Answer to Question #159587 in Electric Circuits for Rashinda

Question

A 2.00 x 10^(-9) F capacitor with an initial charge of 5.10 x 10^(-6) C is discharged through a 1.30 Kohm resistor.
(i) Calculate the current in the resistor 9.00 s after the resistor is connected across the terminals of the capacitor.
(ii) What charge remains on the capacitor after 8.00s? Hence calculate the maximum current in the resistor.

Answer 1

(i) Let's first find the time constant:

$$\tau=RC=1.3\cdot10^3\ \Omega\cdot2.0\cdot10^{-9}\ F=2.6\cdot10^{-6}\ s=2.6\ \mu s.$$

The discharge current in the circuit as a function of time can be found as follows:

$$I=\dfrac{Q_0}{RC}e^{-\dfrac{t}{\tau}},$$ $$I=\dfrac{5.1\cdot10^{-6}\ C}{2.6\cdot10^{-6}\ s}\cdot e^{-\dfrac{9.0\ \mu s}{2.6\ \mu s}}=0.06\ A.$$

(ii) The charge on the capacitor discharging through a resistance as a function of time can be found as follows:

$$Q=Q_0e^{-\dfrac{t}{\tau}},$$ $$Q=5.1\cdot10^{-6}\ C\cdot e^{-\dfrac{8.0\ \mu s}{2.6\ \mu s}}=0.235\ \mu C.$$

The current in the circuit is the maximum at $t=0$:

$$I=\dfrac{Q_0}{RC}e^{-\dfrac{0}{\tau}}=\dfrac{Q_0}{RC},$$ $$I=\dfrac{5.1\cdot10^{-6}\ C}{2.6\cdot10^{-6}\ s}=1.96\ A.$$