Solving an Electric Circuit: Potential Difference & Capacitor Charge

In summary, the question is about a circuit with an EMF source, two capacitors, and a switch. When the switch is open, the potential difference between points a and b is 0V. When the switch is closed, the charge on the capacitors remains the same and the potential difference between points a and b is still 0V. This is because the capacitors are in series and the charges and voltages across them do not change.
  • #1
Jay9313
40
0

Homework Statement


I have this circuit here.. http://s1167.photobucket.com/albums/q639/jay9313/?action=view&current=EM.jpg
The question says that when switch S is open in the circuit shown above, the potential difference between points a and b would be..
A) 0V
B) 6V
C) 12V
D) 24V
E) 36V

Part 2 says when switch S is closed, the carge on the 3 microfarad capacitor would be..
A) 0 μF
B) 12 μF
C) 18 μF
D) 36 μF
E) 72 μF

Homework Equations





The Attempt at a Solution



My professor ways that the answer is A. I need somebody to help me understand why.
 
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  • #2
Jay9313 said:

Homework Statement


I have this circuit here.. http://s1167.photobucket.com/albums/q639/jay9313/?action=view&current=EM.jpg

The question says that when switch S is open in the circuit shown above, the potential difference between points a and b would be..
A) 0V
B) 6V
C) 12V
D) 24V
E) 36V

Homework Equations



The Attempt at a Solution



My professor ways that the answer is A. I need somebody to help me understand why.
Where's the switch ?

Also, the rules for Homework Help in this Forum state:
You MUST show that you have attempted to answer your question in order to receive help.​

So, what do you think the answer should be --- and why ?
 
  • #3
I redrew it and added the switch, and I don't know how he got 0V. My logic was that all of the current would go through the capacitors, so there would be a potential difference of 36V, the same as the EMF source.
 
  • #4
Jay9313 said:
I redrew it and added the switch, and I don't know how he got 0V. My logic was that all of the current would go through the capacitors, so there would be a potential difference of 36V, the same as the EMF source.

Addressing the first part, A is the correct answer.

You may assume that the capacitors charge up very quickly and both capacitors are fully charged.

Take the negative terminal of the power source as being ground (0V).

What is the potential at a? Hint: voltage divider.

What is the potential at b? Hint: what do you know about capacitors in series? Compare the charges across the plates of each capacitor. Now compare the voltages across the plates of each capacitor.
 
  • #5
Oh, ok. It's 0v because both drop 24V in the first resistor or capacitor! Nnow what about the second part, where do I even start for that?
 
  • #6
Is it just 72 μC?
 
  • #7
Jay9313 said:
Is it just 72 μC?

Yes, since a and b are equipotentials. Closing that switch makes no difference. The capacitors don't discharge and the charges (and voltages) across both of them don't change.
 
  • #8
Curious3141 said:
Yes, since a and b are equipotentials. Closing that switch makes no difference. The capacitors don't discharge and the charges (and voltages) across both of them don't change.

Oh thank you so much, you helped me understand it so much =D
 

1. What is potential difference in an electric circuit?

Potential difference, also known as voltage, measures the difference in electric potential energy between two points in a circuit. It is measured in volts (V) and is responsible for the flow of electric current.

2. How do you calculate potential difference in a circuit?

Potential difference can be calculated by dividing the amount of energy transferred (in joules) by the amount of charge (in coulombs). This can be represented by the equation V = W/Q, where V is voltage, W is work or energy, and Q is charge.

3. What is a capacitor and how does it affect potential difference in a circuit?

A capacitor is an electronic component that stores electric charge. It consists of two conductive plates separated by an insulating material. When a potential difference is applied to a capacitor, it stores the charge on its plates, thus affecting the potential difference in the circuit.

4. What is the relationship between potential difference and current in a circuit?

According to Ohm's Law, potential difference is directly proportional to current in a circuit. This means that as the potential difference increases, the current also increases, and vice versa. This relationship is represented by the equation V = IR, where V is voltage, I is current, and R is resistance.

5. How can potential difference be controlled in a circuit?

Potential difference can be controlled by using different components such as resistors, capacitors, and voltage regulators. These components can either increase or decrease the potential difference in a circuit, depending on their properties and arrangement. Additionally, the voltage source can also be adjusted to control the potential difference.

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