Relationship between volts and charge

AI Thread Summary
The discussion revolves around calculating the capacitance of a capacitor in a circuit with known resistors and a 9-volt battery. The initial challenge was determining the charge (Q) needed for the capacitance formula C=Q/V, as the user lacked sufficient information. A key insight was provided regarding the time constant and the exponential behavior of an RC circuit, which can help derive the capacitance from the current decay over time. The user ultimately resolved the issue by analyzing the decay graph to find the time constant, allowing for the calculation of capacitance. This highlights the importance of understanding the relationship between voltage, charge, and time in electrical circuits.
Jacques
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Hi,

I have been given an assignment question detailing a circuit with a few resistors in as well as a capacitator.

Further more the information given was the resitance of the resitors and that the battery had had a potential difference of 9 volts.

The question's part a was workign out the current in the circuit - no problem

The problem however comes in at part b with a question on what the capacitance of the capacitator is.

Now from my understanding of capacitance you need to know the charge and use the forumal C=Q/V but I'm missing Q.

I know V = joules per coulomb but I'm not seeing the relationship to find out what Q is.

Any pointers would be greatly appreciated.

Thanks

Jacques
 
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If the problem is referring to the DC situation after all voltages and currents have settled to their final values, then I agree that it sounds like you don't have enough information to solve for C. If they give you some time-related information, like you close a switch at time t=0 and by time t=5us the voltage across the capacitor has risen to half of its final value... or something like that, then you can use the exponential rise property of an RC circuit to figure out C.

Could you post the exact question, along with the circuit diagram?
 
current is the rate of flow of charge
 
Thanks Berkeman,

In fact they do give me a chart showing the how the current falls from an arbitary 10 amps to 2 amps in 2.5 miliseconds. the question based on the this asks for the time constant.
This question then is followed by the capacitance question.

thanks j
 
Berkeman,

Thanks for your input, I've got it now, I can read the time constant from the decay graph (time where 36.8% is left) and then I can work out the capacitance

thanks, again

j
 
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