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Theory of LCR Experiment
1 The Components
Resistor A resistor obeys Ohms law V = IR. Capacitor
A charge Q is stored on the plates. In order to hold the charge on the plates a voltage V is required
Capacitance C =
Q
V
or re-arranged V =
Q
C
Now current is charge per unit time so Q =
∫ Idt
Hence VC =
C
∫ Idt
Inductor
An inductor consists of “coils” through which a current I flows and causes a magnetic field. Now a changing magnetic field through a set of coils causes a “back-emf”, i.e. a voltage given by Faradays/Lenzs law
V ∝ dφ dt where φ is the magnetic flux through the coils. The amount of flux is given by φ = LI where L is the inductance. By the appropriate definition of units
VL = L dI dt
2 Kirchoffs Laws
- First law — conservation of charge — currents at any point in a circuit should sum to zero (incoming are positive, outgoing are negative).
- Second law — conservation of energy — sum of voltages around a closed “mesh” must equal total voltage applied to the mesh.
3 Differential Equations for Current
For a simple (series) mesh of a resistor, an inductor and a capacitor Kirchoff’s second law gives
E = IR + L
dI dt
C
∫ Idt
which if we differentiate w.r.t. time gives us
dE dt
= L
d^2 I dt^2
+ R
dI dt
C
I
Note for DC we have dE dt = 0, which is also true on the flat part of a square wave.
4 The RC Circuit
If we examine the series RC circuit on its own we have
E = VR + VC = IR +
C
∫ Idt dE dt
= R
dI dt
C
I
0 = R
dI dt
C
I because dE dt
dI dt
RC
I
∫ (^) dI I
RC
∫ dt
works perfectly well for this differential equation. However it is not all of the solution. If we can find a solution to the equation
0 = IR + L dI dt
then we can add it to IS and it will also be a solution to the differential equation. This last differential can be solved by
∫ (^) dI I
R
L
∫ dt
⇒ loge I = loge A −
R
L
t
I = A exp
( − t τ
)
where τ = L/R is the time constant. Thus the full solution
I =
E 0
R
( − t τ
)
will satisfy the differential equation. However we still need to know what the value of A is. If we look at the situation at t = 0 then we must have
I(t = 0) = 0 =
E 0
R
+ A
A = −
E 0
R
⇒ I(t) =
E 0
R
( 1 − exp
( − t τ
))
Now VL(t) = L dI dt
= −L
E 0
R
( −
τ exp
( − t τ
))
= E 0 exp
( − t τ
)
