How Does Closing a Switch Affect Capacitors and Resistors in a Circuit?

[Clever-Name]

Homework Statement

In the circuit show below, C1 = 5 microfarads, C2 = 10 microfarads and R = 1000 ohms. Initially, the switch is open, C1 is charged to 20 volts, and C2 is uncharged. At time t=0 the switch is closed.
(a) Calculate the voltage across C1 at a much later time. Hint: consider charge conservation.
(b) The energy stored in a charged capacitor is given by U = CV2/2. Calculate the energy stored in C1 and C2 before and after closing the switch.
(c) Derive an expression for the power dissipated in R as a function of time for t>0.
(d) Integrate your expression from (c) to find the total energy dissipated by R. Compare with your answer from (b).

*see attachment for image of circuit*

Homework Equations

\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_2}

U = \frac{1}{2}CV^{2}

P = I^{2}R

Q = CV

The Attempt at a Solution

a) Q = C_{1}V = 100{\mu}F
I initially tried adding the capacitances in series but wound up with 30V as my answer, so i figured that must be wrong (how can the voltage go up? :S) So then I did this:

V = \frac{Q}{C_{1} + C_{2}} = \frac{100{\mu}F}{15{\mu}F} = 6.67V

That's 6.67 V across both capacitors, therefore 6.67 across C1

b) Assuming my answer from a is right:

U_{1}_{i} = \frac{1}{2}CV^{2} = 0.001 J
U_{1}_{f} = 1.11 {\times} 10^{-4} J
U_{2}_{i} = 0
U_{2}_f} = 2.22 {\times} 10^{-4} J

c) Here's where I get confused, I honestly have no clue where to start. I've never dealt with a circuit like this before, so I don't know how to set up KVL or KCL to solve for a time-varying current or voltage.

Help please!

 

[gneill]

When all is said and done, you've got an RC circuit. So it'll have a time constant. The effective capacitance will be C1 and C2 in series.

Can you figure out the time constant? How about the current with respect to time? What's the initial current?

 

[Clever-Name]

{\tau} = RC_{eq} = (1000{\Omega})(3.33333 \times 10^-6 F) = 3.333 \times 10^-4

Finding the current is what confuses me. The input source of 20V is not a constant 20V source. Do I just consider it to be a battery source of 20V and then lump the capacitors together? i.e

20V - iR - \frac{Q}{C_{eq}} = 0

??

 

[gneill]

The trick to dealing with RC or LC circuits that aren't being driven by a changing source (such as a varying supply voltage like a sinewave) is to realize that their voltage and current waveforms always follow an exponential function. If you can pin down their starting and ending values, you can write an expression for the whole thing.

In this case the initial current when the switch is closed will be 20V/R, since C2 is initially at zero volts and "looks" like a short. Since we expect the current to eventually die down to zero, we can write:

I(t) = \frac{20V}{R}e^{-\frac{t}{\tau}}

If its decaying from one level to another (such as the voltage across C1 going from 20V down to 6.67V as in this problem) then you've got an expression like:

V(t) = V_0 - (V_0 - V_f)e^{-\frac{t}{\tau}}

It makes life easier if you can spot these cases and just write down the formula without fiddling about with differential equations.

As for the initially charged capacitor, yes, you can represent it as an uncharged capacitor of the same value in series with a constant voltage supply of the same voltage as the initial voltage on the capacitor.

 

[SammyS]

Homework Statement

In the circuit show below, C1 = 5 microfarads, C2 = 10 microfarads and R = 1000 ohms. Initially, the switch is open, C1 is charged to 20 volts, and C2 is uncharged. At time t=0 the switch is closed.
(a) Calculate the voltage across C1 at a much later time. Hint: consider charge conservation.
(b) The energy stored in a charged capacitor is given by U = CV²/2. Calculate the energy stored in C1 and C2 before and after closing the switch.
(c) Derive an expression for the power dissipated in R as a function of time for t>0.
(d) Integrate your expression from (c) to find the total energy dissipated by R. Compare with your answer from (b).

*see attachment for image of circuit*

Homework Equations

\frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_2}

U = \frac{1}{2}CV^{2}

P = I^{2}R

Q = CV

The Attempt at a Solution

a) Q = C_{1}V = 100{\mu}F

Q is charge, so the units should be \mu C

I initially tried adding the capacitances in series but wound up with 30V as my answer, so i figured that must be wrong (how can the voltage go up? :S) So then I did this:

V = \frac{Q}{C_{1} + C_{2}} = \frac{100{\mu}F}{15{\mu}F} = 6.67V

Why does this formula work when C₁ & C₂ are in series, not parallel?

Let Q₁ be the charge on C₁ at the much later time. Then the charge on C₂ becomes Q-Q₁.

V_1=V_2\quad\rightarrow\quad \frac{Q_1}{C_1}=\frac{Q-Q_1}{C_2}\quad\rightarrow\quad Q_1\frac{C_1+C_2}{C_1C_2}=\frac{Q}{C_2}<br /> \quad\rightarrow\quad Q_1=\frac{Q\,C_1}{C_1+C_2}<br /> \quad\rightarrow\quad V_1=\frac{Q_1}{C_1}=\frac{Q}{C_1+C_2}

b) Assuming my answer from a is right:

U_{1}_{i} = \frac{1}{2}CV^{2} = 0.001 J
U_{1}_{f} = 1.11 {\times} 10^{-4} J
U_{2}_{i} = 0
U_{2}_f} = 2.22 {\times} 10^{-4} J

c) Here's where I get confused, I honestly have no clue where to start. I've never dealt with a circuit like this before, so I don't know how to set up KVL or KCL to solve for a time-varying current or voltage.

Help please!

 

[Clever-Name]

Ahhhh it makes so much more sense now, I was so convinced that my assumption of the first capacitor being the same as a voltage supply was completely wrong and that I couldn't use it. Good to know that works.

I followed through with the rest of the calculations and got the power disspated to be the energy loss from part b, so I'm quite satisfied with my answer now.

Thank you very much for your help!

edit - Sammy, thanks for the input! The F is just a typo, lol, i realize charge is C.