Capacitor charge and Discharge

1. Graphical representation of charging and discharging of capacitors:

  • The circuits in Figure 1 show a battery, a switch and a fixed resistor (circuit A), and then the same battery, switch and resistor in series with a capacitor (circuit B). The capacitor is initially uncharged.
  • Figure 1 Circuit diagrams for a battery, resistor and capacitor network.
  • The graphs underneath the circuit diagrams show how the current varies with time from the moment that the switches are closed.
  • In the case of the resistor alone (circuit A), the current immediately jumps to a value 1, where I = V/R  and stays at that value regardless of how long it has been since, the switch was closed.
  • Time is not a factor in this circuit. In the case of circuit B, where an initially uncharged capacitor is connected in the circuit, the current also immediately rises to the same value, I, determined by I = V/R but it then starts to decay away with time, eventually reaching zero. The series capacitor limits the way that current flows through the resistor.
  • If the capacitor is initially uncharged, the amount of charge that can be stored on it per second, [math] \frac{\Delta Q}{\Delta V} =t [/math] is initially determined by I = V/R.
  • As the capacitor starts to store charge, so a p.d. is developed across the capacitor, [math] V_c = \frac{Q}{C} [/math]
  • As the e.m.f. of the battery, ε , remains constant, then the potential difference, [math]V_R [/math] across the fixed resistor, R, reduces because
  • [math] \varepsilon = V_R + V_C [/math]
  • Reducing [math] V_R [/math] reduces the current, 1, flowing. The initial current flowing onto R the capacitor gradually decays away as the capacitor stores more charge, increasing [math] V_C [/math].

  • Figure 2 graph of Q or V and I against t, for charging and discharging capacitor

  • Graphs of V (the p.d. across the capacitor) against t follow the same pattern as the graph of Q against t, because Q ∝ V (from Q = VC).
  • When current-time graphs are plotted, you should remember that current can change direction and will flow one way on charging the capacitor and in the other direction when the capacitor is discharging.
  • The size of the current is always at a maximum immediately after the switch is closed in the charging or discharging circuit, because the charging current will be highest when the capacitor is empty of charge, and the discharging current will be highest when the capacitor is full of charge.
  • This is shown in the graphs in Figure 2.

2. Interpretation of gradients and areas under graphs:

  • Graphs of charge (Q) stored on the capacitor with time are shown in Figure 3, one representing the capacitor charging, and one discharging.
  • Figure 3 Graph of Q against t for a capacitor (a) charging and (b) discharging
  • The charging graph (Figure 3(a)) shows that, initially, the capacitor is uncharged, and the gradient of the graph, [math] \frac{\Delta Q}{ \Delta t} [/math] (equal to the current, I), is at a maximum and is determined by [math] I = \frac{\Delta V_R}{ R} [/math].
  • As more charge is stored on the capacitor, so the gradient (and therefore the current) drops, until the capacitor is fully charged and the gradient is zero.
  • As the capacitor discharges (Figure 3(b)), the amount of charge is initially at a maximum, as is the gradient (or current). The amount of charge then drops, as does the gradient of the graph. This is described by
  • [math] \frac{\Delta Q}{ \Delta t} \propto Q [/math]
  • The shape of the discharging graph is an exponential decay, meaning that the rate of decay of the charge (or the gradient or the current) depends on the amount of charge stored at any given time.
  • For a discharging capacitor, the current is directly proportional to the amount of charge stored on the capacitor at time t.

3. Time constant RC:

  • The time constant RC is the product of the resistance (R) and capacitance (C) in a circuit.
  • It represents the time it takes for a capacitor to charge or discharge by approximately 63.2% of its final value.
  • The unit of τ is seconds (s).
  • Mathematical Equation:
  • [math] \tau = R * C [/math]
  • Where:
    – [math] \tau [/math] (tau) is the time constant
    – R is the resistance
    – C is the capacitance
  • Key Points:
    – The time constant RC determines the rate of charging and discharging of a capacitor.
    – A smaller τ means faster charging and discharging, while a larger [math] \tau [/math] means slower charging and discharging.
    – The time constant RC is a critical parameter in designing and analyzing electrical circuits.
  • Applications:
    – RC circuits
    – Filters
    – Oscillators
    – Power systems
    – Electrical engineering
  • Some examples of time constants include:
    – RC = 1ms (millisecond)
    – RC = 1μs (microsecond)
    – RC = 1s (second)

4. Calculation of time constants:

  • let’s calculate the time constant of an RC circuit with a resistance of [math] 2 \, \text{k}\Omega [/math] and a capacitance of [math] 3 \, \mu\text{F} [/math]
  • Given data:
    Circuit resistance [math] = R = 2 \, \text{k}\Omega = 2 * 10^3 \, \Omega [/math]
    Capacitance [math] = C = 3 \, \mu\text{F} = 3 * 10^{-6} \, \text{F} [/math]
    Find data:
    Time constant = [math] \tau [/math] = ?
    Formula:
  • [math] \tau = RC [/math]
  • Solution:
  • [math] \tau = RC \\ \tau = (2 *10^3)(3 *10^{-6})\\ \tau = 6 *10^{-3} s \\ \tau = 6ms [/math]
  • So, the time constant of this RC circuit is 6 milliseconds.

6. Time to halve:

  • The half- life of capacitor discharge as the time taken for the charge stored on the capacitor (or the current or the voltage) to halve the half- life of capacitor discharge as the time taken for the charge stored on the capacitor (or the current or the voltage) to halve.
  • When [math] t = t_{1/2}, \quad Q = \frac{Q_0}{2} [/math], so
  • [math] \frac{Q_0}{2} = Q_0 e^{\frac{t_{1/2}}{RC}} [/math]
    [math] \text{Cancelling } Q_0 \text{ from each side gives} [/math]
    [math]\frac{1}{2} = e^{\frac{t_{1/2}}{RC}} [/math]
    [math] \text{Taking the natural logarithm of both sides:} [/math]
    [math]\ln 0.5 = -\frac{t_{1/2}}{RC} \\
    t_{1/2} = 0.69RC [/math]

6. Discharging a capacitor:

  • Consider the circuit shown in Figure 6.21.

    Figure 4  A capacitor discharge circuit

  • When switch S is closed, the capacitor C immediately charges to a maximum value given by Q = CV.
  • As switch S is opened, the capacitor starts to discharge through the resistor R and the ammeter.
  • At any time t, the p.d. V across the capacitor, the charge stored on it and the current (I), flowing through the circuit and the ammeter are all related to each other by two equations.
  • Appling Kirchhoff’s second circuit law around the capacitor-resistor loop.
  • [math] V_R + V_C = 0 [/math]
  • Where [math] V_C [/math] and [math] V_R [/math] are the p.d. across the capacitor and the resistor, respectively. Substituting for both using Q = CV and V = IR gives
  • [math] V_C = IR \\ V_R = \frac{Q}{C} \\ \frac{Q}{C} + IR = 0 [/math]
  • But
  • [math] I = \frac{\Delta Q}{\Delta t} [/math]
  • So,
  • [math] \frac{Q}{C} + IR = 0 \\ \frac{Q}{C} = -IR \\ \frac{Q}{C} = -\frac{\Delta Q}{\Delta t} R [/math]
  • Rearranging
  • [math] \Delta Q = -\frac{Q}{RC} \Delta t [/math]
  • When integrating a differential equation, such as the one here, we let , and then change to calculus notation ( and group like terms on each side.
  • [math] \frac{dQ}{Q} = -\frac{1}{RC} \, dt [/math]
  • Showing a constant ratio, and
  • [math] \int_{Q_0}^Q \frac{dQ}{Q} = – \int_0^t \frac{1}{RC} \, dt \\
    \left[ \ln Q \right]_{Q_0}^Q = – \left[ \frac{t}{RC} \right]_0^t \\
    \text{and so} \\
    \ln Q – \ln Q_0 = – \frac{t}{RC} [/math]

  • Then using the rules of logs

  • [math] \ln \frac{Q}{Q_0} = – \frac{t}{RC} [/math]

  • So, finally by exponential 

  • [math] Q = Q_0 e^{-\frac{t}{RC}} [/math]

7. Use of the corresponding equations for V and I:

[math]  Q = Q_0 e^{-\frac{t}{RC}} [/math]

  • As Q = CV and V=IR, at any time during the discharge, Q ∝ V  and V ∝ I, so there are corresponding equations for the p.d. across the capacitor and the current flowing in the circuit during discharge:
  • [math] V = V_0 e^{-\frac{t}{RC}} [/math]
  • and
  • [math] I = I_0 e^{-\frac{t}{RC}} [/math]

  • According to this equation

  • [math] \ln \frac{Q}{Q_0} = – \frac{t}{RC} [/math]
  • Consider the current-time equation:
  • [math] \ln \frac{I}{I_0} = – \frac{t}{RC} [/math]
  • So
  • [math] \ln I – \ln I_0 = – \frac{t}{RC} [/math]
    [math]\ln I = \ln I_0 – \frac{t}{RC} [/math]
  • Or
  • [math]  \ln I = – \frac{t}{RC} + \ln I_0 [/math]

  • This is the equation of a straight line of the form

  • [math] y = mx + c [/math]
  • Figure 5 Graph of the straight line with gradient form  
  • Where m is the gradient and c are the y-axis intercept of the line. Plotting Inl on the y-axis against t on the x-axis produces a gradient of [math] – \frac{t}{RC} [/math] and a y-axis intercept equal to  as determine the figure

8. Charging a capacitor:

  • A capacitor’s charging portion of a circuit is meant to be as rapid as possible, the resistance inside is kept to a minimum (Figure 6).
  • The charging time must be considered, though, if the charging procedure is a component of a circuit that needs a greater resistance.
  • Consider a circuit shown in figure 6. Kirchhoff’s second circuit law tells us that the sum of the e.m.f. in the circuit equals the sum of p.d. in the circuit, so
  • [math] V_R + V_C = V_s [/math]
  • Where [math] V_s [/math] is the e.m.f. of the source, [math] V_R [/math]  is the p.d. across the resistor and [math] V_C [/math]  is the p.d. across the capacitor. Using Ohm’s law and the basic capacitor equation, this becomes
  • [math] V_s = IR + \frac{Q}{C} [/math]
  • During a small time, interval  when the capacitor is charging, [math] V_s [/math] and C do not change, [math] \frac{\Delta V_s}{\Delta t} = 0 [/math] , unlike I and Q, which do change. So in the small time interval
  • [math]0 = \frac{\Delta I}{\Delta t} R + \frac{1}{C} \frac{\Delta Q}{\Delta t} [/math]

  • But [math] \frac{\Delta Q}{\Delta t} = I [/math], so

  • [math] 0 = \frac{\Delta I}{\Delta t} R + \frac{1}{C} [/math]
  • So,
  • [math] \frac{\Delta I}{\Delta t} R = – \frac{1}{C} \\ \frac{\Delta I}{\Delta t}  = – \frac{1}{RC} [/math]
  • [math] Rearranging and replacing [/math]
  • [math] \frac{dI}{dt} = -\frac{I}{RC} \\
    \frac{dI}{I} = -\frac{1}{RC} \, dt [/math]
  • Integrating in a similar way to before gives
  • [math] I = I_0 e^{- \frac{t}{RC}} [/math]

  • At [math] t = 0,V_C = 0 \, \text{and}\, I = I_0 = \frac{V_s}{R}, [/math] and so at any time (t),

  • [math] \text{we can write } \\
    I = \frac{V_s}{R} e^{-\frac{t}{RC}} \\
    \text{or} \\
    IR = V_s e^{-\frac{t}{RC}} \\
    V_R = V_s e^{-\frac{t}{RC}} [/math]

     

    [math]\text{According to Kirchhoff’s second circuit  law,} [/math]
    [math] V_C = V_R – V_s [/math]

    [math] \text{Substituting for } V_R \text{ gives:} [/math]
    [math] V_C = V_s – V_s e^{-\frac{t}{RC}} [/math]

    [math] \text{Factorizing this gives:} [/math]
    [math] V_C = V_s (1 – e^{-\frac{t}{RC}}) [/math]

    [math] \text{Then using the capacitor equation:} [/math]
    [math] Q = CV [/math]

  • So,[math] V = \frac{Q}{C} [/math] ,then we get

  • [math] Q = CV_s \left( 1 – e^{-\frac{t}{RC}} \right)
    [/math]
  • But [math] CV_s [/math] is the maximum charge that can be stored on the capacitor when it is fully charged. That is equal to [math] Q_0 [/math], the initial charge on the capacitor when it is about to discharge. So finally
  • [math] Q = Q_0 \left( 1 – e^{-\frac{t}{RC}} \right) [/math]
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