# Capacitor Charge and Discharge Equations ## The Time Constant - The time constant of a capacitor discharging through a resistor measures how long it takes for the capacitor to discharge - It's a characteristic property of the RC circuit that determines the rate of charge or discharge ### Definition for Discharging - **The time taken for the charge, current or voltage of a discharging capacitor to decrease to 37% of its original value** ### Definition for Charging - **The time taken for the charge or voltage of a charging capacitor to rise to 63% of its maximum value**

τ = R C, where τ is the time constant (s), R is resistance (Ω), and C is capacitance (F).

- The 37% value comes from $\frac{1}{e} \approx 0.37$ where $e$ is the exponential constant - When $t = \tau$, the quantity has decreased to $\frac{1}{e}$ of its original value ($I_0$, $Q_0$ or $V_0$) - The time constant provides an easy way to compare the rate of change of similar quantities (charge, current, p.d.) ### Time to Half (Half-Life) - **The time taken for the charge, current or voltage of a discharging capacitor to reach half of its initial value** - Can be expressed in terms of the time constant: $$t_{1/2} = \ln(2) \tau \approx 0.69 \tau = 0.69RC$$ - Think of it like a battery that loses half its power every fixed interval—except here it's a precise mathematical relationship - For a capacitor discharging through a smartphone circuit, if $\tau = 2$ s, then $t_{1/2} \approx 1.4$ s ## Capacitor Discharge Equations - The time constant is used in exponential decay equations for current, charge, or potential difference for a capacitor discharging through a resistor - These equations determine the **amount remaining** after a certain time ### Current Decay $$I = I_0e^{-\frac{t}{RC}}$$ **Where:** - $I$ = current (A) - $I_0$ = initial current before discharge (A) - $e$ = exponential function - $t$ = time (s) - $RC$ = resistance (Ω) × capacitance (F) = time constant $\tau$ (s) - The smaller the time constant $\tau$, the quicker the exponential decay of current - If $I_0$ is large, the capacitor will take longer to discharge - **Note:** During discharge, $I_0$ is always larger than $I$ because current is always decreasing ### Charge Decay $$Q = Q_0e^{-\frac{t}{RC}}$$ **Where:** - $Q$ = charge on the capacitor plates (C) - $Q_0$ = initial charge on the capacitor plates (C) ### Voltage Decay $$V = V_0e^{-\frac{t}{RC}}$$ **Where:** - $V$ = p.d. across the capacitor (V) - $V_0$ = initial p.d. across the capacitor (V)

Current, p.d., and charge are directly proportional during discharge, so they follow identical exponential decay patterns

### Exponential Decay Characteristics - No matter how much charge is initially on the plates, the time it takes for that charge to halve is the same - This is the **constant-ratio property** of exponential decay - Like pouring water through a funnel—the flow rate depends on how much water remains above | Time | Fraction Remaining | |------|-------------------| | $0$ | $1.00$ (100%) | | $\tau$ | $0.37$ (37%) | | $2\tau$ | $0.14$ (14%) | | $3\tau$ | $0.05$ (5%) | | $4\tau$ | $0.02$ (2%) | | $5\tau$ | $0.007$ (0.7%) | The chart below shows how voltage (or charge or current) decays exponentially during capacitor discharge:

## Capacitor Charge Equations - When a capacitor is charging, the charge $Q$ and potential difference $V$ increase but still follow exponential patterns - Over time, they continue to increase but at a slower rate (approaching maximum values asymptotically) ### Charge Growth $$Q = Q_0(1 - e^{-\frac{t}{RC}})$$ **Where:** - $Q$ = charge on the capacitor plates (C) - $Q_0$ = maximum charge stored on capacitor when fully charged (C) - $e$ = exponential function - $t$ = time (s) - $RC$ = time constant $\tau$ (s) ### Voltage Growth $$V = V_0(1 - e^{-\frac{t}{RC}})$$ **Where:** - $V$ = p.d. across the capacitor (V) - $V_0$ = maximum potential difference across the capacitor when fully charged (V) The chart below shows how voltage (or charge) increases exponentially during capacitor charging, approaching the maximum value asymptotically:

### Current During Charging $$I = I_0e^{-\frac{t}{RC}}$$ - The charging equation for current $I$ is **the same** as its discharging equation - Current still decreases exponentially because the rate of charge flow decreases as the capacitor fills - Imagine filling a pressurised tank—the flow rate decreases as back-pressure builds up

For charging equations, Q₀ and V₀ represent final (maximum) values, not initial values like in discharge equations.

**Key difference between charging and discharging:** - **Discharging:** $Q_0$ and $V_0$ are initial values at $t = 0$ - **Charging:** $Q_0$ and $V_0$ are final (maximum) values when fully charged ## Worked Example: Calculating Resistance from Time Constant **Question:** A capacitor of 7 nF is discharged through a resistor of resistance $R$. The time constant of the discharge is $5.6 \times 10^{-3}$ s. Calculate the value of $R$. **Step 1: Write the known quantities** Capacitance, $C = 7$ nF $= 7 \times 10^{-9}$ F Time constant, $\tau = 5.6 \times 10^{-3}$ s **Step 2: Write down the time constant equation** $$\tau = RC$$ **Step 3: Rearrange for R** $$R = \frac{\tau}{C}$$ **Step 4: Substitute values and calculate** $$R = \frac{5.6 \times 10^{-3}}{7 \times 10^{-9}} = 800 \text{ kΩ}$$ ## Worked Example: Discharge Time Calculation **Question:** 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. **Step 1: Write out the known quantities** Initial current before discharge, $I_0 = 0.6$ A Current, $I = 0.4$ A Resistance, $R = 450$ Ω Capacitance, $C = 620$ µF $= 620 \times 10^{-6}$ F **Step 2: Write down the equation for the exponential decay of current** $$I = I_0e^{-\frac{t}{RC}}$$ **Step 3: Rearrange for t** $$\frac{I}{I_0} = e^{-\frac{t}{RC}}$$ The exponential is removed by taking the natural log (ln) of both sides: $$\ln\left(\frac{I}{I_0}\right) = -\frac{t}{RC}$$ $$t = -RC\ln\left(\frac{I}{I_0}\right)$$ **Step 4: Substitute in the values** $$t = -450 \times (620 \times 10^{-6}) \times \ln\left(\frac{0.4}{0.6}\right) = 0.1131 = 0.1 \text{ s}$$ ## Worked Example: Charging Time Calculation **Question:** 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. **Step 1: Write down the known values** Final potential difference, $V_0 = 12$ A Potential difference, $V = 10$ V Time constant, $RC = 0.5$ s **Step 2: Write down the potential difference charging equation** $$V = V_0(1 - e^{-\frac{t}{RC}})$$ **Step 3: Rearrange for t** $$\frac{V}{V_0} = 1 - e^{-\frac{t}{RC}}$$ Rearrange so the exponential is on its own on the right-hand side: $$\frac{V}{V_0} - 1 = -e^{-\frac{t}{RC}}$$ Divide by -1 on both sides: $$1 - \frac{V}{V_0} = e^{-\frac{t}{RC}}$$ The exponential is removed by taking the natural log (ln) of both sides: $$\ln\left(1 - \frac{V}{V_0}\right) = \ln(e^{-\frac{t}{RC}})$$ $$\ln\left(1 - \frac{V}{V_0}\right) = -\frac{t}{RC}$$ $$t = -RC\ln\left(1 - \frac{V}{V_0}\right)$$ **Step 4: Substitute in the values** $$t = -(0.5) \times \ln\left(1 - \frac{10}{12}\right) = 0.89588 = 0.9 \text{ s}$$ ## Using the Exponential Constant - Knowledge of the exponential constant $e \approx 2.718$ is crucial for mastering charge and discharge equations - On a calculator, it's shown by the button $e^x$ - The inverse function of $e^x$ is $\ln(y)$, known as the **natural logarithmic function** - This is because if $e^x = y$, then $x = \ln(y)$ **Calculating the Time Constant from 37%:** - The 0.37 in the definition of time constant arises from the exponential constant - True definition: Time constant = The time taken for charge of a capacitor to decrease to $\frac{1}{e}$ of its original value - Where $\frac{1}{e} = 0.3678 \approx 0.37$