# Complicated looking circuit

Gold Member
[SOLVED] complicated looking circuit

1. Homework Statement

Consider circuit the below.
(a) How much energy is delivered to the 10 Ω resistor from the time at which S2 is closed (Note: all other switches are open) to t = 30 µs?

(b) How much energy is stored in the 2 µF capacitor if S1 and S3 are also closed and the circuit has reached steady-state? Express your results in electron volts.

(c) What is the current in each branch of the circuit when S1, S2, and S3 are closed and the circuit has reached steady-state?

(d) What is the reading on the voltmeter connected across the 6.0 Ω resistor when all of the switches are closed and the circuit has reached steady-state?

http://img114.imageshack.us/img114/2714/picture3ii5.th.jpg [Broken]

2. Homework Equations

3. The Attempt at a Solution

(a) How much energy is delivered to the 10 Ω resistor from the time at which S_2 is closed (Note: all other switches are open) to t = 30 µs?

hm...what equation would I use for this? (I can't seem to find a section discussing the energy "delivered" to a resistor within a time period, unless I calculate this some other way)
energy would be in Joules as well, right?

I was thinking that it would be: 0.5CV^2 but that is the energy that is stored in a capacitor and not resistor.

(b) How much energy is stored in the 2 µF capacitor if S1 and S3 are also closed and the circuit has reached steady-state? Express your results in electron volts.
I can't figure out which direction is the current going in the first place.
And I think I need to write the sum equation for current, and sum equation for voltage since that's what my prof said, but I'm not sure how to do this for a circuit complicated looking like this one.

Yes, I really think I need help on this problem.

Last edited by a moderator:

Related Introductory Physics Homework Help News on Phys.org
(a) How much energy is delivered to the 10 Ω resistor from the time at which S_2 is closed (Note: all other switches are open) to t = 30 µs?

hm...what equation would I use for this? (I can't seem to find a section discussing the energy "delivered" to a resistor within a time period, unless I calculate this some other way)
energy would be in Joules as well, right?

I was thinking that it would be: 0.5CV^2 but that is the energy that is stored in a capacitor and not resistor.
Remember that resistive elements cannot store energy, they only dissipate it. That said, there is a formula for the energy dissipated by a resistor per unit time, and I'm sure you've used it many times before (hint: power = ?)

(b) How much energy is stored in the 2 µF capacitor if S1 and S3 are also closed and the circuit has reached steady-state? Express your results in electron volts.
I can't figure out which direction is the current going in the first place.
And I think I need to write the sum equation for current, and sum equation for voltage since that's what my prof said, but I'm not sure how to do this for a circuit complicated looking like this one.
So your tools are KCL and KVL, right? More complex circuits simply mean more equations, but the base ideas are all the same. Try writing out all the equations you can and see where that gets you.

Also, what's the current through a capacitor when you reach steady state?

Doc Al
Mentor
3. The Attempt at a Solution

(a) How much energy is delivered to the 10 Ω resistor from the time at which S_2 is closed (Note: all other switches are open) to t = 30 µs?

hm...what equation would I use for this? (I can't seem to find a section discussing the energy "delivered" to a resistor within a time period, unless I calculate this some other way)
energy would be in Joules as well, right?

I was thinking that it would be: 0.5CV^2 but that is the energy that is stored in a capacitor and not resistor.
If you knew the current going through a given resistor, could you tell me the power it "consumes"? (I hope the answer is yes.) The trick is to find an expression for the current--and thus power--as a function of time, then integrate.

Hint: Since S1 & S3 are open, all you have to worry about is the upper loop. Treat it as a standard RC circuit.

(b) How much energy is stored in the 2 µF capacitor if S1 and S3 are also closed and the circuit has reached steady-state? Express your results in electron volts.
I can't figure out which direction is the current going in the first place.
And I think I need to write the sum equation for current, and sum equation for voltage since that's what my prof said, but I'm not sure how to do this for a circuit complicated looking like this one.
To find the energy stored in a capacitor, you'll need to find the voltage across it.

Do it systematically. Assign currents and write Kirchoff's laws for loop voltage and current. Hint: Once steady-state has been reached, what's the current through any segment containing a capacitor?

Gold Member
Remember that resistive elements cannot store energy, they only dissipate it. That said, there is a formula for the energy dissipated by a resistor per unit time, and I'm sure you've used it many times before (hint: power = ?)
well I found that$$P= \frac{\Delta q V} {\Delta t}= IV$$
So your tools are KCL and KVL, right? More complex circuits simply mean more equations, but the base ideas are all the same. Try writing out all the equations you can and see where that gets you.
alright.
Also, what's the current through a capacitor when you reach steady state?
I=0

If you knew the current going through a given resistor, could you tell me the power it "consumes"? (I hope the answer is yes.) The trick is to find an expression for the current--and thus power--as a function of time, then integrate.

Hint: Since S1 & S3 are open, all you have to worry about is the upper loop. Treat it as a standard RC circuit.
yes because if you have the Resistance ($$\Omega$$) and current (A)
then I could use this equation to find the power: $$P= I^2 R$$ right?

To find the energy stored in a capacitor, you'll need to find the voltage across it.

Do it systematically. Assign currents and write Kirchoff's laws for loop voltage and current. Hint: Once steady-state has been reached, what's the current through any segment containing a capacitor?
well, when a steady state is reached, the current in the segment is zero since the capacitor is fully charged. (I= 0).

For part a)
well... I have to find an expression for the current.
I am thinking that the current goes from right to left after the circuit is closed.
And the function for current when a capacitor is charging is
$$I(t)= \frac{ \epsilon} {R} e^{-t/RC}$$

so would I use that to find the current? (but that wouldn't give me the collective current from t=0 to t= 30) so would this involve integration....(I suppose it would but I really don't know how to set up equations and I find it different to integrate in physics than it was to do in actual calculus so I get sort of lost with that, unfortunately)

$$R_1= 10 \Omega$$
$$R_2= 6.0 \Omega$$
$$C= 3.0x10^-6F$$
$$V= 2.0V$$

$$P=I^2R$$
$$I(t)= \frac{ \epsilon} {R} e^{-t/RC}$$

I know the R and the current if that is the right equation but as to how to use that to integrate...

I'm going to work on B and post that.

Doc Al
Mentor
yes because if you have the Resistance ($$\Omega$$) and current (A)
then I could use this equation to find the power: $$P= I^2 R$$ right?
Right.

well, when a steady state is reached, the current in the segment is zero since the capacitor is fully charged. (I= 0).
Right.

For part a)
well... I have to find an expression for the current.
I am thinking that the current goes from right to left after the circuit is closed.
And the function for current when a capacitor is charging is
$$I(t)= \frac{ \epsilon} {R} e^{-t/RC}$$
Good. That's the current as a function of time, so what's the power as a function of time?

To get the energy from the power, you'll have to integrate: $\int P(t) dt$.

(Luckily an exponential function is easy to integrate. )

Gold Member
Good. That's the current as a function of time, so what's the power as a function of time?

To get the energy from the power, you'll have to integrate: $\int P(t) dt$.

(Luckily an exponential function is easy to integrate. )

Hm...lets see if I understand what your saying.

$$\int P(t) dt= \int I^2 R= R \int (\frac{ \epsilon} {R} e^{-t/RC})^2= \frac{\epsilon^2} {R} \int e^{-2t/RC}$$

b)How much energy is stored in the 2 µF capacitor if S1 and S3 are also closed and the circuit has reached steady-state? Express your results in electron volts.

assuming what I think is correct.

since it says the system reaches a steady state I think that the current in that segment would be zero so it would be like it was cut, and the only segment that would be left is what I drew below.

http://img99.imageshack.us/img99/7941/77664909qx5.th.jpg [Broken]

Is it right or not?

Thanks

Last edited by a moderator:
Doc Al
Mentor
Hm...lets see if I understand what your saying.

$$\int P(t) dt= \int I^2 R= R \int (\frac{ \epsilon} {R} e^{-t/RC})^2= \frac{\epsilon^2} {R} \int e^{-2t/RC}$$
Good! (Don't drop off the "dt".) Now do the integration.

b)How much energy is stored in the 2 µF capacitor if S1 and S3 are also closed and the circuit has reached steady-state? Express your results in electron volts.

assuming what I think is correct.

since it says the system reaches a steady state I think that the current in that segment would be zero so it would be like it was cut, and the only segment that would be left is what I drew below.

http://img99.imageshack.us/img99/7941/77664909qx5.th.jpg [Broken]

Is it right or not?
I think you've got the right idea. The loop you outlined in yellow is the only one with a non-zero current. (But you can still find and analyze a loop that contains the capacitor.) Find the current and you'll be able to figure out the voltage drops across the resistors (the ones with current through them).

Last edited by a moderator:
Gold Member
Good! (Don't drop off the "dt".) Now do the integration.
$$\int P(t) dt= \int I^2 R= R \int (\frac{ \epsilon} {R} e^{-t/RC})^2 dt= \frac{\epsilon^2} {R} \int e^{-2t/RC}dt=$$

that's where I got stuck. I tried to look in my book for a hint (cal book) but no luck there.

I was thinking of the chain rule but not sure if that is correct. (I think I could change the
$$-2t/RC$$ to $$-2t(RC)^{-1}$$ then use product rule on that)

I think you've got the right idea. The loop you outlined in yellow is the only one with a non-zero current. (But you can still find and analyze a loop that contains the capacitor.) Find the current and you'll be able to figure out the voltage drops across the resistors (the ones with current through them).
how can I analyze the one with the capacitor if there is no current through it at the time that they say I have to analyze it at? (steady state= I= 0)

what I'm thinking (even though the circuit looks funny if I eliminate the 2 capacitors since the voltage sources are + =>+ and -=> - and I don't think that is possible.

what I'm thinking is : (going from left to right on the circuit with current

$$I_1 = I_2$$ since in each resistor there is the same ammount of current

$$\Sum V= \epsilon_1 - IR_1 - \epsion _2 - IR_2= 0$$

$$I= \frac{ \epsilon _1 - \epsilon_2} {R2+ R_2}$$

but I don't know the $$\epsilon_1$$ (not given) so how do I find the voltage?
________________________________________________
I've just noticed that it says "how much E is stored in 2micro F capacitor at the time that circuit reaches steady state and I've been thinking of finding something else.

AND I'm going in a loop right now.

Help please? (not sure where to start)

Gold Member
Help anyone?

I haven't looked all of it. But you are just complicating it too much.
1) Reduce your thing to Thevian (two times - intial and final)
2) use x(t) = x(f) - (x(i)-x(f))*e^ .. equation for 'i' through your thing
3) find V as a function of time
3) use 1/2*C*v^2
4) find change in energy

no messy integrals.

Doc Al
Mentor
$$\int P(t) dt= \int I^2 R= R \int (\frac{ \epsilon} {R} e^{-t/RC})^2 dt= \frac{\epsilon^2} {R} \int e^{-2t/RC}dt=$$

that's where I got stuck. I tried to look in my book for a hint (cal book) but no luck there.

I was thinking of the chain rule but not sure if that is correct. (I think I could change the
$$-2t/RC$$ to $$-2t(RC)^{-1}$$ then use product rule on that)
This is much easier than you think. For one thing, RC is a constant.

Hint: What's the derivative of $e^x$?

how can I analyze the one with the capacitor if there is no current through it at the time that they say I have to analyze it at? (steady state= I= 0)
You can always take any loop and add up the voltage drops--even a loop that has a capacitor. (Hint: If a resistor has no current through it, there's no voltage drop across it.)

what I'm thinking (even though the circuit looks funny if I eliminate the 2 capacitors since the voltage sources are + =>+ and -=> - and I don't think that is possible.

what I'm thinking is : (going from left to right on the circuit with current

$$I_1 = I_2$$ since in each resistor there is the same ammount of current

$$\Sum V= \epsilon_1 - IR_1 - \epsion _2 - IR_2= 0$$

$$I= \frac{ \epsilon _1 - \epsilon_2} {R2+ R_2}$$

but I don't know the $$\epsilon_1$$ (not given) so how do I find the voltage?
To find the current going through the only loop that has current, just add up the voltage drops and set equal to zero. Let's say you start just below the left battery and move clockwise. (I can't read the voltage on that battery, so I don't know which battery is bigger--I'll just assume the current goes clockwise.) Here's what I get: +V - (10)I -2 -(20)I = 0. Solve for I.
________________________________________________
I've just noticed that it says "how much E is stored in 2micro F capacitor at the time that circuit reaches steady state and I've been thinking of finding something else.

AND I'm going in a loop right now.
Once you've got the current you know what the voltage drops are across the 10 ohm and 20 ohm resistors. To find the voltage across the capacitor (which will tell you the energy), pick a loop that contains the capacitor and add up the voltages, as usual. For example, pick the loop that contains the 2.0V battery, the 20 ohm resistor, the 5 ohm resistor (no current here!), and the capacitor.

Gold Member
This is much easier than you think. For one thing, RC is a constant.
Hint: What's the derivative of $e^x$?

I know it's e^x but if something is e^2x it would be 1/2e^2x right?

BUT if that is right in the above derivative then wouldn't

$$\int P(t) dt= \int I^2 R= R \int (\frac{ \epsilon} {R} e^{-t/RC})^2 dt= \frac{\epsilon^2} {R} \int e^{-2t/RC}dt=$$
be equal to $$\frac{RC} {2} e^{-2t/RC}$$??

I don't think that's right though.

Last edited:
Gold Member
You can always take any loop and add up the voltage drops--even a loop that has a capacitor. (Hint: If a resistor has no current through it, there's no voltage drop across it.)
okay.
To find the current going through the only loop that has current, just add up the voltage drops and set equal to zero. Let's say you start just below the left battery and move clockwise. (I can't read the voltage on that battery, so I don't know which battery is bigger--I'll just assume the current goes clockwise.) Here's what I get: +V - (10)I -2 -(20)I = 0. Solve for I.
But that battery voltage is not given. But solving for I anyway, I get,

$$+V - 10 \Omega (I) - 2V - 20 \Omega (I)= 0$$

$$-V -30 \Omega (I)= 0$$

$$I= -V/30 \Omega$$

Once you've got the current you know what the voltage drops are across the 10 ohm and 20 ohm resistors. To find the voltage across the capacitor (which will tell you the energy), pick a loop that contains the capacitor and add up the voltages, as usual. For example, pick the loop that contains the 2.0V battery, the 20 ohm resistor, the 5 ohm resistor (no current here!), and the capacitor.
not really since I don't know the current since the other V was not given.

but subsituting anyway I get

$$V= -10 \Omega (-V/30 \Omega)= V/3$$
$$V= - 20 \Omega (-V/30 \Omega)= 2V/3$$

I think I have to solve for the V..

Thanks Doc.

Last edited:
Doc Al
Mentor
Hint: What's the derivative of $e^x$?

I know it's e^x but if something is e^2x it would be 1/2e^2x right?
The derivative of e^(ax) is ae^(ax); the integral is (1/a)e^(ax)

BUT if that is right in the above derivative then wouldn't

$$\int P(t) dt= \int I^2 R= R \int (\frac{ \epsilon} {R} e^{-t/RC})^2 dt= \frac{\epsilon^2} {R} \int e^{-2t/RC}dt=$$
be equal to $$\frac{RC} {2} e^{-2t/RC}$$??
You left out the minus sign and the $\epsilon^2/R$ factor.

okay.

But that battery voltage is not given.
Before we go any further, please double check that that battery voltage isn't given. (It looks like the diagram may have been truncated.) It doesn't make sense that they would not give that voltage.

Gold Member
The derivative of e^(ax) is ae^(ax); the integral is (1/a)e^(ax)
You left out the minus sign and the $\epsilon^2/R$ factor.
$$-\frac{\epsilon^2} {R} (\frac{RC} {2}) e^{-2t/RC} = \frac{\epsilon^2 C} {2} e^{-2t/RC}$$

Before we go any further, please double check that that battery voltage isn't given. (It looks like the diagram may have been truncated.) It doesn't make sense that they would not give that voltage.

Doc Al
Mentor
$$-\frac{\epsilon^2} {R} (\frac{RC} {2}) e^{-2t/RC} = \frac{\epsilon^2 C} {2} e^{-2t/RC}$$
Good, but don't drop the minus sign. You'll need to evaluate this over the interval from t = 0 to t = 30 µs, once you have epsilon.
That makes no sense to me as you are given no data (such as a current or voltage somewhere). I'd say he's wrong. (Are you sure he wasn't referring to the voltage across the 6.0 Ω resistor, as asked in part d?)

Is this taken from your text? Or is it a problem your professor created himself?

(Sorry I've been slow in responding; I've been away for a few days.)

Gold Member
Good, but don't drop the minus sign. You'll need to evaluate this over the interval from t = 0 to t = 30 µs, once you have epsilon.
$$-\frac{\epsilon^2} {R} (\frac{RC} {2}) e^{-2t/RC} = -\frac{\epsilon^2 C} {2} e^{-2t/RC}$$

so I would have to find the voltage drop across the resistor, correct?

That makes no sense to me as you are given no data (such as a current or voltage somewhere). I'd say he's wrong. (Are you sure he wasn't referring to the voltage across the 6.0 Ω resistor, as asked in part d?)
no, other people asked the same question about the voltage and whether he forgeot to give it to us, but no. My professor actually drew the circuit on the board and said that it could be done.

Is this taken from your text? Or is it a problem your professor created himself?
I really don't know. It was part of a set of problems given to us to do. He may have created it himself, or he may have gotten it from somewhere.

(Sorry I've been slow in responding; I've been away for a few days.)
yes, I noticed. And the assignment was do a few days ago, and I'm not sure what I wrote, but I think I stopped at the unknown.

Doc Al
Mentor
$$-\frac{\epsilon^2} {R} (\frac{RC} {2}) e^{-2t/RC} = -\frac{\epsilon^2 C} {2} e^{-2t/RC}$$

so I would have to find the voltage drop across the resistor, correct?
I think I messed you up a bit on this one by not distinguishing total resistance from the 10 ohm resistance. Let's start over.

You need to find the current as a function of time:
$$I = \frac{ \epsilon} {R} e^{-t/RC}$$

Note that R is the total resistance in that RC circuit; R = 10 + 6 = 16 ohms. Epsilon is the battery voltage; epsilon = 2 V.

To find the power through the 10 ohm resistor, use $P = I^2 r$ where r is now just 10 ohms. Integrate that over the time interval.

$$\int P(t) dt= \int I^2 r= r \int (\frac{ \epsilon} {R} e^{-t/RC})^2 dt= \frac{r \epsilon^2} {R^2} \int e^{-2t/RC}dt= -\frac{r \epsilon^2} {R^2} (\frac{RC} {2}) e^{-2t/RC} = -\frac{r \epsilon^2 C} {2 R} e^{-2t/RC}$$

Sorry about that! (You'd better double check it. )

Note that you have all the data needed to get the actual answer for this part.
no, other people asked the same question about the voltage and whether he forgeot to give it to us, but no. My professor actually drew the circuit on the board and said that it could be done.
I'd love to see that solution.

Gold Member
Sorry for the late reply Doc, I just saw this now.
I think I messed you up a bit on this one by not distinguishing total resistance from the 10 ohm resistance. Let's start over.

You need to find the current as a function of time:
$$I = \frac{ \epsilon} {R} e^{-t/RC}$$

Note that R is the total resistance in that RC circuit; R = 10 + 6 = 16 ohms. Epsilon is the battery voltage; epsilon = 2 V.

To find the power through the 10 ohm resistor, use $P = I^2 r$ where r is now just 10 ohms. Integrate that over the time interval.

$$\int P(t) dt= \int I^2 r= r \int (\frac{ \epsilon} {R} e^{-t/RC})^2 dt= \frac{r \epsilon^2} {R^2} \int e^{-2t/RC}dt= -\frac{r \epsilon^2} {R^2} (\frac{RC} {2}) e^{-2t/RC} = -\frac{r \epsilon^2 C} {2 R} e^{-2t/RC}$$
So the current as a function of time, R refers to the total resistance..okay, and since I'm trying to find the energy delivered to only 1 resitor it would mean that the r in power is only refering to the resistance in that one resistor.
Sorry about that! (You'd better double check it. )
It's fine, I think it's right.
Note that you have all the data needed to get the actual answer for this part.
yes, since I have
r= 10 Ohms
R= 16 Ohms
C= 3x10^-6 F
and it would be solved from t= 0 to t= 30
is this correct?
I'd love to see that solution.
I would give you the solution but I don't think my professor put it up yet.

Thanks Doc Al

Doc Al
Mentor
Looks like you got it.

Gold Member
Looks like you got it.
okay.

(c) What is the current in each branch of the circuit when S1, S2, and S3 are closed and the circuit has reached steady-state?

well I do know that the branch with the circuits have 0 current at steady states but for the rest, not so sure.

but if I don't have the V unknown I assume I cannot find it...

(d) What is the reading on the voltmeter connected across the 6.0 Ω resistor when all of the switches are closed and the circuit has reached steady-state?

I think that would be the voltage drop across the resistor and that I would use Ohms law I think to find that but as to the voltage how would I find that...or do I need that unknown voltage again?

Thanks

Doc Al
Mentor
(c) What is the current in each branch of the circuit when S1, S2, and S3 are closed and the circuit has reached steady-state?

well I do know that the branch with the circuits have 0 current at steady states but for the rest, not so sure.
Every segment with a capacitor will have 0 current at steady state.

but if I don't have the V unknown I assume I cannot find it...
Right.

The only path with current is the one containing the unknown battery, the 2 V battery, and the 10 and 20 ohm resistors.

(d) What is the reading on the voltmeter connected across the 6.0 Ω resistor when all of the switches are closed and the circuit has reached steady-state?

I think that would be the voltage drop across the resistor and that I would use Ohms law I think to find that but as to the voltage how would I find that...or do I need that unknown voltage again?
Use the result from part (c).

Gold Member
Every segment with a capacitor will have 0 current at steady state.

The only path with current is the one containing the unknown battery, the 2 V battery, and the 10 and 20 ohm resistors.

Use the result from part (c).
but I don't know how to find that since the loop with the voltage meter is across the loop with the unknown batery as well as being connected to the battery which is known.

Thank you

Doc Al
Mentor
but I don't know how to find that since the loop with the voltage meter is across the loop with the unknown batery as well as being connected to the battery which is known.
What's the current through the 6 ohm battery? Hint: It's in series with a capacitor.

Gold Member
What's the current through the 6 ohm battery? Hint: It's in series with a capacitor.
well if it's in series with a capacitor wouldn't that make it's current 0 as well?

so would the voltage be the same as that from the battery? 2V?