(c) equations of the form x = x₀e^(−t/CR) and x = x₀(1 − e^(−t/CR)) for…

The idea

In a capacitor–resistor circuit, the time constant \(\tau = CR\) determines the rate of charging and discharging. For a discharging capacitor, charge \(Q\), potential difference \(V\), and current \(I\) all decay exponentially according to \(x = x_0 e^{-t/CR}\), where \(x_0\) is the initial value. During charging, the charge and potential difference rise towards a maximum steady value according to \(x = x_0(1 - e^{-t/CR})\). However, the charging current does not rise; it decays exponentially from an initial maximum as the capacitor opposes further charge flow, following \(I = I_0 e^{-t/CR}\). The exponent \(-t/CR\) must be dimensionless, requiring \(t\) and \(CR\) to be in seconds.

For example: After \(t = CR\), a discharging quantity falls to \(x_0 e^{-1} \approx 0.37 x_0\), while a charging \(V\) or \(Q\) rises to \(x_0(1 - e^{-1}) \approx 0.63 x_0\).

What earns the marks

• Select and apply the discharging equation \(x = x_0 e^{-t/CR}\) for charge, potential difference, or current.
• Select and apply the charging equation \(x = x_0(1 - e^{-t/CR})\) for charge and potential difference only.
• Recognise that current decays exponentially as \(I = I_0 e^{-t/CR}\) during both charging and discharging processes.
• Ensure the exponent \(-t/CR\) is dimensionless by using consistent units of seconds for both \(t\) and \(CR\).

Common mistakes and how to put them right

• Assuming current rises during charging.

  Put it right: Current always decays exponentially in both charging and discharging circuits, following \(I = I_0 e^{-t/CR}\).

• Using the charging equation for a discharging process or vice versa.

  Put it right: Use \(x = x_0 e^{-t/CR}\) for decreasing quantities and \(x = x_0(1 - e^{-t/CR})\) for increasing quantities.

• Treating \(x_0\) as the final value in the discharging equation.

  Put it right: In \(x = x_0 e^{-t/CR}\), \(x_0\) is the initial value at \(t = 0\).

Exam tips

• Always check which quantity is being asked for during charging; use the \((1 - e^{-t/CR})\) form for \(Q\) and \(V\), but the \(e^{-t/CR}\) form for \(I\).
• To find \(\tau\) from a graph, plot \(\ln x\) against \(t\) for a decaying quantity; the gradient is \(-1/CR\).
• Remember that after one time constant, a decaying quantity is about 37% of its initial value, and a rising quantity is about 63% of its final value.