# Capacitor in Circuit.

1. May 20, 2006

### willydavidjr

Consider the electrical circuit as shown on the website I provided below(and an attachment I provided too). Consisting of E=6[V] battery, two switches S1 and S2, two resistors R1=4ohms and R2=2ohms, and a capacitor C=2 microFarad. The internal resistance of the battery maybe ignored. Initially the switches are both open and the capacitor has no charge. Close the S1 at a certain time. At a sufficiently long time after the S1 is closed, the capacitor is fully charged and the circuit becomes steady.

Question:
1.) Just after the S1 is closed, what is the current flowing through R1.
2.) How much charge is stored in the capacitor C?
3.) During the period in which the capacitor is charged, how much work done by the battery.

1.)1.5 amperes
2.)From q=CV
so 2*6= 12 micro Coulombs

3.)From E=1/2CV^2
so 1/2(2)(6^2)
=36 micro joules.

Am I correct?

This is the website: www.geocities.com/willydavidjr/circuit.html

#### Attached Files:

• ###### circuit.jpg
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2. May 20, 2006

### DaMastaofFisix

I'm not so sure those are right. Though those are the right formulas and such, remember that in the setup where only switch 1 is closed as you say, once the capacitor is charged, there is no more current. Come to think of it, part two and three are both correct, but re-look the first part. Remeber that if the second swith is open, the second loop of the circuit is functionally "not there".

3. May 20, 2006

### maverick280857

The voltage across a capacitor cannot change instantaneously (because if it would, an infinite current would flow through it according to $i=Cdv/dt$). Your capacitor is initially uncharged. At the instant you close switch S1, the capactor behaves as a short circuit and so, you are correct when you say that the initial current is (6V/4ohm)=1.5A. Your answer for the second part seems correct as well. (It looks like S2 has no role to play?)

For the work done, you can think of a differential deposition of charge on the capacitor plates (as the charge gets deposited on the capacitor). Please try and work this out (by considering the work done by the battery, the energy dissipated in the resistor and that stored in the capactor) yourself. Alternatively, you can say that the work done by the battery in the complete charging process is equal to the energy dissipated in the resistor plus the energy stored on the capacitor.

Note: $\frac{1}{2}{CV^2}$ is the energy stored in the capacitor.

4. May 20, 2006

### maverick280857

Are you sure S2 doesn't come up anywhere in the question?

5. May 20, 2006

### willydavidjr

So maverick, my first and second answer are correct. Number three is wrong because I use the potential energy formula and it is not the work done?Am i right?

The fourth question is: How much thermal heat is emitted from the resistor R1? (I have no idea about this)

The fifth question is: Keeping the switch S1 closed, the switch S2 is also closed. At a sufficiently long time after the switch S2 is closed, the circuit becomes steady again. How much charge is stored in the capacitor C long after the S2 is closed. (Am I right if the answer to this is the same as my answer in number two?)

6. May 20, 2006

### maverick280857

Lets calculate the expression for the work done by the battery:

When S2 is closed, Kirchoff's Loop rule gives

$$E = iR_{1} + \frac{q}{C}$$

The current in the circuit is $i = dq/dt$. Multiplying both sides of this equation by $dq=idt$ gives

$$Edq= i^{2}R_{1}dt + \frac{q}{C}dq$$

The term on the left hand side is the work done by the battery (with an emf E and zero internal resistance) in time $dt$. As you can see, part of the work in getting a charge across to the circuit is dissipated as heat in the resistor (Joule heating, first term on the right hand side) and some of it is stored as electrostatic energy in the capacitor (second term on the right hand side).

So for your problem, the work done by the battery is simply $E\Delta Q$ where $\Delta Q$ is the total charge that leaves the battery in the charging period (from when the switch S1 is closed to steady state).

If a current $i$ passes through a resistor $R$ in time $dt$, the heat dissipated is equal to $Vdq = (iR)(idt) = i^{2}R dt$.

I am assuming that you are closing S2 after the steady state due to the left part of the circuit (not involving R2 and S2) has been achieved first. If so, the steady state conditions computed earlier are initial conditions when you close S2. Once you have closed S2, you now have two loops so you can either solve the circuit by assuming two currents in the two loops and writing the d.e. of the circuit (Kirchoff's Loop law really) or by using Thevenin's Theorem (if you know this, otherwise don't bother).

What is the physics of the fifth problem then? Before you closed S2, capacitor C was an open circuit (due to steady state in the left half of the circuit). Now on closing S2, you provide the emf source E a conducting path (and also a path for the capacitor to discharge from $Q=CE$). How does the charge evolve in the circuit as a function of time (you don't need to set up the d.e.'s and solve them for your question, but it would give you good insight.)

Try solving the problem with these hints and feel free ask more

Last edited: May 20, 2006
7. May 21, 2006

### willydavidjr

I am now getting close at it but I am bothering myself with the derivatives. How can I get the value of dt and dq? There is no even time given on the problem.

8. May 21, 2006

### maverick280857

You don't need the 'dt'. We were deriving an expression from first principles, hence using the differential approach.

$\Delta Q$ is the total charge transferred to the capacitor. This means it is the charge transferred from $t=0$ to $t=\infty$. The phrase "sufficiently long" means greater than 5 time constants so effectively steady state has been achieved.

9. May 22, 2006

### willydavidjr

I think the answer on number 5 is the same as in number 2. After the S2 is closed, the charge in Capacitor is still the same. We only need the capacitor value and the value of the battery, in short the answer is also 12 microCoulombs.

10. May 23, 2006

### maverick280857

$E\Delta Q=CE^{2}$ gives the right answer, but I don't know how much that is numerically.