Question

A 10.0 µF capacitor initially charged to 30.0 µC is discharged through a 1.50 kO resistor. How long does it take to reduce the capacitor's charge to 5.00 µC?

 

Answer 1

Concepts and reason: The concept is used to charge and discharge a capacitor in an RC circuit. First, find the equation of discharging of the capacitor is the RC circuit. Then, the rearrange equation for time. With the help of this equation of time, the time required by the capacitor to discharge can be calculated.

Fundamentals

The RC circuit consists of a resistor and a capacitor, and a voltage source. When the switch is closed, the capacitor gets charged:

\(Q=Q_{0}\left(1-e^{\frac{-t}{R C}}\right)\)

Here \(Q\) is the charge on the capacitor at time \(t, Q_{0}\) is the maximum or final charge that the capacitor can hold, \(R\) is the resistance of the circuit, and \(C\) is the capacitance of the circuit. At time \(t=0\), the charge in the capacitor is zero, and when \( t \) tends to infinity, the capacitor is fully charged. When the voltage source is removed, the capacitor will discharge. The discharging of the capacitor is expressed by the equation as follows; \(Q=Q_{0} e^{-t / R C}\)

Here \(Q\) is the charge on the capacitor at time \(t Q_{0}\) is the initial charge on the capacitor, \(R\) is the resistance of the circuit, and \(C\) is the capacitance of the circuit. So here, at time \(t=0\) the charge in the capacitor is \(Q\), and when \( t \) tends to infinity, the capacitor is fully discharged. 

Calculate the relation between initial charge, the final charge on the capacitor, and time as follows; The equation of discharging of the capacitor is expressed as follows; \(\frac{Q}{Q_{0}}=e^{-t / R C}\)

Take natural logarithm on both sides in equation (1) and solve.

\(\ln \left(\frac{Q}{Q_{0}}\right)=-\frac{t}{R C}\)

Since the capacitor is discharging, that is why here equation \(Q=Q_{0} e^{-t / R C}\) is used. Here \(Q\) is the charge on the capacitor at

time \(t, Q_{0}\) is the initial charge on the capacitor, \(R\) is the resistance of the circuit, and \(C\) is the capacitance of the circuit.

Calculate the time required by the capacitor to discharge to \(5.00 \mu \mathrm{C}\) as follows; Rearrange the equation (2) for \(t\) as follows; \(t=-R C \ln \left(\frac{Q}{Q_{0}}\right)\)

\(t=-R C \ln \left(\frac{Q}{Q_{0}}\right)\)

Substitute \(30 \mu \mathrm{C}\) for \(Q_{0}, 5 \mu \mathrm{C}\) for \(Q, 1.5 \mathrm{k} \Omega\) for \(R\) and \(10 \mu \mathrm{F}\) for \(C\) in the equation \(Q_{0}\) to find the value of time.

\(t=-\left(1.5 \mathrm{k} \Omega\left(\frac{10^{3} \Omega}{1 \mathrm{k} \Omega}\right)\right)\left(10 \mu \mathrm{F}\left(\frac{10^{-6} \mathrm{~F}}{1 \mu \mathrm{F}}\right)\right) \ln \left(\frac{5 \mu \mathrm{C}}{30 \mu \mathrm{C}}\right)\)

\(=0.027 \mathrm{~s}\)

Hence the time required by the capacitor to discharge to \(5.00 \mu \mathrm{C}\) is \(0.027 \mathrm{~s}\).

s exponentially. The equation of discharging of the capacitor is \(Q=Q_{0} e^{-t / R C}\). The equation \(t=-R C \ln \left(\frac{Q}{Q_{0}}\right)\) is the modified form of the equation of discharging of the capacitor.

The time required by the capacitor to discharge to \(5.00 \mu \mathrm{C}\) is \(0.027 \mathrm{~s}\).