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- Separation of variables
- Functions of a linear function of x

For any circuit current = steady state current + transient current

i = I

The first thing to do is establish an equation that describes the circuit so we can manipulate it. We know the voltage across the supply = the sum of voltages across the other components.

The volt-drop across
the inductor V_{L} = L |
di dt |

So the supply voltage E = iR + L | di dt |

Separation of variables

I am going to manipulate this so I get di on one side (with any expressions with 'i' in it) and dt on the other so I can integrate both sides:

E = iR + L | di dt |

E - iR = L | di dt |

E - iR L |
= | di dt |
now I will transpose this so di is on one side with the 'i' part and dt with L |

1 E - iR |
di = | dt L |
I can put integral signs on both |

∫ | 1 E - iR |
di = | ∫ | 1 L |
dt |

Functions of a linear function of x (or in this case, 'i')

Notice that | ∫ | 1 E - iR |
di is like | ∫ | 1 i |
di and the E - iR part is linear (a line). |

I am going to set u = E - iR so I can integrate 1/u quite easily. I have to now change the di part. I can find this by differentiating u:

u = E - iR so | du di |
= -R |

re-arranging this means that - | du R |
= di |

So, to find the solution to | ∫ | 1 E - iR |
di, we swap E-iR with u and di with - | du R |

This gives us | ∫ | 1 u |
du -R |
which I can
re-arrange to put all the 'u' expressions on one side and the constants outside of the integral (to make it a bit easier) |

- | 1 R |
∫ | 1 u |
du,
this is integrated to become: |

ln(u) -R |
+ K, where K is just some constant. We can substitute back E - iR for u and get: |

ln(E
- iR) -R |
+ K |

Let's remind ourselves of this equation again:

∫ | 1 E - iR |
di = | ∫ | 1 L |
dt |

- | ln(E - iR) R |
+ K = | ∫ | 1 L |
dt, and we can find the second part to be |

- | ln(E - iR) R |
= | t L |
+ K (this is the general solution) |

Finding the unknown constant

To find this, we need boundary conditions. We know that when the circuit is first switched on (at time = 0 or t = 0), the current must also be zero (i = 0) because the inductor starts off with an equal and opposite voltage. We can put these in to the general solution:

i = 0 when t = 0

- | ln(E - 0R) R |
= | 0 L |
+ K |

- | ln(E) R |
= | K so we can put this back in to the general solution, in replacement for K |

- | ln(E - iR) R |
= | t L |
- | ln(E) R |

Transposing for 'i'

ln(E) R |
- | ln(E
- iR) R |
= | t L |

1 R |
ln | ( | E E - iR |
) | = | t L |

ln | ( | E E - iR |
) | = | Rt L |

i = | E |
( | 1 - e | - | Rt L | ) |

R |

Notice that if you multiply out the bracket, this shows us the transient and steady state currents:

i = I

Steady state current = | E | and the transient current = - e | - | Rt L |

R |

E - iR - L | di dt | = 0 |

E - R•i(t) - L | d dt | i(t) = 0 |

E s |
- Ls•i(s) -R•i(s) = 0 |

E s |
= i(s)(Ls - R) |

i(s) = | E s(Ls - R) |

i(t) = | E |
( | 1 - e | - | Rt L | ) |

R |

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information to these pages? You can contact me at the bottom of the home page. |