Analyzing an LRC circuit
For an object oscillating on a spring we added a resistive force
proportional (and opposite) to the velocity. This gave us:
ma = -kx -bv
Where b is known as the damping parameter.
We can write this as equation as:
| m |
| d2x
|  |
| dt2
|
|
= |
-kx |
- |
b |
| dx
| |
| dt
|
|
The solution to this equation is:
x(t) = A e-bt/2m cos(wt)
If there is no damping the solution reduces to the simple harmonic motion solution. Turn the damping on and the oscillations decrease as time goes by. The stronger the damping the faster the decrease.
| The angular frequency is given by |
w2 |
= |
| k
| |
| m
|
|
– |
| b2
| |
| 4m2
|
|
For an LRC circuit we follow a similar process. The equation we get is:
| L |
| dI
| |
| dt
|
|
= |
– |
| Q
| |
| C
|
|
– |
IR |
Write this as:
| L |
| d2Q
| |
| dt2
|
|
= |
– |
| Q
| |
| C
|
|
– |
R |
| dQ
| |
| dt
|
|
The solution is:
Q(t) = Qo e-Rt/2L cos(wt)
| w2 |
= |
| 1
| |
| LC
|
|
– |
| R2
| |
| 4L2
|
|
What happens to the current in the circuit when the inductance is increased? The equivalent question for an object oscillating on a spring is: what happens to the velocity when the mass is increased?
- The frequency of the current increases
- The frequency of the current decreases
- The amplitude of the current increases
- The amplitude of the current decreases
- Both 1 and 3
- Both 1 and 4
- Both 2 and 3
- Both 2 and 4
Increasing the inductance decreases the natural oscillation frequency of the circuit, and decreases the amplitude of the current much like increasing the mass of an oscillating object would slow down the oscillations.