Drop Capacitor
Drop Capacitor
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Can I find the charge in a capacitor like this?
All I am given is the current. Can I just find the charge in the capacitor like this:
I(t) = C dv/dt
integrate [ I(t) dt ] = C dv
Since I am not given the voltage drop across the capacitor can I just totally ignore the right hand side of the equation and just integrate the left hand side to find the charge stored in the capacitor?
Or is there some way to find the voltage drop across it without knowing the charge?
I would not have thought so, but maybe you are not expected to find numerical values for it but simply an expression for it!
I(t) = C dV/dt
integral of [ I(t) dt ] = C intergral of [dV]
Integral of [dQ/dt dt]= C x integral of [dV]
Integral of [dQ]= C x integral of [dV]
Q = CV