# Circuits: Current and Energy of Capacitor

In summary: Because the voltage on the capacitor is greater than the voltage on the generator, the current through the circuit is positive.

## Homework Statement

I am just looking for someone to verify that my approach is correct.
i(t) is the current through the capacitor and w(t) is the energy stored in the capacitor.

## Homework Equations

For a capacitor, the current through it can be related to the voltage drop across the capacitor:

$$i_c = C\frac{dv_c}{dt}$$

where C is the capacitance (a constant in this case).

## The Attempt at a Solution

The source voltage is given (as function of time) pictorially in figure 1.2.21. From Kirchoff's voltage law, we know that $v_c(t) = -v_{source}(t)$ Hence we can find the slope $dv_c/dt = -dv_{source}/dt$.

From the figure, from t = -1 to t =0, the slope of the source voltage is 2, hence the capacitor voltage's slope is -2. So the current through the capacitor is -2*C from t = -1 to 0.

I agree with that answer and approach.
It does seem that there are two extra minus signs in there, though. The voltage on the capacitor is the same as the voltage on the generator - no negative. In fact, when you put your voltmeter across the generator, you are also putting it across the capacitor. I guess there would be a negative sign necessary if you are switching the leads when you measure the voltage on the capacitor.

The slope on the graph is negative 2, not positive 2.

Delphi51 said:
I agree with that answer and approach.
It does seem that there are two extra minus signs in there, though. The voltage on the capacitor is the same as the voltage on the generator - no negative. In fact, when you put your voltmeter across the generator, you are also putting it across the capacitor. I guess there would be a negative sign necessary if you are switching the leads when you measure the voltage on the capacitor.

The slope on the graph is negative 2, not positive 2.

Hmmm... The slope on the source voltage graph is -2. That is why I think that the slope of the voltage of the capacitor is positive 2.

KVL implies that vsource + vcap = 0 yes?

We are looking at things slightly differently, both correct!
I would naturally put the red lead of the voltmeter on the top of the capacitor and the black on the bottom. Same for the source. You are going around the circuit in a clockwise direction, keeping the black lead ahead of the red, as you would when using the rule that the sum of the voltages around a closed loop is zero.

Anyway, the top of the capacitor is positive and the (positive) current flows down through it.

From t = 0 to t = 1, the slope of the source voltage is -2, hence the capacitor voltage's slope is 2. So the current through the capacitor is 2*C from t = 0 to 1.Your approach looks correct. Just a few notes:

- Make sure to specify the units for the current and voltage (e.g. amps and volts).
- It may be helpful to include a diagram or figure to illustrate the circuit and voltage/current changes over time.
- You can also include the equation for the energy stored in a capacitor: E = 1/2 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.
- It may also be useful to mention any assumptions or simplifications made in your approach, such as assuming ideal components or neglecting any resistances in the circuit.

## 1. What is a capacitor and how does it work?

A capacitor is an electronic component that stores energy in the form of an electric charge. It consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied to the capacitor, an electric field is created between the plates, causing one plate to accumulate a positive charge and the other to accumulate a negative charge. This stored charge can then be released when the capacitor is connected to a circuit, providing a surge of energy.

## 2. What is the difference between current and energy in a capacitor?

Current refers to the flow of electric charge through a circuit, while energy refers to the ability to do work. In a capacitor, current is the flow of charge into or out of the capacitor, while energy is stored in the capacitor's electric field. Current and energy are closely related, as the flow of current into a capacitor results in an increase in energy stored in the capacitor.

## 3. How does the amount of charge and voltage affect the energy of a capacitor?

The energy stored in a capacitor is directly proportional to the amount of charge that it can hold and the square of the voltage applied. This means that increasing either the amount of charge or the voltage will result in a higher energy storage capacity for the capacitor. Conversely, decreasing the charge or voltage will decrease the energy stored.

## 4. What is the formula for calculating the energy stored in a capacitor?

The energy stored in a capacitor can be calculated using the formula E = 1/2 * C * V^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts. This formula shows the direct relationship between energy and capacitance and the square of the voltage.

## 5. How does a capacitor affect the flow of current in a circuit?

A capacitor can affect the flow of current in a circuit by storing and releasing electrical energy. When the capacitor is fully charged, it blocks the flow of current in the circuit. However, when it is discharged, it can provide a surge of current. This can be useful in smoothing out fluctuations in a circuit's current and maintaining a steady flow of energy.

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