capacitors

Thread Starter

kaz21

Joined Jun 11, 2011
5
How to derive a mathematical relationship for the voltage across a capacitor in the transient stage of charging the capacitor??The voltage source is a dc voltage source.
 

ErnieM

Joined Apr 24, 2011
8,377
A device that holds a charge Q and has voltage V across has a capacitance of:

\(C = \fra{Q(t)}{V(t)} \)

Solving for Q(t):

\(Q(t) = C V(t)\)

Noting that the total charge is the integral of current, or

\( \int I(t) dt = Q(t)\)

We can substitute for Q(t):

\( Q(t) = \int I(t) dt = C V(t)\)

If we differentiate by t we get the traditional relationship between voltage and current:

\( I(t) = C \fra {dV(t)} {dt}\)

Solving for \(dV(t) \):

\( dV(t) = \fra {1} {C} I(t) dt\)

Integrate both sides:

\( V(t) = \fra{1}{C}\int I(t) dt\) (limits of -∞ to t)

(I somehow suspect this is not the answer you were looking for.)
 
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