Charge in a RC Circuit

By Isabella Hoskins

This page gives a quantitative analysis of how to obtain the charge of a capacitor in a series RC Circuit with time.

The Main Idea

Below is a RC circuit in series, which contains an ideal battery with a emf, a light bulb (which has a resistance R), an uncharged capacitor with a capacitance C, and a switch. When the switch is closed, electrons flow from the negative end of the battery to the capacitor, where they accumulate on one of the plates of the capacitor, thus causing the plate to acquire a negative charge. This charge causes an electrostatic field that pushes the electrons on the second plate away from the second plate, which not only results a positive charge on the second plate, but also allows the current to move throughout the rest of the circuit. As a result, the initial brightness of the bulb would be the same brightness of the bulb in a circuit without a capacitor.

However, as capacitor becomes charge (i.e. more electrons pile up on the negative plate), an opposing electric field from the negative plate of the capacitor slows the current and the bulb begins to dim. Once the capacitor is fully charge, current stops flowing in the circuit and the bulb no longer shines.

Our goal is to obtain an equation that shows how the charge of the capacitor changes with time.

A Mathematical Model

Below are some important equations that will help achieve this equation:

[math]\displaystyle{ {V}_{round trip} = 0 }[/math]

[math]\displaystyle{ {V} = IR }[/math], where I is the current of the circuit and R is the resistance of the resistor.

[math]\displaystyle{ {Q} = CV }[/math], where Q is the charge of the capacitor, C is the capacitance of the capacitor, and V is the change in potential difference across the capacitor.

In a circuit loop, the change of potential difference has to be zero. The equation below describes the change of potential difference of the circuit above:

[math]\displaystyle{ {V}_{round trip} = emf-RI-Q/C = 0 }[/math]

Initially, the capacitor is not charge ([math]\displaystyle{ {Q} = 0 }[/math]), so the loop equation becomes:

[math]\displaystyle{ {V}_{round trip} = emf-RI = 0 }[/math] (for initial state)

[math]\displaystyle{ {I}_{0} = {\frac{emf}{R}} }[/math]. This current [math]\displaystyle{ {I}_{0} }[/math] resembles a current in a circuit that has no capacitor.

When the capacitor becomes fully charged, the current stops in the circuit. Thus, the RI becomes zero and the loop equation is now the following:

[math]\displaystyle{ {V}_{round trip} = emf-Q/C = 0 }[/math] (for final state)

[math]\displaystyle{ {Q} = emf*C }[/math]

This shows that the resistor depends on the emf of the battery and the capacitance. The resistor, then, determines the amount of time it takes to reach the circuit's final state.

Knowing this, it is possible to assume that the rate of charge Q of the plate is equal to [math]\displaystyle{ {I} = dQ/dt }[/math]. By taking the derivative of all the terms in the closed loop equation, the following equation is achieved:

[math]\displaystyle{ {V}_{round trip} = emf-RI-Q/C = 0 }[/math]

[math]\displaystyle{ {\frac{d(emf)}{dt}-R\frac{I}{dt}-\frac{dQ}{dt}/C = 0} }[/math]

[math]\displaystyle{ {R\frac{dI}{dt} = -I/C } }[/math] , since [math]\displaystyle{ {\frac{d(emf)}{dt}} }[/math] goes to zero since emf does not change with time and [math]\displaystyle{ {I}= \frac{dQ}{dt} }[/math]

[math]\displaystyle{ {R\frac{dI}{dt}(1/I) = -1/C } }[/math]

By integrating the right side of the equation from [math]\displaystyle{ {I}_{0} }[/math] to [math]\displaystyle{ {I} }[/math] and the right side of the equation from time 0 to t (the amount of time to get the capacitor fully charged), we get the following equation:

[math]\displaystyle{ R\int_{{I}_{0}}^I \frac{dI}{I}\,\mathrm{d}t=-(1/C) \int_0^t\,\mathrm{d}t }[/math]

[math]\displaystyle{ {lnI-ln{I}_{0} = {\frac{-1}{RC}}dt} }[/math]

[math]\displaystyle{ {ln(IR/V) = {\frac{-1}{RC}}dt} }[/math], since [math]\displaystyle{ {I}_{0} = {\frac{V}{R}} }[/math]

[math]\displaystyle{ {I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}} }[/math]

If both sides are exponentiate, the following is achieved

[math]\displaystyle{ {\frac{IR}{emf}} = {e^{\frac{-t}{RC}}} }[/math]

[math]\displaystyle{ {I} = {\frac{emf}{R}e^{\frac{-t}{RC}}={\frac{dQ}{dt}}} }[/math]

[math]\displaystyle{ {dQ} = \int_0^t{\frac{emf}{R}e^{\frac{-t}{RC}}\,\mathrm{d}t} }[/math]

[math]\displaystyle{ {Q} = {C(emf)(1-e^{\frac{-t}{RC}})} }[/math]. This is the charge in a series RC Circuit with respect to time.

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

The charge equation achieved above can also be achieved by thinking analytically. At [math]\displaystyle{ {t} = {0} }[/math], the capacitor acquires a new charge [math]\displaystyle{ {dQ}_{1} }[/math], which can be calculated as [math]\displaystyle{ {dQ}_{1}= {Q}_{0} + dQ = (emf/R)dt }[/math], since [math]\displaystyle{ {Q}_{0} = {0} }[/math]. After another small increment has passed, the capacitor gains a new charge [math]\displaystyle{ {dQ}_{2} }[/math], which is calculate as [math]\displaystyle{ {dQ}_{2} = {{Q}_{1} + \frac{dQ}{dt}}={{Q}_{1}+\frac{emf-{Q}_{1}/C}{R}} }[/math]. Notice that [math]\displaystyle{ {\frac{dQ}{dt}} }[/math] is smaller than the first time increment. In fact, if the charge of the capacitor is graphed, it looks like an exponential function.

[graph here]

Additionally, since it is known as Q increases, I should decrease due to the growing fridge field of a capacitor. As a result, its graph should look something like this:

[graph here]

By looking at the graph, one could make an educated guess that I should be given to a similar equation to [math]\displaystyle{ {I} = {{I}_{0}e^{-at}} }[/math]. The following is done to find the value of a:

[math]\displaystyle{ {I} = {\frac{dQ}{dt}}={\frac{emf-Q/C}{R}} }[/math]

[math]\displaystyle{ \frac{dI}{dt} = \frac{-1}{RC}\frac{dQ}{dt} = \frac{-1}{RC}I }[/math]

Because [math]\displaystyle{ \frac{d({I}_{0}e^{-at})}{dt} = {-aI} }[/math], [math]\displaystyle{ {a} = \frac{1}{RC} }[/math], thus resulting in the equation [math]\displaystyle{ {I} = {I}_{0}e^{\frac{-t}{RC}} }[/math], which is the same equation that was derived in the mathematical model section.

Examples

Be sure to show all steps in your solution and include diagrams whenever possible

Connectedness

1. How is this topic connected to something that you are interested in?
2. How is it connected to your major?
3. Is there an interesting industrial application?

History

Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.

See also

Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?

Further reading

Books, Articles or other print media on this topic

External links

Internet resources on this topic

References

This section contains the the references you used while writing this page