Energy of Capacitor in RC Circuit

buttermellow7

1. The problem statement, all variables and given/known data
The circuit consists of a battery, switch, resistor, and capacitor connected in a loop in series. When the switch is closed at time t=0, a time-dependent current flows out of the battery and charges the capacitor. The capacitor is fully charged at time t=infinity. Show that only half of this energy supplied by the battery ends up being stored in the fully charged capacitor.

2. Relevant equations
[tex] V(t)=V_0(1-e^{-t/\tau} [/tex]
[tex] U=1/2CV^2 [/tex]
[tex] P=IV=V^2/R=\dot{E} [/tex]

3. The attempt at a solution
[tex]U(\infty)=1/2CV_0^2=\frac{QV_0}{2}=\frac{V_0^2}{2} [/tex]

I'm confused about where my extra factor of V₀ is coming from. Do I need to use the power equation?

willem2

[tex] \frac { Q V_0 } {2} [/tex]

is the correct expression for the energy stored in the capacitor. The next step is wrong.

Try to show that the total energy delivered by the battery is twice as large.

cupid.callin

Energy supplied by battery up to any point is qV_o
here q is total charge supplied up to that point

use this and qV_o/2 for capacitor at t = ∞

creillyucla

The instantaneous power delivered to a component is given by

I * V

where the I is the current through the component, and V is the voltage across the component.

Note that the current through the resistor and capacitor are the same, because of Kirchoff's current law:

[tex]
I(t) = \frac{dQ(t)}{dt} = C \frac{dV(t)}{dt}
[/tex]

where Q(t) is the charge on the capacitor at a time t, and the voltage [tex]V_R[/tex] across the resistor plus the voltage [tex]V_C[/tex] across the resistor is:

[tex]
V(t) = V_R + V_C = V_o
[/tex]

Since the voltage across the capacitor and resistor always sums to [tex]V_o[/tex] (Kirchoff loop law).

The total energy delivered to the component is given by the time integral of this expression, since power is the time derivative of the energy. So the total energy delivered to the RC circuit is
[tex]
C \int_0^{t_o} \frac{d V(t)}{d t} V_o d t
[/tex]

The fundamental law of calculus says this is equal to:

[tex]
C V_o \left( V(t_o) - V(0) \right)
[/tex]

There is no initial voltage across the capacitor so [tex] V(0) = 0[/tex]. [tex]t_o[/tex] is infinite so [tex]V(t_o) = V_o[/tex]. So the total energy expended to charge the capacitor is:

[tex]
C V_o^2
[/tex]

Which is twice the energy stored in the capacitor.