# 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

Continues below ⇩

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)`

Proof

We start with:

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

Divide through by *R*:

`(di)/(dt)+(1/(RC))i=0`

We recognise this as a first order linear differential equation**.**

Identify *P* and *Q*:

`P=1/(RC)`

Q= 0

Find the integrating factor (our independent variable is *t* and the dependent variable is *i*):

`intP dt=int1/(RC)dt` `=1/(RC)t`

So

`IF=e^(t"/"RC`

Now for the right hand integral of the 1st order linear solution:

`intQe^(intPdt)dt=int0 dt=K`

Applying the linear first order formula:

`ie^(t"/"RC)=K`

Since `i = V/R` when `t = 0`:

`K=V/R`

Substituting this back in:

`ie^(t"/"RC)=V/R`

Solving for *i* gives us the required expression:

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

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**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.

Answer

We will solve this 3 ways, since it has a constant voltage source:

1 and 2: Solving the DE in *q*, as:

- a linear DE and
- variables separable

3. Using the formulas `V_C=V(1-e^(-t"/"RC))` and `i=V/R e^(-t"/"RC`.

#### Method 1 - Solving the DE in *q*

From the formula: `Ri + 1/C int i dt=V`, we obtain:

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

On substituting, we have:

`5(dq)/(dt)+1/0.02q=100`

`5(dq)/(dt)+50q=100`

`(dq)/(dt)+10q=20`

We can solve this DE 2 ways, since it is variables separable or we could do it as a linear DE. The algebra is easier if we do it as a linear DE.

Solving this differential equation as a **linear DE**, we have:

`"IF"=e^(10t`

So `qe^(10t)=int (e^(10t))20 dt` `=2e^(10t)+K`

So `q=2+Ke^(-10t)`

Now, since `q(0)=0`, (that is, when `t=0`, `q=0`) this gives: `K=-2`.

So `q=2(1-e^(-10t))`.

Graph of `q=2(1-e^(-10t))`, solution of a differential equation.

As `t->oo`, `q->2\ "C"`.

Now,

`V_C=1/Cinti dt`

`=1/Cq`

`=1/0.02 2(1-e^(-10t))`

`=100(1-e^(10t)) `

For comparison, here is the solution of the DE using **variables separable**:

`(dq)/(dt)=10(2-q)`

`(dq)/(2-q)=10dt`

` -ln |2-q|=10t+K`

(We could continue and get the same expression as above.)

Since `t=0`, `q=0`, we have `K=-ln 2`.

So

`-ln |2-q|=10t-ln 2`

` -ln 2+ln |2-q|=-10t`

`(2-q)/2=e^(-10t)`

` 2-q=2e^(-10t)`

` q=2(1-e^(-10t))`

#### Method 2:

We use the formulae `V_C=V(1-e^(-t"/"RC))` and `i=V/R e^(-t"/"RC)`.

Now `1/(RC)=1/(5xx0.02)=10`

So:

`V_C=V(1-e^(-t"/"RC))` `=100(1-e^(-10t))`

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

`=100/5e^(-t"/"0.1) `

`=20e^(-10t) `

Now

`q=inti dt`

`=int20e^(-10t)dt`

`=-2e^(-10t)+K_1`

From here, we use `q(0)=0` and obtain: `K_1=2.`

So `q=2(1-e^(-10t))`, as before. Also, as `t->oo`, `q->2\ "C"`.

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### 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.

Answer

We will solve this 2 ways:

1. Solving in *q*.

2. Using Scientific Notebook.

[**NOTE: **We cannot use the formulae `V_C=V(1-e^(-t"/"RC))` and `i=V/re^(-t"/"RC)`, since the voltage source is **not constant**.]

From the formula: `Ri+1/Cint i dt=V`, we obtain:

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

Since `R=10`, `C=4xx10^-3`, and `V=85 cos 150t`, we have:

`10(dq)/(dt)+1/(4xx10^-3)q=85 cos 150t`

`10(dq)/(dt)+250q=85 cos 150t`

`(dq)/(dt)+25q=8.5 cos 150t`

Now, we can solve this differential equation in *q* using the **linear DE** process as follows:

`"IF"=e^(25t)`

`e^(25t)q=inte^(25t)8.5 cos 150t dt` `=8.5inte^(25t)cos 150t dt`

Then we use the integration formula (found in our standard integral table):

`inte^(mt)cos nt dt=` `(e^(mt))/(m^2+n^2)(m cos nt+n sin nt)`

We obtain:

`e^(25t)q=8.5inte^(25t) cos 150t dt`

`=8.5(e^(25t))/23125(25 cos 150t+` `{:150e^(25t) sin 150t)`

`=0.0092e^(25t) cos 150t+` `0.055e^(25t) sin 150t+K`

Dividing throughout by `e^(25t` gives:

`q=0.0092 cos 150t+` `0.055 sin 150t+` `Ke^(-25t)`

We now need to find `K`:

`q(0)=-0.05` means `K=-0.05-0.0092` `=-0.0592`

So this gives us:

`q=0.0092 cos 150t+` `0.055 sin 150t-` `0.0592e^(-25t)`

#### Method Using Scientific Notebook

We set up the differential equation and the initial conditions in a **matrix** (not a table) as follows:

`(dq)/(dt)+25q=8.5 cos 150t`

`q(0)=-0.05`

Choosing **Solve ODE - Exact** from the **Compute** menu gives:

Exact solution is:

`q(t)=0.0092 cos 150t+` `0.055 sin 150t-` `0.059e^(-25t)`

The graph for `q(t)` is as follows:

Graph of current `q` at time `t`. It's in steady state by around `t=0.12`.

We are also asked to find the current. We simply differentiate the expression for *q*:

`i` `=d/(dt)(0.0092 cos 150t+` `{:0.055 sin 150t-0.0592e^(-25t))` `=-1.38 sin 150t+` `8.25 cos 150t+` `1.48e^(-25t)`

The graph for *i*(*t*):

Graph of current `i` at time `t`. It's also in steady state by around `t=0.12`.

### 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.

Answer

At *t* = 0**, **the switch is at **Position 1**.

We note that `q(0)=0`.

`R(dq_1)/(dt)+1/Cq_1=V`

`500(dq_1)/(dt)+1/(0.5xx10^-6)q_1=20`

`(dq_1)/(dt)+4000q_1=0.04`

Using SNB to solve this differential equation, we have:

`q_1(t)=1/(100,000)(1-e^(-4,000t))`

**NOTE:** By differentiating, this gives us:

`i_1(t)=` `d/(dt)1/100000(1-e^(-4000t))=` `0.04e^(-4000t`

We need to find `tau`:

`tau=RC=500xx0.5xx10^-6=` `0.00025`.

Now, at `t=0.00025`, the charge will be:

`q_1(0.00025)`

`=[1/(100,000)(1-e^(-4,000t))]_(t=0.00025)`

`=6.3212xx10^-6`

#### Position 2

**At** `t=tau`**, switch at Position 2:**

Applying the formula `R(dq_2)/(dt)+1/Cq_2=V` again:

`500(dq_2)/(dt)+1/(0.5xx10^-6)q_2=-40`

**NOTE:** The negative voltage is because the current will flow in the opposite direction through the resistor and capacitor.

Once again, we solve using Scientific Notebook:

`500(dq_2)/(dt)+1/(0.5xx10^-6)q_2=-40`

` q_2(0)=6.3212xx10^-6`

Exact solution is:

`q_2(t)` `=-0.00002+2.6321xx10^-5e^(-4000t)`

So the current transient will be:

`i_2=(dq_2)/(dt)`

`=d/(dt)(-0.00002+2.6321xx10^-` `{:5e^(-4000t))`

`=-0.10528e^(-4000t)`

This expression assumes that time starts at `t=0`. However, we moved the switch to Position 2 at `t=0.00025`, so we need:

`i_2=-0.10528e^(-4000(t-0.00025))` `=-0.10528e^(1-4000t`

So the complete current transient is:

`i_1(t)=0.04e^(-4000t)` for `0 <= t <= 0.00025`

`i_2(t)=-0.10528e^(1-4000t)` for `t>0.00025`

The graph is quite interesting:

Graph of current `i` at time `t`.

### 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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