Question

Draw two graphs of charge versus time on a capacitor. Draw one for charging an initially uncharged capacitor in series with a resistor, as in the circuit in Figure 21.38, starting from\({\rm{t = 0}}\). Draw the other for discharging a capacitor through a resistor, as in the circuit in Figure 21.39, starting at \({\rm{t = 0}}\), with an initial charge\({{\rm{Q}}_{\rm{0}}}\). Show at least two intervals of\(\tau \).

Short answer

Charging of a capacitor, \({{\rm{Q}}_{\rm{c}}}{\rm{(t) = }}{{\rm{Q}}_{\rm{0}}}\left( {{\rm{1 - }}{{\rm{e}}^{{\rm{ - t}}/\tau }}} \right)\)

Discharging of a capacitor, \({{\rm{Q}}_{\rm{d}}}{\rm{(t) = }}{{\rm{Q}}_{\rm{0}}}{{\rm{e}}^{{\rm{ - t}}/\tau }}\)

Step-by-step solution

 Step 1: Concepts

Consider a\({\rm{RC}}\)circuit, where the capacitor is initially uncharged. When the switch will be closed, the voltage across the capacitor increases initially at a faster pace and then at a slower pace, reaching a maximum value equal to the emf of the voltage source in the circuit.

Formula for charging or discharging of a capacitor

Consider the time dependence of the charge on a capacitor while charging and discharging.

The relation between the capacitance and the voltage across a capacitor is

\({\rm{C = }}\frac{{\rm{Q}}}{{\rm{V}}}{\rm{,}}\)

so to obtain the charge of a capacitor, just multiply the formulas for the voltage by the capacitance, which is fixed.

Charging of a capacitor

For a charging capacitor, the voltage is equal to

\({\rm{V(t) = }}{{\rm{V}}_{\rm{0}}}\left( {{\rm{1 - }}{{\rm{e}}^{ - {\rm{t}}/\tau }}} \right)\)

When the capacitor is completely charged, it will receive all of the energy from the emf (if ignore energy loss on the resistances), which means that

\({{\rm{Q}}_{\rm{0}}}{\rm{ = C}}{{\rm{V}}_{\rm{0}}}{\rm{.}}\)

Therefore the formula for the charge of a charging capacitor

\({\rm{Q(t) = }}{{\rm{Q}}_{\rm{0}}}\left( {{\rm{1 - }}{{\rm{e}}^{ - {\rm{t}}/\tau }}} \right)\)

The plots of charging of a capacitor are shown in the figure below, where \({{\rm{Q}}_{\rm{0}}}{\rm{ = 1}}\)

 Step 4: Charging of a capacitor

When the capacitor is discharging, its charge depends on time as

\({\rm{Q(t) = }}{{\rm{Q}}_{\rm{0}}}{{\rm{e}}^{{\rm{ - t}}/\tau }}\)

where used the formula for the voltage of a discharging capacitor

\({\rm{V(t) = }}{{\rm{V}}_{\rm{0}}}{{\rm{e}}^{{\rm{ - t}}/\tau }}\)

The plots of discharging of a capacitor are shown in the figure below, where \({{\rm{Q}}_{\rm{0}}}{\rm{ = 1}}\)