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

Question

In an RC circuit (refer to my drawing):  
______VVVV(R)____|  
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||  
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||(capacitor)EMF  
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||_____/(switch)_____|
In the steady state, the charge on the capacitor remains constant as no current flows through it.

What is the charge ( q ) on the capacitor in the steady state?
   Express your answer in terms of the given variables: EMF (the electromotive force), ( R ) (resistance), and ( C ) (capacitance).

How much work ( W ) is done by the voltage source by the time the steady state is reached?

I attempted to solve these but couldn't arrive at the correct answers. Can someone help clarify this? Thank you!

Anonymous · Accepted Answer

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.)

Anonymous · Answer

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.

Vella O'Hara

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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