Asked by Buline
Suppose a capacitor having a capacitance of C fully charged to a voltage V and then discharged through a resistor R. Derive the equation (in terms of C,V,and R) that shows how long a time the capacitor will take to lose 45 % of its charge.
Answers
Answered by
Damon
Q = V C
V = i R
i = - dQ/dt
so
i R = Q/C
-dQ/dt = Q/(RC)
-dQ/Q = dt/(RC)
-ln Q = - t/(RC)
Q = k e^[- t/(RC) ] as we all knew
when t = 0 Q = Qi
so
Q = Qi e^[-t/(RC)]
when is Q = .55 Qi ??
.55 = e^[-t/(RC)]
ln .55 = -.5978 = -t/(RC)
so
t = 0.5978 R C
V = i R
i = - dQ/dt
so
i R = Q/C
-dQ/dt = Q/(RC)
-dQ/Q = dt/(RC)
-ln Q = - t/(RC)
Q = k e^[- t/(RC) ] as we all knew
when t = 0 Q = Qi
so
Q = Qi e^[-t/(RC)]
when is Q = .55 Qi ??
.55 = e^[-t/(RC)]
ln .55 = -.5978 = -t/(RC)
so
t = 0.5978 R C
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