The circuit shown in Figure 28.5 is constructed using a 6V battery, a 3 k Omega

Question

The circuit shown in Figure 28.5 is constructed using a 6V battery, a 3 k Omega resistor, and a 6 k Omega resistor. Initially both switches are open and the capacitor is uncharged. Switch A is closed and the capacitor begins to charge. After 1/2 second the potential difference across the capacitor is 3.78 V. What is the capacitance of the capacitor? After several minutes switch A is opened. What is the approximate potential difference across the capacitor? Switch B is now closed and the capacitor begins to discharge. What is the characteristic time for discharging this capacitor? What will the current through the 6 k Omega resistor be after 2 seconds? What will the charge on the capacitor be after 3 seconds?

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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