Capacitor Questions Charging and Discharging

[AN630078]

Homework Statement

Hello, I have been teaching myself about capacitors recently and although I feel I have understood the fundamental principles and equations I am still a little uncertain of pertinent information and the functioning of capacitor circuits. I have been attempting to improve my understanding through some relatively straight-forward questions but I still have a few questions here which have confronted with material I have not learnt about. I have attached a couple of such questions below which I would really appreciate if anyone could explain in further detail and overlook my workings

Question 1: A 22 μF capacitor is charged using a 12 V battery.
a. Find the charge is stored on the capacitor?
b. Calculate how much energy is stored on the capacitor?
c. How much electrical energy has the battery provided?
d. Comment upon the difference between your previous two answers and explain why this occurs?

Question 2:
Two capacitors, of capacitance 22μF and 11μF respectively, are connected in series to a 12 V battery. What is the charge stored on each capacitor and the PD across each capacitor? If the combination is then discharged through a resistor, what total charge flows through the resistor?

Relevant Equations

C=Q/V
E=1/2CV^2

Question 1:

a. I am aware that the general equation for capacitance is C=Q/V thus Q=CV.
22 μF = 0.000022 or 2.2*10^-5 F
Would the charge stored by equal to Q=2.2*10^-5*12
Thus, Q=2.64*10^-4 C

b. The energy stored by a capacitor is given by E=1/2QV=1/2CV^2=1/2Q^2/C
I think with the information provided it would be most appropriate to use E=1/2QV^2 here.
Hence, E=1/2*2.2*10^-5*12^2
E=1.584*10^-3 J

c. I do not know how to find the electrical energy provided by the battery, although I assume this is rather a simple question. I have just not read anything regarding this specific line of inquiry therefore I am at a brick wall in terms of finding a solution.

d. Again, I am uncertain. Would this difference be on account of half of the energy being supplied by the battery being lost to heat in the circuit? I do not really know why the energy provided by the battery and the energy stored by the capacitor are different. Since the capacitor begins with zero stored energy but as the current flows, the capacitor charges requiring greater increases in the voltage input until it is fully charged. At this point there would not be a voltage difference yet the accelerated charges are still moving. Therefore, would half of the energy go into the capacitor and half go into the current in the wire. The current will continue to flow, charging the capacitor until the current stops. So would half of the energy supplied be dissipated in the resistance in the circuit?

Question 2;
22μF=2.2*10^-5 F
11μF=1.1*10^-5 F
To find the charge stored on each capacitor I believe that one must first find the equivalent total capacitance of the circuit, since the charge would be the same on both capacitors. Since the capacitors are said to be in series the total capacitance would be equal to 1/CT=1/C1+1/C2+...1/Cn
Thus, 1/CT=1/22+1/11=3/22
CT=22/3=7.3333...μF or 7.3 *10^-6 F

Then, Q=CV
Q=7.3 *10^-6*12
Q=8.796*10^-5 C ~ 8.8*10^-5 C

To find the p.d. across the first 22μF capacitor
V=Q/C
V1= 8.8*10^-5/2.2*10^-5
V1=3.9981 ~4.00 V

To find the p.d. across the second 11μF capacitor
V=Q/C
V1= 8.8*10^-5/1.1*10^-5
V1=7.9963 ~8.00 V

I am not certain how to find the charge flowing through the resistor, would this be using the discharging charge equation Q=Q0e^-t/RC?
But how would I calculate this without knowing the resistance or the time to discharge? I am sorry I am having great difficulty here.

 

[BvU]

c. I do not know how to find the electrical energy provided by the battery

Do you know the expression for the power that a source provides ?

And do you know how the charging process of a capacitor proceeds as a function of time ?

( check here and possibly under 'time domain considerations' here)

Of course part d) has to wait

$\$

 

[BvU]

But how would I calculate this without knowing the resistance or the time to discharge?

The nice thing here is that you don't have to know: The capacitors discharge until there is no more voltage over the resistor. Takes a short time for a small R and a long time for a big R, but in both cases it finishes when V = 0 and we know what that means for the charge !

$\$

 

[kuruman]

$\dots$ it would be most appropriate to use E=1/2QV^2 here.

That would be most inappropriate. The three equivalent expressions for the stored energy are $$E=\frac{1}{2}QV=\frac{1}{2}CV^2=\frac{1}{2}\frac{Q^2}{C}.$$

 

[Steve4Physics]

c. I do not know how to find the electrical energy provided by the battery, although I assume this is rather a simple question. I have just not read anything regarding this specific line of inquiry therefore I am at a brick wall in terms of finding a solution.

The definition of potential difference (V) is essentially V = E/Q.

This means when a charge (Q) moves through a potential difference (V), the energy transferred is E = QV.

Can you apply this to the charge which has passed through the battery?

Edit -typo's corrected.