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Capacitor question
I'm not too sure about this question (attached is the diagram given)
If a voltage of V sin (omega. t) is applied to each of the components, write down an expression of the current in each.
Is it the derivative?
If a voltage of V sin (omega. t) is applied to each of the components, write down an expression of the current in each.
Is it the derivative?
current is the rate of flow of charge i.e. dQ/dt
We know that Q=CV
therefore
dQ/dt = C . dV/dt
therefore I = C. dV/dt
and we have an expression for V in terms of t.
do you see what you have to do?
We know that Q=CV
therefore
dQ/dt = C . dV/dt
therefore I = C. dV/dt
and we have an expression for V in terms of t.
do you see what you have to do?
dont forget that there are two capacitors in parallel, so take that into account before you do the above for them.
wgeipfm
and is it that
It = CV?
It = CV?
yes, that is correct.
That's wrong ...the differential is omegaVcos(omega t). The V in this expression is the peak voltage. The current is therefore C.omega.Vcos(omega t). This is the full expression for the current in each capacitor, as you have stated the question, where V is the peak voltage. Don't sub CV for I as you introduce another unknown into the answer (I, the peak current).
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