Question

A capacitor is discharged through a \(30{\Omega}\) resistor. The discharge current decreases to27 \({\%}\) of its initial value in \(1.7 \mathrm{~ms}\). What is the value of the capacitor in \(\mu F ?\) Express your answer using two significant figures.

Answer 1

Concepts and reason The concept used to solve this problem discharging a capacitor in an RC circuit. Initially, the instantaneous current can be calculated by using the percentage of the discharged current of the capacitor. Later, the capacitance of the capacitor can be calculated by using the formula for the current for a discharging capacitor circuit.

Fundamentals

The expression for the discharge of current through the capacitor is \(i(t)=i_{0} e^{-t / R C}\)

Here, \(i_{0}\) is the current value, \(t\) is the instantaneous time, \(R\) is the resistance connected in the circuit, and \(C\) is the capacitance connected. The current discharge to \(27 \%\) of its initial value, \(i(t)=(27 \%) i_{0}\)

The expression for the instantaneous current is

$$ i(t)=(27 \%) i_{0} $$

Find the value of the instantaneous current,

$$ \begin{aligned} i(t) &=(27 \%) i_{0} \\ &=\left(\frac{27}{100}\right) i_{0} \\ &=0.27 i_{0} \end{aligned} $$

The value of the instantaneous current is calculated by determining the percentage of the discharged current of the

capacitor circuit. The instantaneous current after a certain time is \(27 \%\) of the maximum current, i.e., \(0.27 i_{0}\).

The expression for the discharge of current through the capacitor is \(i(t)=i_{0} e^{-t / R C}\)

Substitute \(0.27 i_{0}\) for \(i(t)\)

\(0.27 i_{0}=i_{0} e^{-t / R C}\)

\(0.27=e^{-t / R C}\)

Take a natural log on either side to simplify.

$$ \begin{aligned} \ln (0.27) &=\frac{-t}{R C} \\ C &=\frac{-t}{\ln (0.27) R} \end{aligned} $$

Substitute \(30 \Omega\) for \(\mathrm{R}\) and \(1.7 \mathrm{~ms}\) for \(\mathrm{t}\) in the above expression.

$$ \begin{aligned} C &=\frac{-\left[(1.7 \mathrm{~ms})\left(\frac{1 \times 10^{-3} \mathrm{~s}}{1 \mathrm{~ms}}\right)\right]}{[\ln (0.27)](30 \Omega)} \\ &=43 \times 10^{-6} \mathrm{~F} \\ &=43 \mu \mathrm{F} \end{aligned} $$

As time passes, the current in the capacitor exponentially decreases on discharging the capacitor. The capacitance value is calculated using the formula of discharged current in terms of resistance, capacitance, and instantaneous time. The capacitance connected in the circuit depends on the instantaneous current, resistance, and instantaneous time.

The capacitance connected in the circuit is \(43 \mu \mathrm{F}\).