Question

In an RCseries circuit, emf $\mathbf{\epsilon }\mathbf{=}\mathbf{12}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{V}\mathbf{}$, resistance $\mathbf{R}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{M\Omega }$, and capacitance $\mathbf{C}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{\mu F}$. (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to $\mathbf{16}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\mu C}$?

Short answer

1. The time constant is $2.52\mathrm{s}$.
2. The maximum charge that will appear on the capacitor during charging of the capacitor is $21.6\mathrm{\mu C}$.
3. It takes $3.40\mathrm{s}$a long for the charge to build up to $16\mathrm{\mu C}$.

Step-by-step solution

The given data

Emf of the RC circuit,$\mathrm{\epsilon }=12\mathrm{V}$

The resistance value is given,$\mathrm{R}=1.40\mathrm{M\Omega }$

The given capacitance value, $\mathrm{C}=1.80\mathrm{\mu F}$

Understanding the concept of time constant

The RC time constant of an RC circuit is equal to the product of the circuit resistance (in ohms) and the circuit capacitance. Using the given relation of the voltage, the time constant of the RC circuit can be calculated. As the time approaches infinity, the exponential goes to zero, and the charge approaches the maximum charge.

Formulae:

The voltage equation for the resistor and capacitor connection,$\mathrm{V}={\mathrm{V}}_{\mathrm{o}}{\mathrm{e}}^{-\mathrm{t}/\mathrm{RC}}$ (1)

The charge between the capacitor plates, $\mathrm{Q}=\mathrm{CV}$ (2)

The time constant of the RC circuit, $\mathrm{t}=\mathrm{RC}$ (3)

a) Calculation of the time constant

Using the given data in equation (3), the time constant of the RC circuit can be given as follows:

$\begin{array}{rcl}\mathrm{t}& =& \left(1.40\times {10}^6\mathrm{\Omega }\right)\left(1.80\times {10}^{-6}\mathrm{F}\right)\\ & =& 2.52\mathrm{s}\end{array}$

Hence, the time constant is 2.52 s.

b) Calculation of the maximum charge

Using the given data in equation (2), the maximum charge that will appear on the capacitor during charging can be given as follows:

$\begin{array}{rcl}\mathrm{Q}& =& \left(1.80\times {10}^{-6}\mathrm{F}\right)\left(12.0\mathrm{V}\right)\\ & =& 21.6\mathrm{\mu C}\end{array}$

Hence, the value of the maximum charge is $21.6\mathrm{\mu C}$.

c) Calculation of the time taken for the given charge to build up

Now, for the given charge ${\mathrm{Q}}^{\text{'}}=16\mathrm{\mu C}$to accumulate, the time taken for the given charge to build between the capacitor plates can be given using equation (1) and equation (2) as follows:

$\begin{array}{rcl}{\mathrm{Q}}^{\text{'}}& =& \mathrm{Q}\left(1-{\mathrm{e}}^{\mathrm{t}/\mathrm{RC}}\right)\\ \mathrm{t}& =& \mathrm{RCln}\left(\frac{\mathrm{Q}}{\mathrm{Q}-{\mathrm{Q}}^{\text{'}}}\right)\\ \mathrm{t}& =& \left(2.52\mathrm{s}\right)\ln \left(\frac{21.6\mathrm{\mu C}}{21.6\mathrm{\mu C}-16\mathrm{\mu C}}\right)\\ \mathrm{t}& =& 3.40\mathrm{s}\end{array}$

Hence, the value of time is 3.40 s.