# Circuit problems with capacitors and parallel emf sources

1. Apr 8, 2014

### MathewsMD

I have quite a few questions so I've attached the associated images instead of making multiple threads. I'll share my attempts/reasoning for the problems and please tell me if my logic is flawed or if I am doing something completely wrong.

Question 41: Bulb brightness is related to power output, correct? The bulbs must then have the same resistance and P = i2R so then the answer must be D since the total current is V/(R/2) = 2V/R and since the two bulbs are in parallel and equal, the each have current V/R, thus the power output is also V2/R in these bulbs. Any problems? Also, for P = iV, is this expression only true for the energy output from an emf source and P = i2R = V2/R only true for loads? Please correct me if I am mistaken. An explanation for why this is true would be greatly appreciated!

Question 43: P = 500 W and I = 15 A to ensure each bulb has minimum potential since they are in series and potential is added. (500/15) = potential for each bulb, 120 / (500/15) = 3.6 so must round down to 3. Therefore only 3 bulbs can be operated t full brightness from this line. Right?

Question 49: I was slightly confused on this one and believe my reasoning is fallacious. If I'm not mistaken, by connecting the two emf sources from cathode - cathode and anode - anode, this makes the total output emf ε2 - ε1 since ε2 > ε1. If it was not connected in this arrangement, but instead cathode-anode, would it be ε2 + ε1? Regardless, I found V = ε2 - ε1 because of this. Then current in the battery 1 is, I = ε1/r so P = iV = (ε2 - ε1)(ε1/r) but this is not the answer. I'm uncertain on how they arrived at their answer and am not sure if I've done anything wrong. If you could please tell mw what I did wrong and how to arrive at the proper conclusion, that would be much appreciated!

Question 62: I know: ## i(t) = ε/R (e−t/RC) when discharging and i(t) = 1 - ε/R (e−t/RC) when discharging. τa = ReqC is the discharging time constant where the loads are in series so Req = 2R. τb = RC/2 since Req = R/2 since the loads are in parallel. When doing τab = 4, but this is incorrect. Any help please? I've done question 63 and seem to have gotten it right, but knowing 62 will surely help in understanding it better.

Any help with these problems and additional advice or links to help understand circuit more comprehensively would be great! Also, is it possible to put two emf sources with different emf values in parallel? If so, will their emf values become equivalent very quickly? Is there a specific equation for this?

2. Apr 8, 2014

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3. Apr 8, 2014

### SammyS

Staff Emeritus
I don't see that these questions are all that closely related.

I would suggest posting each in its own thread.

4. Apr 8, 2014

### CWatters

Q41: Correct. In circuit D each bulb will be as bright as the bulb in circuit X. I think it's slightly easier to understand this if you use P=V2/R rather than P=I2R.

Q43: Three is the right answer but your explanation is wrong/confused. To achieve maximum brightness each bulb must see the full line voltage so the bulbs need to be in parallel not series.

Q49: The voltage drop across the combined internal resistance 2R is indeed (ε2-ε1). This allows you to write an equation for the current flowing around the loop I = (ε2-ε1)/2R . Then use P = IV to work out the power being converted in the battery and you get P = ((ε2-ε1)/2R) * ε1 which is answer D.

Q62:

When the switch is opened you have a capacitor C being discharged through two resistors in series. So the time constant is proportional to C*2R.

When the switch is made (conducting) the capacitor is being charged from the voltage source via a single resistor R (the top one). The other resistor is in parallel with the ideal voltage source so it has no effect on the circuit. You could remove it. So the charging time constant is proportional to C * R

The ratio of these time constants is C*2R/C*R = 2.

You can try but it's not usually recommended! What happens depends on how "ideal" the voltage sources are. All real world voltage sources have some internal resistance but it can be very small. What happens to the current if the internal resistance is very small? Go back to problem 49 and work out how much current would flow if ε2 = 12V, ε1=6V and R = 0.01Ω.

5. Apr 8, 2014

### MathewsMD

Thanks so much for the help!

With regards to q. 62, we actually never really discussed multiple resistors in a circuit when charging the capacitor, in class but you're saying the one in the middle branch does not affect the equivalent resistance of the circuit when charging the capacitor? Would not current still be going through this branch and affecting the time constant? Trying to grasp the theory behind this...

6. Apr 8, 2014

### SammyS

Staff Emeritus
Have you learned Kirchhoff's Loop Law?

7. Apr 8, 2014

Yes....

8. Apr 8, 2014

### SammyS

Staff Emeritus
The voltage drop across the resistor in middle the same as the voltage source.

That's also the same as the voltage drop in the left hand branch.

The voltage drop in the left hand branch is independent of the middle branch.

9. Apr 9, 2014

### MathewsMD

Okay. Sorry if I'm grossly mistaken here, but regardless of the potential being constant when in parallel, the time constant is RC. C in this case is a constant since it is proportional to A/d, which is not changing (or do we assume this is not necessarily true?). Now, R is the equivalent resistance in the circuit, is it not? Don't resistors in parallel still affect the total resistance and therefore the middle resistor should be considered? Are you trying to convey R = V/I, in which case the potential is the same, yes, and I (current) cannot be determined, right? The capacitor and resistor are in parallel, so how would this be taken into account? Since we approximate capacitance resistance as ~0, do we consider all current to be going through the capacitor and the only resistor affecting the capacitor in this case is the one in series with it?

Also, if there was a resistor on the branch of the capacitor, how would this affect things (i.e., there is a capacitor in series with capacitor on the branch that is parallel to the middle resistor)?
An explanation would be really appreciated!

Last edited: Apr 9, 2014
10. Apr 9, 2014

### Staff: Mentor

For Q62, redraw the circuit and color in the complete loop taken by the current which puts charge onto the capacitor while the switch is closed. Write equations for the voltage at every point along that path. Do any of those equations involve the vertical resistor, Rp?

Now, color in the loop taken by current while the switch is open and as the plates discharge. Write equations for the voltage at every point along that path. Do any of those equations involve Rp?

11. Apr 14, 2014

### MathewsMD

Okay, maybe this is my cause for misunderstanding, but when finding the time constant, RC, you ONLY look at the loop in which the capacitor is connected? Not the resistance of the entire circuit?

12. Apr 14, 2014

### Staff: Mentor

You take into account all the resistors that affect the charging/discharging of the capacitor. There may not be a neat single loop path.

13. Apr 14, 2014

### MathewsMD

I see the voltage drops is independent and that both branches have the same R, so yes, they should have the same current. I'm just failing to grasp why the two resistors are not in parallel...

The total current is affected by both resistors, though, even though they both have the same current.

Q/V = C

V/I = R

RC = Q/I and from what I see, the current is affected...i've tried colouring the loops but fail to recognize how the middle resistor does not affect R since there are 2 loops through which current goes, the far one (with the capacitor) and the short loop (with the middle resistor).

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