Can voltage appear out of the blue?

  • Thread starter flyingpig
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In summary: , the problem is that the capacitors were not initially uncharged, which caused the incorrect calculations.
  • #1
flyingpig
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Homework Statement



This is a lab problem, I sort of got an idea, but the instructions are holding me back from answering it properly

Instructions in manual said:
If two capacitors with different nominal capacitances, C1 and C2 (both initially uncharged) are connected in seris with a power supply set to some 10.0V. What will the voltage across each capacitor be?

Alright I can do the math, but in reality something is wrong with my data

flyingpig's data said:
C1 = 3300uF
C2 = 2200uF

V1 = 4.4V
V2 = 5.6V

Predictions using math

V1 = 4V
V2 = 6V

So what is the problem? The capacitors were not charged, how did I manage to get 4.4V and 5.6V? I must have charged it up right? I know this is a stupid question but I need some confidence.
 
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  • #2
flyingpig said:

Homework Statement



This is a lab problem, I sort of got an idea, but the instructions are holding me back from answering it properly



Alright I can do the math, but in reality something is wrong with my data



So what is the problem? The capacitors were not charged, how did I manage to get 4.4V and 5.6V? I must have charged it up right? I know this is a stupid question but I need some confidence.

What method did you use? You need to use the equation that relates capacitance, voltage and charge...
 
  • #3
What do you mean "method", do you mean the prediction or the actual hands on measuring?
 
  • #4
flyingpig said:
What do you mean "method", do you mean the prediction or the actual hands on measuring?

Yeah, I just wanted to see the math you used for the predictions. If the math is right, then something must have been going on with the measurements.
 
  • #5
Series, so [tex]\Delta V = \Delta V_1 + \Delta V_2[/tex] and [tex]Q_{net} = Q_1 + Q_2[/tex]

[tex]C_{eq} = \frac{1}{C_1} + \frac{1}{C_2}[/tex]

[tex]C_{eq} = \frac{1}{3300} + \frac{1}{2200} = 1320uF[/tex]

[tex]Q_net = 1320uF/10V = 132uC[/tex]

[tex]\Delta V_1 = 132uC/3300uF = 0.04[/tex]

Wait...one second.. something is wrong with my units...
 
  • #6
It's not clear what it is you're problem is. The procedure said to charge the series connected capacitors via a 10V supply, and you're complaining because they got charged? :confused:
 
  • #7
gneill said:
It's not clear what it is you're problem is. The procedure said to charge the series connected capacitors via a 10V supply, and you're complaining because they got charged? :confused:

But they are just connected, it doesn't mean they are charged right...? Or in fact they are charged indirectly...?
 
  • #8
What procedure did you follow to make sure that the capacitors were initially uncharged? Capacitors can hold a charge for quite some time, and may have a residual charge from some other experiment(er).

Did you take note of the capacitor's value tolerance? It's usually marked on the body of the capacitor in some fashion.

Electrolytic capacitors have been known to self-charge to some extent over time (it's a quirk of the chemistry of the electrolyte and the plates).
 
  • #9
We touched the leads of the wires to uncharge it
 
  • #10
flyingpig said:
Series, so [tex]\Delta V = \Delta V_1 + \Delta V_2[/tex] and [tex]Q_{net} = Q_1 + Q_2[/tex]

Should be: [tex] Q_{net} = Q_1 = Q_2 [/tex]

[tex]C_{eq} = \frac{1}{C_1} + \frac{1}{C_2}[/tex]

[tex]C_{eq} = \frac{1}{3300} + \frac{1}{2200} = 1320uF[/tex]

Should be: [tex] \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} [/tex]

[tex]Q_net = 1320uF/10V = 132uC[/tex]

[tex]\Delta V_1 = 132uC/3300uF = 0.04[/tex]

Wait...one second.. something is wrong with my units...

Not just your units...
 
  • #11
[tex]C_1 \Delta V_1 = C_2 \Delta V_2[/tex]

[tex]C_1 (10V - \Delta V_2) = C_2 \Delta V_2 [/tex]

[tex]\frac{3300uF}{2200uF} = \frac{V_2}{10V - V_2}[/tex]

[tex]1.5(10V - \Delta V_2) = \Delta V_2[/tex]

[tex]\Delta V_2 = 6V[/tex]

[tex]10V - 6V = \Delta V_1 = 4V[/tex]

There we go
 

1. What is voltage and how does it appear?

Voltage is a measure of electrical potential difference between two points in an electrical circuit. It is created by the movement of charged particles, such as electrons, which carry energy and can be used to power devices.

2. Can voltage truly appear out of the blue?

No, voltage cannot appear out of the blue. It is always created by the flow of charged particles through a conductor, such as a wire or circuit. However, it may seem like it appears suddenly if there is a sudden change in the flow of charged particles, such as from a lightning strike or a power surge.

3. Is voltage always present in a circuit?

Yes, voltage is always present in a circuit as long as there is a source of electrical energy, such as a battery or power outlet. Even when a device is turned off, there is still voltage present in the circuit, but it is not actively being used.

4. Can voltage be dangerous?

Yes, voltage can be dangerous if not handled properly. High voltage can cause electric shock, burns, and even death. It is important to follow safety precautions when working with electricity and to never touch live wires or electrical equipment without proper training and protection.

5. How can voltage be measured?

Voltage can be measured using a device called a voltmeter, which is specifically designed to measure electrical potential difference. It is important to use proper safety measures and to follow instructions when using a voltmeter to avoid electric shock.

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