Energy of Capacitor in RC Circuit

[buttermellow7]

Homework Statement

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.

Homework Equations

$$V(t)=V_0(1-e^{-t/\tau}$$
$$U=1/2CV^2$$
$$P=IV=V^2/R=\dot{E}$$

The Attempt at a Solution

$$U(\infty)=1/2CV_0^2=\frac{QV_0}{2}=\frac{V_0^2}{2}$$

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

 

[willem2]

$$\frac { Q V_0 } {2}$$

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:

$$I(t) = \frac{dQ(t)}{dt} = C \frac{dV(t)}{dt}$$

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

$$V(t) = V_R + V_C = V_o$$

Since the voltage across the capacitor and resistor always sums to $$V_o$$ (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
$$C \int_0^{t_o} \frac{d V(t)}{d t} V_o d t$$

The fundamental law of calculus says this is equal to:

$$C V_o \left( V(t_o) - V(0) \right)$$

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

$$C V_o^2$$

Which is twice the energy stored in the capacitor.

 

[rwisz]

I would like to clarify a few things about the energy of a capacitor in an RC circuit.

Firstly, the energy stored in a capacitor is given by the equation U=1/2CV^2, where C is the capacitance and V is the voltage across the capacitor. This energy is proportional to the square of the voltage, not just the voltage itself.

In an RC circuit, when the switch is closed, the capacitor starts to charge and the voltage across it starts to increase. As time goes by, the voltage across the capacitor approaches the voltage of the battery, but it never reaches it completely. This is because as the capacitor charges, the current through the circuit decreases and eventually becomes zero when the capacitor is fully charged. Therefore, the voltage across the capacitor never reaches the full voltage of the battery.

Now, let's consider the energy supplied by the battery. At t=0, the battery supplies a voltage V0 and the current through the circuit is at its maximum. As time goes by, the voltage across the capacitor increases and the current decreases. At t=infinity, the voltage across the capacitor is equal to the voltage of the battery, but the current is zero. This means that the energy supplied by the battery is given by U=IV0t, where I is the initial current and t is the time it takes for the capacitor to charge.

Now, if we compare the energy stored in the capacitor (U=1/2CV^2) to the energy supplied by the battery (U=IV0t), we can see that they are not equal. This is because the voltage across the capacitor never reaches the full voltage of the battery. Therefore, only a fraction of the energy supplied by the battery is stored in the capacitor.

To calculate this fraction, we can use the equation U=1/2CV0^2 and substitute V0 with the voltage across the capacitor at t=infinity, which is equal to V0(1-e^-t/tau). This gives us U=1/2CV0^2(1-e^-t/tau)^2. If we divide this by the energy supplied by the battery (U=IV0t), we get U/U=1/2CV0^2(1-e^-t/tau)^2/IV0t. Simplifying this expression, we get U/U=1/2(1-e^-t/tau)^2/t. At