Workdone in a Capacitor
The work done in a capacitor is the workforce required to move a unit positive charge from infinity to that point.
W = QV
W = F x d
Fd = Qv
Therefore F =\( \frac{Qv}{d} \)
But E = \( \frac{Qv}{d} \; \times \; \frac{1}{Q}\)
E = \( \frac{v}{d} \)
EnergyEnergy is the ability to do work. Energy exists in several forms such as heat, kinetic or mechanical energy, light, potential energy, and electrical energy. Units of Energy: The SI unit... More Stored in a Capacitor
When charging a capacitor, the average value of the potential difference is given by:
\( \frac{0 + v}{2} \)= \( \frac{1}{2} \scriptsize v \)
The workdone in charging the capacitor through an average potential difference = \( \frac{1}{2} \scriptsize v \) is
W= \( \frac{1}{2} \scriptsize v \; \times \; q \)
W = \( \frac{1}{2} \scriptsize qv \) ___________(1)
But c = \( \frac{q}{v} \) or v = \( \frac{q}{c} \)
W = \( \frac{1}{2} \scriptsize qv = \normalsize \frac{1}{2} \scriptsize q \; \times \normalsize \frac{q}{c} \)
W = \( \frac{1}{2} \frac{q^2}{c} \)
But q = cv
W = \( \frac{1}{2} \frac{(cv)^2}{c} \)
W = \( \frac{1}{2} \scriptsize CV^2 \)
Examples
A capacitor of 100µF is charged to a voltage of 16v. What is the energy stored in the capacitor.
Solution
C = 100µF = 1.0 x 10-4 F
V = 16v
W = \( \frac{1}{2} \scriptsize CV^2 \)
=\( \frac{1}{2} \scriptsize \; \times \; 1.0 \times 10^{-4} \; \times \; 16^2 \)
= 1.28 x 10-3J
Responses