Question

In the circuit shown in Fig. E26.49, C = 5.90 mF, Ԑ = 28.0 V, and the emf has negligible resistance. Initially, the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2 so that the capacitor begins to charge. (a) What will be the charge on the capacitor a long time after S is moved to position 2? (b) After S has been in position 2 for 3.00 ms, the charge on the capacitor is measured to be 110 mC What is the value of the resistance R? (c) How long after S is moved to position 2 will the charge on the capacitor be equal to 99.0% of the final value found in part (a)?

Short answer

(a) the charge on the capacitor a long time after S is moved to position 2 is $165.2\mathrm{\mu C}$

(b) the value of the resistance R is 463 Ω

(c) the charge on the capacitor is equal to 99.0% of the final value found in part (a) is 12.6 ms.

Step-by-step solution

Concept

A capacitor is a device that stores electrical energy in an electric field. It is a passive electronic component with two terminals. The effect of a capacitor is known as capacitance.

Given data

Capacitance C = 5.90 mF

Ԑ = 28.0V

Emf has negligible resistance.

Calculate the charge on the capacitor for different positions of switches.

Calculation for charge

(a)

for the charge on the capacitor, a long time after S is moved to position 2.

The capacitor is charged by the battery in a series resistor when the switch is in position 2.

Charge on the capacitor, a long time after S moves,

$$\begin{gathered} Q_f=\left(28.0\mathrm{V}\right)\left(5.90\times {10}^{-6}\mathrm{F}\right) \\ {Q}_f=165.2\times {10}^{-6}\mathrm{C}\left(\frac{{10}^{-6}\mathrm{C}}{1\mathrm{\mu C}}\right) \\ {\mathrm{Q}}_{\mathrm{f}}=165.2\mathrm{\mu C} \end{gathered}$$

Calculation of resistance

(b)

forthe value of the resistance R.

$$\begin{gathered} q=Q_F\left(1-e^{-t/RC}\right) \\ =\frac{q}{Q_f}=1-e^{-t/RC} \end{gathered}$$ (1)

$$\begin{gathered} {\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}=1-\frac{\mathrm{q}}{{\mathrm{Q}}_{\mathrm{f}}} \\ \mathrm{Substituting}\mathrm{the}\mathrm{given}\mathrm{data}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{we}\mathrm{get}, \\ \frac{-\mathrm{t}}{\mathrm{RC}}=\mathrm{IN}\left(1-\frac{\mathrm{q}}{{\mathrm{Q}}_{\mathrm{f}}}\right) \\ \mathrm{R}=\frac{-\mathrm{t}}{\mathrm{IN}\left(1-{\displaystyle \frac{\mathrm{q}}{{\mathrm{Q}}_{\mathrm{f}}}}\right)} \\ \mathrm{R}=\frac{-\left(3.00\times {10}^{-3}\mathrm{s}\right)}{\left(5.90\times {10}^{-6}\mathrm{F}\right)\mathrm{IN}\left(1-\left({\displaystyle \frac{110\mathrm{\mu C}}{165\mathrm{\mu C}}}\right)\right)} \\ \mathrm{R}=463\mathrm{\Omega } \\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{resistance}\mathrm{R}\mathrm{is}463\mathrm{\Omega } \end{gathered}$$

Calculation of time constant

$$\begin{gathered} \frac{q}{Q_f}=0.99 \\ \mathrm{Substitute}\mathrm{these}\mathrm{values}\mathrm{in}\mathrm{equation}\left(2\right),\mathrm{and}\mathrm{we}\mathrm{get} \\ \mathrm{t}=-\left(463\mathrm{\Omega }\right)\left(5.90\times {10}^{-6}\mathrm{F}\right)\mathrm{IN}\left(1-0.99\right) \\ \mathrm{t}=0.0126\mathrm{s}\left(\frac{1\mathrm{ms}}{{10}^{-3}\mathrm{s}}\right) \\ \mathrm{t}=12.6\mathrm{ms} \\ \mathrm{thus},\mathrm{the}\mathrm{charge}\mathrm{on}\mathrm{the}\mathrm{capacitor}\mathrm{is}\mathrm{equal}\mathrm{to}99.0\%\mathrm{of}\mathrm{the}\mathrm{final}\mathrm{value}\mathrm{found}\mathrm{in}\mathrm{part}\left(\mathrm{a}\right)\mathrm{is}12.6\mathrm{ms} \end{gathered}$$(c)

for the charge on the capacitor to be equal to 99.0% of the final value found in part (a).

Time is given as,

$\mathrm{t}=-\mathrm{RC}\times \mathrm{IN}\left(1-\frac{\mathrm{q}}{{\mathrm{Q}}_{\mathrm{f}}}\right)$ (2)

for the capacitor value = 99.0% ,the value of part (a) is,