Current in a RC circuit

When a capacitor is connected in a circuit, it varies the current. This wiki will discuss the varying current in a circuit while the capacitor is charging.

The Main Idea

Now that we have an understanding of steady state current, we can begin to examine the current in a RC circuit.

The current in a RC circuit differs from the current in a simple circuit because the capacitor acquires and releases charge; this varies the current.

This a graphical representation of the changing current and voltage on a capacitor with respect to time.

A Mathematical Model

The graph presented in the previous section is representative of a exponential equation. The current across a capacitor falls of like [math]\displaystyle{ I = ({\frac{V}{R}})e^{\frac{-t}{RC}} }[/math] where V is the voltage driving the current, R is the resistance of the circuit, t is time, and C is the capacitance of the capacitor.

As can be seen, as t approaches infinity, I approaches 0. In plain english, if the circuit is closed for a "very long time" the current in the circuit will approach zero.

A Computational Model

To aid in visualization, I will provide a "way of thinking".

When the switch is closed, the circuit is complete and the current begins to run. Remember current is just mobile electrons. The excess of electrons on one plate (along with the the consequential deficit of electrons on the other plate) provide a surface charge that repels the incoming electrons. Decreasing the flow of electrons decreases the current at the rate provided in the previous section.

Examples

[math]\displaystyle{ V_s }[/math] = 5 V
R = 2 Ohms
C = 1 Farad

Simple

What is the current in the circuit immediately after the switch is closed?

SOLUTION

We can solve this problem mathematically or analytically.

Mathematically:

Looking at the equation for current in an RC circuit, [math]\displaystyle{ I = ({\frac{V}{R}})e^{\frac{-t}{RC}} }[/math] , and with the knowledge that [math]\displaystyle{ t \approx 0 }[/math], we can see that the term [math]\displaystyle{ e^{\frac{-t}{RC}} \approx 1 }[/math] . Therefore, the current in the circuit is the same as it would be as if the capacitor wasn't there. [math]\displaystyle{ I = {\frac{V}{R}} }[/math] .

Analytically:

Think about it. What is it about an RC circuit that causes the current to decrease with time? It's simply the buildup of surface charge on the plate of the capacitor. This creates an electric field that opposes the electric field thats driving the original current. This buildup will go on until the opposing electric fields are equal and [math]\displaystyle{ E_{net} = 0 }[/math].

We will consider the current immediately after we close the switch. Electrons move quickly, but not instantaneously. Because of this, immediately after the switch is closed, surface charge hasn't had the time to buildup. So the electrons are only being driven by the original electric field. So we can see that [math]\displaystyle{ I = {\frac{V}{R}} }[/math]

Middling

Originally, the connecting wires in the circuit have electron mobility [math]\displaystyle{ \mu = .00006 }[/math].

What is the new electron mobility, [math]\displaystyle{ \mu_{0} }[/math] , when [math]\displaystyle{ t = 3 }[/math]?

Solution

[math]\displaystyle{ \mu = \mu_{0} }[/math]

The inclusion of a capacitor in a circuit does cause current to vary with time but not by affecting the properties of the connecting wires. The capacitor changes the current by providing a growing electric field that opposes the original electric field.

Difficult

What is the current in the circuit 3 seconds after closing the switch?

SOLUTION

We know that [math]\displaystyle{ I = ({\frac{V}{R}})e^{\frac{-t}{RC}} }[/math], therefore;

[math]\displaystyle{ I = ({\frac{5}{2}})e^{\frac{-3}{2*1}} }[/math]

[math]\displaystyle{ I = ({\frac{5}{2}})e^{-1.5} }[/math]

[math]\displaystyle{ I = .557 A }[/math]

See also

Steady State

Surface Charge Distributions

RC