# In the steady state, what is the charge q of the capacitor? How much work W is done by the voltage source…?

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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.

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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.)

• 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.