# 6. Application: Series RC Circuit

An RC series circuit

In this section we see how to solve the differential equation arising from a circuit consisting of a resistor and a capacitor. (See the related section Series RL Circuit in the previous section.)

In an RC circuit, the **capacitor** stores energy between a pair of plates. When voltage is applied to the capacitor, the charge builds up in the capacitor and the current drops off to zero.

## Case 1: Constant Voltage

The voltage across the resistor and capacitor are as follows:

`V_R= Ri`

and

`V_C=1/Cinti\ dt`

Kirchhoff's voltage law says the total voltages must be zero. So applying this law to a series RC circuit results in the equation:

`Ri+1/Cinti\ dt=V`

One way to solve this equation is to turn it into a
**differential equation**, by differentiating throughout with
respect to *t*:

`R(di)/(dt)+i/C=0`

Solving the equation gives us:

`i={V}/Re^(-t"/"RC)`

**Important note:** We are assuming that the circuit has a **constant voltage** source, *V*. This equation does not apply if the voltage
source is **variable**.

The **time constant** in the case of an RC circuit is:

τ =

RC

The function

`i=V/Re^(-t"/"RC)`

has an **exponential
decay** shape as shown in the graph. The current stops
flowing as the capacitor becomes fully charged.

Graph of `i=V/R e^(-(t"/"RC))`, an exponential decay curve.

Applying our expressions from above, we have the following expressions for the voltage across the resistor and the capacitor:

`V_R=Ri=Ve^(-t"/"RC)`

`V_C=1/Cinti\ dt=V(1-e^(-t"/"RC))`

While the voltage over the resistor drops, the voltage over the capacitor rises as it is charged:

Graphs of `V_R=Ve^(-t"/"RC)` (in green) and `V_C=V(1-e^(-t"/"RC))` (in gray).

## Case 2: Variable Voltage and 2-mesh Circuits

We need to solve variable voltage cases in *q*,
rather than in *i*, since we have an integral to deal
with if we use *i*.

So we will make the substitutions:

`i=(dq)/(dt)`

and

`q=int i\ dt`

and so the equation in *i* involving an integral:

`Ri+1/Cinti\ dt=V`

becomes the differential equation in
*q***:**

`R(dq)/(dt)+1/Cq=V`

### Example 1

A series RC circuit with *R* = 5 W and *C* = 0.02 F
is connected with a battery of *E* = 100 V. At *t* = 0,
the voltage across the capacitor is zero.

(a) Obtain the subsequent voltage across the capacitor.

(b) As *t* → ∞, find the charge in the
capacitor.

### Example 2

Find the charge and the current for *t* > 0 in a
series RC circuit where *R* = 10 W, *C* = 4 × 10^{-3} F and *E* = 85 cos 150*t*
V.

Assume that when the switch is closed at *t* = 0, the
charge on the capacitor is -0.05 C.

### Example 3

In the RC circuit shown below, the switch is closed on
position 1 at *t* = 0 and after 1 τ is moved to position 2.
Find the complete current transient.

### Do not try this next one at home!

Here's a great Java-based RLC simulator (on an external site). He is actually making a coil gun. You can play with each of V, R, L and C and see the effects. Play and learn :-)

Sorry, it won't work on a mobile device.

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