Consider the RC circuit Find the time constant T What is the

Consider the RC circuit Find the time constant T. What is the maximum charge on the capacitor after the switch is closed? Find the current in the resistor 10s after the switch is closed. Consider the following circuit: A charged capacitor is connected to a resistor and a switch. The circuit has a time constant of 1.50 s. Soon after the switch is closed the charge on the capacitor is 75% of its original value Find the interval required for the capacitor to reach this charge.

Solution

We need to know the following concepts so as to solve the given problem:

1.) For a charging RC cicuit, the time constant is given as T = RC and the charge on the capacitor after time t is given as: Q(t) = Qo(1 - e^-t/RC)

2.) For a discharging RC circuit, the time constant is equalt to RC and the charge on the capacitor after time t is given as: Q(t) = Qoe^-t/RC

We will make use of the above to solve the given problems as follows:

1a.) For the given circuit we have, R = 1 MOhms, C = 5 uF and V = 30 V

So, the time constant = T = RC = 1 x 10^6 x 5 x 10^-6 = 5 seconds

1b.) The maximum charge for emf of V and capacitance of C is given as Q = VC

Hence for the given circuit we have: Q(max) = 30 x 5 x 10^-6 = 1.5 x 10^-4 C

1c.) The voltage in the capacitor after 10 seconds would be given as:

V(t) = 30 (1- e^-10/5) = 25.939 Volts

2a.) For the given discharging circuit, we have the equation for charge as Q(t) = Qoe^-t/RC

Now, the time constant = RC = 1.5

Hence the equation for 75% charge would be:

0.75 Q = Q e^-t/1.5

or, -t/1.5 = -0.2877

or, t = 0.4315

Therefore the required time at which the charge on the capacitor is down to 75% is t = 0.4315 seconds