Voltage Drop Across Capacitor? EK physics exam 7 #158 &159

[dreamer67]

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158 The capacitor is full charged. Both resistors have a 2 ohm resistance. When the switch is open what is the initial current through A?

159 What is the charge on the capacitor after the circuit has been on a long time?

I was never taught capacitance, so I am having a hard time understanding these problems, and the solutions aren't going into the depth I need. How did they determine the voltage drop across the capacitor to be 6 V?

Also, for 159, the solution says charge is equal to voltage times capacitance. I did 12 V x 1x10^-6 F, but this answer was incorrect. The correct answer is 6 x 10^-6C. Are they using the voltage drop instead of the voltage to calculate this? If so, how do you know when to use voltage drop rather than voltage?

 

[indianjatt]

Do you have the exact question or can you post a similar question?

 

[deleted393595]

Do you have the exact question or can you post a similar question?

Agreed. We need a lot more info to solve your query.

 

[colberag]

158 The capacitor is full charged. Both resistors have a 2 ohm resistance. When the switch is open what is the initial current through A?

159 What is the charge on the capacitor after the circuit has been on a long time?

I was never taught capacitance, so I am having a hard time understanding these problems, and the solutions aren't going into the depth I need. How did they determine the voltage drop across the capacitor to be 6 V?

Also, for 159, the solution says charge is equal to voltage times capacitance. I did 12 V x 1x10^-6 F, but this answer was incorrect. The correct answer is 6 x 10^-6C. Are they using the voltage drop instead of the voltage to calculate this? If so, how do you know when to use voltage drop rather than voltage?

If you are trying to find the capacitance or charge on a capacitor plate then you use the voltage drop across that capacitor in the equation q=VC.

As for your initial question I cant help without a picture. Sorry

 

[dreamer67]

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[dreamer67]

I added the image as an attachment. Hope this helps!

 

[Temperature101]

158 The capacitor is full charged. Both resistors have a 2 ohm resistance. When the switch is open what is the initial current through A?

159 What is the charge on the capacitor after the circuit has been on a long time?

I was never taught capacitance, so I am having a hard time understanding these problems, and the solutions aren't going into the depth I need. How did they determine the voltage drop across the capacitor to be 6 V?

Also, for 159, the solution says charge is equal to voltage times capacitance. I did 12 V x 1x10^-6 F, but this answer was incorrect. The correct answer is 6 x 10^-6C. Are they using the voltage drop instead of the voltage to calculate this? If so, how do you know when to use voltage drop rather than voltage?

Since then two resistors are in series and both of them are two ohms. The across the resistors is I = 12/2+2 =3 amps. So the voltage drop across the second resistor is the same with the capacitor since they are in parallel, and this voltage drop is V = RI = 2x3= 6 volts, which is also the voltage drop across the capacitor. Hope I am right.

 

[bambam02]

Fully charged capacitors act as broken wires initially. so treat it as a broken wire.... no current will go through the capacitor.

Initially, the same amount of current will be going through both 2 ohm resistors. Treat this as a 12V voltage source, combine both resistors in series to make a 4 ohm resistance circuit, and solve for I using V=IR, so V/R = I = 12/4= 3 A.


Therefore the voltage drop over each resistor will be 6 ohms a piece.

Now to find the voltage drop over a fully charged capacitor.. you have to do the loop rule around the entire circuit.

So, going around the outside of the circuit: 12V-Voltage on capacitor - Voltage on resistor = 0.... solve for Vol. on Cap., and find it is 6V.

Q = V*C so = 6V * 1x10^-6 = 6x10^-6 C.

 

[dreamer67]

I'm not seeing how you can just go around the outside of the circuit to find the vol. on the capacitor. Why don't you have to take the other resistor into consideration?

Fully charged capacitors act as broken wires initially. so treat it as a broken wire.... no current will go through the capacitor.

Initially, the same amount of current will be going through both 2 ohm resistors. Treat this as a 12V voltage source, combine both resistors in series to make a 4 ohm resistance circuit, and solve for I using V=IR, so V/R = I = 12/4= 3 A.


Therefore the voltage drop over each resistor will be 6 ohms a piece.

Now to find the voltage drop over a fully charged capacitor.. you have to do the loop rule around the entire circuit.

So, going around the outside of the circuit: 12V-Voltage on capacitor - Voltage on resistor = 0.... solve for Vol. on Cap., and find it is 6V.

Q = V*C so = 6V * 1x10^-6 = 6x10^-6 C.

 

[Temperature101]

I'm not seeing how you can just go around the outside of the circuit to find the vol. on the capacitor. Why don't you have to take the other resistor into consideration?

See my explanation above...Whenever a resistor is in parallel with another resistor or a capacitor, the voltage drop across them is the same. The capacitor is in parallel with one of the two resistors; therefore, finding the voltage drop across this resistor would give you the voltage drop across the capacitor ...There are many ways you can solve this problem; I think using loop for this particular problem is unecessary.

 

[Majik]

Yeah, I'm assuming this question is already answered. I made a mistake like this recently, and I'm glad I caught it. The voltage of a battery does NOT always equal the voltage of a capacitor (the assumption I made). This question is a prime example of that. Here one resistor is in parallel to the capacitor. Both have the same voltage drop. However, we see from the diagram there's an additional resistor in the circuit diagram that is NOT in parallel to either the resistor we just considered and the capacitor. You must find the equivalent resistance first and then the current flowing through the circuit to answer this question.

Both resistors are in series so simply add them up. Using the handy dandy equation, V=IR, you can find net current by simply dividing the battery voltage by equivalent resistance: V/R = I ---> 12V/4ohms = 3 Amps of current.

Now that we know the current, we use V=IR to find the voltage DROP of that resistor. Rather than plugging in the equivalent resistance of the whole circuit (which would be wrong if you did do that), to find the voltage drop for a given resistor, we plug in the resistance of the resistor we are interested in. The resistor we are interested in is the one in parallel to the capacitor. Why is this important? Because anything in parallel has the same voltage drop. So let's solve for this: V=IR --> (3A)(2ohm), so the voltage drop = 6V.

This is the voltage you use in the capacitance equation: Q = CV, NOT the voltage of the battery. Hopefully this clears things up.

 

[Majik]

Another way they could trick you is by giving you the following:

- Current in the circuit
- Resistance of Resistor 2 (the one closer to the battery)
- Voltage of Battery

Using this information alone, you can also find the voltage drop of the other resistor, since its the only other one in the circuit. All voltage drops by each resistor in the circuit must add up to the voltage of the battery itself. This is an oversimplification of one of Kirkoff Loop Rules.

Anyways, we know the battery voltage is 12V. Subtracting off the voltage of the second resistor, we can find the remaining voltage drop that must be consumed by the remaining resistor in the circuit diagram. This voltage drop is also shared with the capacitor because they are in parallel.

Here's the calculaton:

Total Voltage - Voltage Drop (of Resistor 2) = Voltage Drop (Resistor 1)
12V (Battery) - (3A)(2ohm) = 6V