*Notes to a video lecture on http://www.unizor.com*

__Series LC Circuit__

Consider a circuit that contains an AC generator, an inductor of inductance

*and a capacitor of capacity*

**L***in a series.*

**C**The current

*going through an inductor is the same as the current*

**I**_{L}(t)*going through a capacitor. So, we will use an expression*

**I**_{C}(t)*for both.*

**I(t)**The electromotive force (EMF) generated by an AC generator depends only

on its own properties and can be described as a function of time

**t**

**E(t) = E**_{0}·sin(ωt)where

*is a peak voltage on the terminals of a generator,*

**E**_{0}*is an angular velocity of a rotor in radians per second.*

**ω***Inductance*

*of an inductor and*

**L***capacity*

*of a capacitor produce voltage drops*

**C***and*

**V**_{L}(t)*correspondingly.*

**V**_{C}(t)As we know, the voltage drop on an inductor is causes by self-induction

and depends on the rate of change (that is, the first derivative by

time) of a magnetic flux

*going through it*

**Φ(t)**

**V**d_{L}(t) =**Φ(t)/**d**t**Magnetic flux, in turn, depends on a current going through a wire of an inductor

*and the inductor's inductance*

**I(t)**

**L**

**Φ(t) = L·I(t)**Therefore, the voltage drop on an inductor equals to

**V**d_{L}(t) = L·**I(t)/**d**t**As we know, the amount of electricity

*accumulated in a capacitor is proportional to voltage*

**Q(t)***applied to its plates (that is, voltage drop on a capacitor). The constant*

**V**_{C}(t)*capacity*of a capacitor

**C**is the coefficient of proportionality (see lecture "Electric Fields" -

"Capacitors" in this course) that depends on a type of a capacitor

**C = Q(t)/V**_{C}(t)Therefore,

**Q(t)=C·V**_{C}(t)Knowing the amount of electricity

*accumulated in a capacitor as a function of time*

**Q(t)***, we can determine the electric current*

**t***in a circuit, which is a rate of change (that is, the first derivative by time) of the amount of electricity*

**I(t)**

**I(t) =**d**Q(t)/**d**t = C·**d**V**d_{C}(t)/**t**The sum of voltage drops on an inductor and a capacitor is supposed to be equal to EMF produced by an AC generator

*, which is the final equation in our system:*

**E(t)=E**_{0}·sin(ωt)

**V**d_{L}(t) = L·**I(t)/**d**t**

**I(t) = C·**d**V**d_{C}(t)/**t**

**E**_{0}·sin(ωt) = V_{L}(t) + V_{C}(t)To solve this system of three equations, including two differential ones, let's resolve the third equation for

*and substitute it in the second one.*

**V**_{C}(t)

**V**_{C}(t) = E_{0}·sin(ωt) − V_{L}(t)*[*

**I(t)=C·**d**]**

*E*_{0}·sin(ωt)−V_{L}(t)

*/**d*

**t**In the last equation we can differentiate each component and, using symbol

*for a derivative, obtain*

**'**

**I(t)=CωE**_{0}·cos(ωt)−C·V'_{L}(t)Together with the first equation from the original system of three

equations above we have reduced the system to two equations (again, we

use symbol

*for brevity to signify differentiation)*

**'**

**V**_{L}(t) = L·I'(t)

**I(t)=CωE**_{0}·cos(ωt)−C·V'_{L}(t)Substituting

*from the first of these equations into the second, we obtain one equation for*

**V**_{L}(t)*, which happens to be a differential equation of the second order (we will use symbol*

**I(t)***to signify a second derivative of*

**"***for brevity)*

**I(t)**

**I(t)=CωE**_{0}·cos(ωt)−CL·I"(t)or in a more traditional for differential equation form

**a·I"(t) + b·I(t) = E**_{0}·cos(ωt)where

**a = L/ω**

**b = 1/(Cω)**Without getting too deep into a theory of differential equations, notice that the one and only

**known**function in this equation that depends on time

*is*

**t***. It's second derivative also contains*

**cos(ωt)***. So, if*

**cos(ωt)***is proportional to*

**I(t)***, its second derivative*

**cos(ωt)***will also be proportional to*

**I"(t)***and we can find the coefficient of proportionality to satisfy the equation.*

**cos(ωt)**Let's try to find such coefficient

*that function*

**K***satisfies our equation.*

**I(t)=K·cos(ωt)**

**I'(t) = −ωK·sin(ωt)**

**I"(t) = −ω²K·cos(ωt)**Now our differential equation for

*is*

**I(t)**

**−a·ω²K·cos(ωt) + b·K·cos(ωt) = E**_{0}·cos(ωt)From this we can easily find a coefficient

**K**

**K = E**_{0}/(b−aω²)Since

*and*

**a=L/ω**

**b=1/(Cω)***{[*

**K = E**_{0}/*]*

**1/(Cω)***}*

**− Lω**In the previous lectures we have introduced concepts of

*capacitive reactance*

*and*

**X**_{C}=1/(Cω)*inductive reactance*

*. Using these variables, the expression for coefficient*

**X**_{L}=Lω*is*

**K**

**K = E**_{0}/(X_{C}−X_{L})Therefore,

**I(t) = E**_{0}·cos(ωt)/(X_{C}−X_{L})or

**I(t) = I**_{0}·cos(ωt)where

**I**_{0}= E_{0}/(X_{C}−X_{L})The last equation brings us to a concept of a

*reactance*of the LC circuit

**X**_{C}−X_{L}that is similar to

*resistance*of regular resistors.

Using a concept of

*reactance*, the last equation resembles the Ohm's Law.

Let's determine the voltage drops on a capacitor

*and an inductor*

**V**_{C}(t)*using the expression for the current*

**V**_{L}(t)*.*

**I(t)**Since

*and*

**Q(t)=C·V**_{C}(t)*, we can find*

**I(t)=Q'(t)***by integrating*

**V**_{C}(t)*.*

**I(t)/C**

**V**d_{C}(t) = ∫_{[0,t]}I(t)·**t/C =**

= I

= X= I

_{0}·sin(ωt)/(C·ω) == X

_{C}·E_{0}·sin(ωt)/(X_{C}−X_{L})

**V**

= −L·E

= −X_{L}(t) = L·I'(t) == −L·E

_{0}·sin(ωt)·ω/(X_{C}−X_{L}) == −X

_{L}·E_{0}·sin(ωt)/(X_{C}−X_{L})Let's check that the sum of voltage drops in the circuit

*and*

**V**_{L}(t)*is equal to the original EMF generated by a source of electricity.*

**V**_{C}(t)Indeed,

**V**

=(X

= E_{L}(t) + V_{C}(t) ==(X

_{C}−X_{L})·E_{0}·sin(ωt)/(X_{C}−X_{L})== E

_{0}·sin(ωt)*Summary*

EDF generated by a source of electricity

**E(t) = E**_{0}·sin(ωt)where

*is a peak voltage on the terminals of a generator,*

**E**_{0}*is an angular velocity of a rotor in radians per second.*

**ω**Alternating electric current in the circuit

**I(t) = I**_{0}·cos(ωt)where

**I**_{0}= E_{0}/(X_{C}−X_{L})

**X**_{C}= 1/(ω·C)

**X**_{L}= ω·LVoltage drop on a capacitor

**V**_{C}(t) = X_{C}·E_{0}·sin(ωt)/(X_{C}−X_{L})Voltage drop on an inductor

**V**_{L}(t) = −X_{L}·E_{0}·sin(ωt)/(X_{C}−X_{L})*Phase Shift*

Notice that

*cos(x)=sin(x+π/2)*. Graph of function

*y=sin(x+π/2)*is shifted to the left by

*π/2*relative to graph of

*y=sin(x)*.

Therefore, oscillations of the current

*in the LC circuit are ahead of oscillation of the EMF generated by a source of electricity*

**I(t)***by a*

**E(t)***phase shift*of

*π/2*.

Oscillations of the voltage drop on a capacitor

*in the LC circuit are synchronized (*

**V**_{C}(t)*in phase*) with generated EMF.

Notice that

*−sin(x)=sin(x+π)*. Graph of function

*y=sin(x+π)*is shifted to the left by

*π*relative to graph of

*y=sin(x)*.

Therefore, oscillations of the voltage drop on an inductor

*in the LC circuit are ahead of oscillation of the EMF generated by a source of electricity*

**V**_{L}(t)*by*

**E(t)***π*(this is called

*in antiphase*).