor charging or discharging

For a resistor R and capacitor C, τ=RC. After one time constant in a discharge, the relevant quantity has fallen to e⁻¹≈37% of its initial value.

Larger R or C makes the response slower. Identify the effective resistance seen by the capacitor, not every resistor in the diagram.

R=2.0 kΩ and C=100 μF gives τ=0.20 s.

τ is not the time to reach exactly zero; exponential decay approaches zero asymptotically.

Use exponential equations for capacitor discharge current, charge or voltage

Quantity Discharge equation Initial value
charge Q=Q0e^(-t/RC) Q0=CV0
p.d. V=V0e^(-t/RC) V0=Q0/C
current magnitude I=I0e^(-t/RC) I0=V0/R

At t=0, e^0=1 so x=x0. At t=RC, x/x0=e^-1=0.368. As t becomes large, x approaches zero. These limits quickly expose a wrong sign or charging formula.

x/x0=e(−t/RC)ln(x/x0)=−t/(RC)t=−RCln(x/x0)x/x0=e^(-t/RC) ln(x/x0)=-t/(RC) t=-RC ln(x/x0)

For τ=RC=3.6 s, the time for current to fall to 15% is t=-(3.6)ln(0.15)=6.83 s.

lnx=lnx0−t/(RC):agraphoflnxagainsttisstraight,withinterceptlnx0andgradient−1/(RC).HalvingRdoublesthegradientmagnitude.ln x = ln x0 - t/(RC): a graph of ln x against t is straight, with intercept ln x0 and gradient -1/(RC). Halving R doubles the gradient magnitude.

Use consistent time units so t/(RC) is dimensionless. These equations describe discharge; do not substitute the charging form x0(1-e^(-t/RC)). For signed current, attach the direction sign separately from the decaying magnitude.