28.1An 18 HF capacitor has been charged to 100 V. A 15 kn resistor and a 5 k2 re

Question

28.1An 18 HF capacitor has been charged to 100 V. A 15 kn resistor and a 5 k2 resistor, are connected in series with the capacitor. (a) What is the time constant of this circuit? (b) Approximately how long (in terms of T) will it take for the charge stored on the resistor to drop to 0.1% of its original charge? Once the capacitor is fully discharged the 5 k2 resistor is removed and replaced with a 12 V battery (c) What is the time constant of the circuit now? (d) Approximately how long will it take for the charge stored on the resistor to rise to 95% of its maximum charge (in terms of T)?

Explanation / Answer

Hi,

In this case you should remember Ohm's Law, as well as certain electric definitions such as resistance, potential, capacitance and what happens once you have resistors in series.

First of all, the time constant (T) is a parameter that is equal to R*C and therefore depends on the equivalent capacitance and equivalent resistance of a circuit.

(a) For the first part we have that C = 18 uF and two resistors in series, so the equivalent resistance is:

Req = R1 + R2 = (15 + 5 ) k = 20 k

Therefore the time constant for this circuit is: T = Req*C = 20*103 * 18*10-6 F = 0.36 s

(c) In the same way, when the capacitor is fully discharged and one of the resistances has been removed we have a new constant time, which is equal to:

T = R1*C = 15*103 * 18*10-6 F = 0.27 s

(b) The function that tell us how the charge is changing when the capacitor is being discharged is:

Q(t) = Q0*exp(-t/T) ; where Q is the charge at certain moment, Qo is the initial charge and t is the time.

They are asking us the value of t is terms of T when Q = 0.01*Qo so:

t = -T*Ln(0.01) = 4.61*T ; remember that this T is the first we calculated (T = 0.36 s)

(d) In this case we do something similar that in the previous step, but the function Q(t) is different:

Q(t) = Qo*[1 - exp(-t/T)]

For this case they are asking us the value of t in terms of T when Q = 0.95 Qo so:

t = -T*Ln(1 - 0.95) = 3.0*T (remember that in this case T is the second one we calculated, T = 0.27 s)

I hope it helps.

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