Solving for voltage and current on an RC circuit

[DODGEVIPER13]

Homework Statement

Uploaded problem 5

Homework Equations

i1=(R2/(R1+R2))
i2=(R1/(R1+R2))

The Attempt at a Solution

Are my answers for problem 5 ok?

 

[gneill]

The problem statement is a bit ambiguous since the switch in the circuit appears to have three separate "positions" as it moves from a to b. Each position creates a different network. Since no information is given about the length of time the switch remains in each position as it makes the transition, it's not possible to give a definitive answer for the voltage or current functions with respect to time. Or are we to assume that the switch begins in the center position for time t ≤ 0, and moves to position b at time t = 0?

 

[DODGEVIPER13]

Yah I think it goes from a at t<0 and moves to b at t=0

 

[gneill]

Yah I think it goes from a at t<0 and moves to b at t=0

So you're making the assumption that the switch moves from position a to position b instantaneously, spending no time in the region where the contacts overlap.

In that case, your workings for the capacitor voltage w.r.t. time look okay. But you should take another look at the current through the 40Ω resistor; how does the potential across that resistor vary for time t > 0?

 

[DODGEVIPER13]

at t>0 it is strictly at position b right so that would eliminate the 40 v source and leave me with just the 30 volt source going over the 120 ohm and 40 ohm resistors right?

 

[DODGEVIPER13]

which would give me i=30/160=.1875A hmmmm I got that for the first part to the current at t=0? So I know my logic is wrong

 

[gneill]

At switch position b, there's more to the circuit than just the 30V source and the 120 and 40 Ohm resistors.

 

[DODGEVIPER13]

the 60 ohm resistor and capacitor is as well right?

 

[gneill]

the 60 ohm resistor and capacitor is as well right?

Right.

 

[DODGEVIPER13]

ok so are the 40 ohm and 60 ohm resistors in parallel with each other and then the 120 ohm one is in series with that giving a total R=144 ohm so therefore I= 30/144=.208 A??

 

[gneill]

ok so are the 40 ohm and 60 ohm resistors in parallel with each other and then the 120 ohm one is in series with that giving a total R=144 ohm so therefore I= 30/144=.208 A??

Well, your figuring of total resistance may be correct, but you're looking for the current through the 40Ω resistor alone...

 

[DODGEVIPER13]

could I use my current divider as I did earlier multiplied by the total current in the b portion only so .208(60/100)=.1248A and the current through the 40 ohm resistor is 30/40=.75A then using my charascteristic equation I get .1248+.6252e^(-t/20).

 

[gneill]

could I use my current divider as I did earlier multiplied by the total current in the b portion only so .208(60/100)=.1248A and the current through the 40 ohm resistor is 30/40=.75A then using my charascteristic equation I get .1248+.6252e^(-t/20).

Just scanning your text I can't tell where the numbers are coming from. But if you look at the schematic you can see that the 40Ω resistor parallels the capacitor when the switch is in the (b) position. So if you already have the expression for the capacitor voltage with respect to time, you can trivially write the expression for the resistor current...

 

[DODGEVIPER13]

so is the capacitor voltage 30 volts because 40(60/(60+20))=30 V or should I use an equation to find for the capacitor voltage?

 

[gneill]

so is the capacitor voltage 30 volts because 40(60/(60+20))=30 V or should I use an equation to find for the capacitor voltage?

? The capacitor voltage will start at some initial value at time t=0 and will evolve to some other value as time progresses. You worked out the initial value for Vc before (you found 30V), and you should be able to find the eventual value (with the switch in the b position) for t >> 0. But you did all this in the workings that you posted previously, right?

 

[DODGEVIPER13]

at t>0 the eventual value would be this 5+25e^(-t/20) you said I already calculated it so I can only assume this is what you are referring to?

 

[gneill]

at t>0 the eventual value would be this 5+25e^(-t/20) you said I already calculated it so I can only assume this is what you are referring to?

The EVENTUAL value is 5V. That's when t → ∞. What you have written, 5+25e^(-t/20), is the function that describes the value Vc(t) for all t > 0.

So, if you have the function for Vc for t > 0, you should be able to write the function for the current i(t) as well, since the same potential Vc(t) appears across the 40Ω resistor.

 

[DODGEVIPER13]

ok well as t approaches infinity I have Vc= 5 V so if this is true 5 volts is what goes across the 40 ohm resistor. therefore the amperage across the 40 ohm is I=5/40=.125A.

 

[gneill]

ok well as t approaches infinity I have Vc= 5 V so if this is true 5 volts is what goes across the 40 ohm resistor. therefore the amperage across the 40 ohm is I=5/40=.125A.

Yes, that would be the final current...

Does the solution require the final current or an expression for the current over time, i(t)?

 

[DODGEVIPER13]

Hmmm it calls for an expression so I know it is of the form I(t)= .125+Ae^(-t/20)

 

[gneill]

Hmmm it calls for an expression so I know it is of the form I(t)= .125+Ae^(-t/20)

Sure. But you ALREADY KNOW the expression for Vc(t), and the resistor parallels the capacitor... so Vc(t) is across the resistor...

 

[DODGEVIPER13]

oh so are you saying that I(t)=Vc(t)/40 which would give i(t)= .125+.625e^(-t/20)

 

[DODGEVIPER13]

am I getting warmer?

 

[gneill]

oh so are you saying that I(t)=Vc(t)/40 which would give i(t)= .125+.625e^(-t/20)

Yes (But be sure to always include the units in your results).

 

[DODGEVIPER13]

ok so the answer should be:
i(t)=.1875A for t<0
i(t)=.125+.625e^(-t/20)A for t>0

 

[gneill]

ok so the answer should be:
i(t)=.1875A for t<0
i(t)=.125+.625e^(-t/20)A for t>0

Sure.

(Although I still think that the question is flawed to the ambiguity introduced by the switch design in the schematic)

 

[DODGEVIPER13]

oh wait nvm I did something wrong I think it should be :
i(t)=.750A for t<0
i(t)=.125+.750e^(-t/20)A for t>0

 

[DODGEVIPER13]

never mind again heh I was correct earlier