# Can voltage appear out of the blue?

#### flyingpig

1. 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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#### berkeman

Mentor
1. 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...

#### flyingpig

What do you mean "method", do you mean the prediction or the actual hands on measuring?

#### berkeman

Mentor
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.

#### flyingpig

Series, so $$\Delta V = \Delta V_1 + \Delta V_2$$ and $$Q_{net} = Q_1 + Q_2$$

$$C_{eq} = \frac{1}{C_1} + \frac{1}{C_2}$$

$$C_{eq} = \frac{1}{3300} + \frac{1}{2200} = 1320uF$$

$$Q_net = 1320uF/10V = 132uC$$

$$\Delta V_1 = 132uC/3300uF = 0.04$$

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

#### gneill

Mentor
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?

#### flyingpig

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?
But they are just connected, it doesn't mean they are charged right...? Or in fact they are charged indirectly...?

#### gneill

Mentor
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).

#### flyingpig

We touched the leads of the wires to uncharge it

#### gneill

Mentor
Series, so $$\Delta V = \Delta V_1 + \Delta V_2$$ and $$Q_{net} = Q_1 + Q_2$$

Should be: $$Q_{net} = Q_1 = Q_2$$

$$C_{eq} = \frac{1}{C_1} + \frac{1}{C_2}$$

$$C_{eq} = \frac{1}{3300} + \frac{1}{2200} = 1320uF$$

Should be: $$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$

$$Q_net = 1320uF/10V = 132uC$$

$$\Delta V_1 = 132uC/3300uF = 0.04$$

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

#### flyingpig

$$C_1 \Delta V_1 = C_2 \Delta V_2$$

$$C_1 (10V - \Delta V_2) = C_2 \Delta V_2$$

$$\frac{3300uF}{2200uF} = \frac{V_2}{10V - V_2}$$

$$1.5(10V - \Delta V_2) = \Delta V_2$$

$$\Delta V_2 = 6V$$

$$10V - 6V = \Delta V_1 = 4V$$

There we go