RC Circuit(see my drawing)
______VVVV(R)____
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| =(capacitor)
EMF |
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|_____/(switch)_____|
In that steady state, the charge of the capacitor is not changing.
In the steady state, what is the charge q of the capacitor?
Express your answer in terms of any or all of EMF, R, and C.
How much work W is done by the voltage source by the time the steady state is reached?
I was unable to answer these two.
2 Answers
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The definition of capacitance is given by Q = V * C where q is the charge stored and C is the capacitance.
So if the the emf is called V and if the capacitor is fully charged then there is no current flowing and the amount of charge in the capacitor is simply VC
In the second question there is a trick intended.
Energy = Emf * Charge
so E= VQ= V * V * Q = V^2 C
This is the energy supplied by the battery. Some has gone into the capacitor and some has been lost in the form of heat in the resistor.
( incidentally the energy IN the capacitor is 1/ C V^2 so half of the energy was lost in the resistor.)
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in case you recommend “stable-state” as that not extra cutting-edge would be flowing into it: The capacitor won’t straight away gain that situation. it is going to require a while till now that situation is finished, on the grounds that’s being charged exponentially and as a function of time.
