Answer to Question #159608 in Electric Circuits for joel

Question

A 2.00 x10^(-9)F capacitor with an initial charge of 5.10 x 10^(-6) C is discharged through a 1.30Kilo ohm resistor.
(a) Calculate the current in the resistor 9.00s after the resistor is connected across the terminals of the capacitor.
(b) What charge remains on the capacitor after 8.00s? (c) What is 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 $t=0$:

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