1. Chapter 7 RL and RC first order circuits
Transient Circuits
Chapter 7
RL and RC first order circuits

2. First Order RL and RC circuits
L and C components have the ability to store energy in magnetic or electric fields
Circuits that have only 1 L or 1 C have a first order response to a sudden influx or release of the energy.
When the energy is suddenly released, the response is called the natural response.
When the energy is suddenly supplied, to the circuit, it is called the step response.
Since there is only 1 element with a differential characteristic the differential equation is first order.

3. RL circuit A switch is drawn as:
Depending on the circuit it either opens or closes to release charge.
Assume a switch is closed/open until the circuit is settled to its eventual steady state. Then it is instantly changed and the charge moves.
𝑅𝑖+𝐿 𝑑𝑖 𝑑𝑡 =0
𝑣 𝑡ℎ = 𝑅 𝑡ℎ 𝑖+𝐿 𝑑𝑖 𝑑𝑡

4. Circuit Natural Response
If a circuit with an inductor is fully charged for a long time and then the charge is suddenly released the circuit response is called the natural response.
The charge dissipates in the resistance of the circuit and decays exponentially to 0.
For example if the circuit shown is charged for a long time prior to t=0, then the switch is open. The natural response as the stored charge flows in R
Write a circuit equation around the loop on the right.
𝑅𝑖+𝐿 𝑑𝑖 𝑑𝑡 =0

5. Natural Response
To solve the equation, separate the variables and Integrate both sides. Then the response is i(t) is a decaying exponential.
𝑅𝑖+𝐿 𝑑𝑖 𝑑𝑡 =0 𝑑𝑖 𝑖 = −𝐿 𝑅 𝑑𝑡 𝑖(𝑡 0 ) 𝑖(𝑡) 𝑑𝑖 𝑖 = −𝑅 𝐿 𝑡 0 𝑡 𝑑𝑡 ln 𝑖 𝑡 𝑖 𝑡 = −𝑅 𝐿 t 𝒊 𝒕 =𝒊( 𝒕 𝟎 ) 𝒆 −𝑹𝒕 𝑳

6. Voltage at t = 0 The voltage at t=0 undergoes an instantaneous change.
Just before we release the switch the inductor voltage (due to no change in current) at t=0- is 0
Once the switch is closed and the circuit releases the energy (natural response) the current is changing the voltage at t= 0+ is v= I0R
The voltage at exactly t = 0 is unknown.
The voltage is also a decaying exponential over time.

7. Power in the RL circuit 𝑝=𝑣𝑖=𝑖2𝑅= 𝑣2 𝑅 𝑝= 𝐼 0 2 𝑅 𝑒 −2𝑅𝑡 𝐿
The power can be expressed in a variety of ways.
The power in the resistor is:
The energy delivered to the resistor after the switch is thrown is:
𝑤= 0 𝑡 𝑝 𝑑𝑥= 0 𝑡 𝐼 0 2 𝑅 𝑒 −2𝑅𝑡 𝐿 𝑑𝑥⇒ 1 2 𝐿 𝑖 − 𝑒 −2𝑅𝑡 𝐿
The second term goes to 0 as t gets larger.
Energy is: 𝑤= 1 2 𝐿 𝑖 0 2
𝑝=𝑣𝑖=𝑖2𝑅= 𝑣2 𝑅
𝑝= 𝐼 0 2 𝑅 𝑒 −2𝑅𝑡 𝐿

8. Time constant for the RL circuit
The time constant term is the reciprocal of the coefficient of t in the exponential term.
The term: 𝑒 −𝑅𝑡 𝐿 relates how the current or voltage falls off as a function of time.
The coefficient of t: 𝑅 𝐿 is the rate at which it falls off.
The reciprocal of this term is called the time constant,  = 𝑳 𝑹

9. Elapsed time vs. fraction of initial value
t/tau
exp(-t/tau) 1 2 3 4 5 10
4.54E-05
Note after ~5 the response is down to of initial max

10. Assessment Problem 7.1a.
What is the initial value of I at t=0+:

11. Assessment Problem 7.1a. “Thevenize” around the inductor.
Use Source transformations from left side and simplify to inductor:
Transform 120V to a 40 A in parallel with 3
Resistors 3  || 30  becomes 2.73 
Another source transformation 109.1V in series with 
Add the 6 becomes  in series with 109.1V
a last source transformation leaves 12.5 A current source
And resistors in parallel 8.73  and 2  1.6 
With the inductor now the initial inductor current is 12.5 A

12. Assessment Problem 7.1a.     

13. Capacitor Natural Response
Similar to the inductor circuit we charge up the capacitor,
Instead of soring the energy in magnetic flux, the capacitor stores energy in electric flux.
The capacitor voltage is stored until a path for he current allow the charge to dissipate. When the capacitor is fully charged it appears as an open circuit, stopping the flow of current until the switch is thrown at time t=0.
Vg appears across the capacitor at t=0-
When the switch is thrown at t=0+ the capacitor starts at Vg and dissipates through the resistor R to 0 V.
The time characteristic is exponential similar to the inductor but the time constant is dependent on R and C.

14. Analysis for the natural response of C
Write a loop equation
𝐶 𝑑𝑣 𝑑𝑡 + 𝑣 𝑅 =0
Use the same techniques for voltage in C as was used for current in L previously
𝐶 𝑑𝑣 𝑑𝑡 =− 𝑣 𝑅
𝑑𝑣 𝑣 = 𝑑𝑡 RC integrate both sides

15. Natural response of the Capacitor
We have the result for voltage response
V(0) = Vg in this case
Now the time consent is  = RC
Now we can devise an expression for current, power and energy
𝑣 𝑡 =𝑣(0) 𝑒 −𝑡 𝑅𝐶

16. Current, Power and Energy
𝑖 𝑡 = 𝑣(𝑡) 𝑅 = 𝑣 0 𝑅 𝑒 −𝑡 𝑅𝐶 𝑡≥0+ current 𝑝 𝑡 =𝑣𝑖= 𝑣 0 2 𝑅 𝑒 −2𝑡 𝑅𝐶 𝑡≥0+ power 𝑤 𝑡 = 0 𝑡 𝑝 𝑥 𝑑𝑥 = 0 𝑡 𝑣 0 2 𝑅 𝑒 −2𝑥 𝑅𝐶 𝑑𝑥 energy 𝑤 𝑡 = 1 2 C 𝑣 0 2 (1− 𝑒 −2𝑡 𝜏 ) = 1 2 C 𝑣 0 2 at t= 0

17. Assessment Problem 7.3
Use source transformations to reduce the LHS up to the capacitor
Initial value of v(t) = ~200V
After the switch is open and the current stored in the capacitor flows through the 50K resistor
time constant is RC =50K*0.4u=20ms

18. Assessment Problem 7.3 Use source transformations as follows -->

19. Step Response - Inductor
A full discharged inductor (L) or Capacitor (C) when subjected to an initial infusion of charge reacts to the sudden change with a “step response”
Take the inductor first: initially i=0 then the switch is closed at t=0 and current flows:
𝑉 𝑠 =𝑅𝑖+𝐿 𝑑𝑖 𝑑𝑡 the loop equation 𝑑𝑖 𝑑𝑡 = −𝑅𝑖+ 𝑉 𝑠 𝐿 rearranging the terms 𝑑𝑖 𝑑𝑡 = −𝑅 𝐿 𝑖− 𝑉 𝑠 𝑅 now multiply both side by dt 𝑑𝑖= −𝑅 𝐿 𝑖− 𝑉 𝑠 𝑅 𝑑𝑡 separate the variables
𝑑𝑖 (𝑖− 𝑉 𝑠 𝑅 ) = −𝑅 𝐿 𝑑𝑡 Then integrate both sides

20. Completing the integration
The LHS has the form:
Where a=1 b= - 𝑉 𝑠 𝑅
So the LHS integral becomes: =ln (𝑥− 𝑉 𝑠 𝑅 ) 𝑖0 𝑖
And the RHS is just =(− 𝑅𝑡 𝐿 ) 0 𝑡
ln 𝑖− 𝑉 𝑠 𝑅 −ln 𝑖 0 − 𝑉 𝑠 𝑅 =− 𝑅𝑡 𝐿
ln 𝑖− 𝑉 𝑠 𝑅 𝑖 0 − 𝑉 𝑠 𝑅 = −𝑅𝑡 𝐿 now raise both sides by e

21. Completing the formula
𝑖− 𝑉 𝑠 𝑅 𝑖 0 − 𝑉 𝑠 𝑅 = 𝑒 −𝑅𝑡 𝐿
Now solve for i(t): 𝑖 𝑡 = 𝑖 0 − 𝑉 𝑠 𝑅 𝑒 −𝑅𝑡 𝐿 + 𝑉 𝑠 𝑅
If initial energy in the inductor is 0 this reduces to
𝑖 𝑡 = 𝑉 𝑠 𝑅 − 𝑉 𝑠 𝑅 𝑒 −𝑅𝑡 𝐿

22. Step response of inductor current
𝑖 𝑡 = 𝑉 𝑠 𝑅 − 𝑉 𝑠 𝑅 𝑒 −𝑅𝑡 𝐿
As t ->  𝑒 −𝑅𝑡 𝐿 -> 0 and the final value -> 𝑉 𝑠 𝑅

23. Voltage across the inductor
𝑣 𝐿 =𝐿 𝑑𝑖 𝑑𝑡 =𝐿 𝑑 𝑉 𝑠 𝑅 + 𝐼 0 − 𝑉 𝑠 𝑅 𝑒 −𝑅𝑡 𝐿 𝑑𝑡
𝑣 𝐿 =𝐿 −𝑅 𝐿 𝐼 0 − 𝑉 𝑠 𝑅 𝑒 −𝑅𝑡 𝐿 multiply by L and –R
𝑣 𝐿 = 𝑉 𝑠 − 𝑅𝐼 0 𝑒 −𝑅𝑡 𝐿
VL falls from Vs exponentially

24. Assessment Problem 7.5

25. Spice solution Prior to t=0, I  12A
After t= 0 switch in position a. The voltage across resistor is -80V

26. After switch ia closed
.ic is initial condition for inductor current I(L1) which yields 200V at V2
(8+12)A*10 =200V
.tran is transient or time simulation

27. Plot of voltage: at V2 starts at -200V-> 0 current IL: starts at 12A and decays to-8 A

28. Step response of a capacitor first order circuit
Before t=0 the switch is open the node at top is : v=is R
When the switch is closed at t=0 the node equation becomes:
𝐼𝑠= 𝑣 𝑅 +𝐶 𝑑𝑣 𝑑𝑡 𝑑𝑣 𝑑𝑡 + 𝑣 𝑅𝐶 = 𝐼 𝑠 𝐶 𝑑𝑣 dt = 𝑅 𝐼 𝑠 𝑅𝐶 − 𝑣 𝑅𝐶 𝑑𝑣 𝑅 𝐼 𝑠 −𝑣 = 𝑑𝑡 𝑅𝐶 v

29. Capacitive Step Response
Integrate both sides a=-1 b= RIs
𝑣 0 = 𝑉 0 𝑣0 𝑣 𝑑𝑥 𝑥−𝑅 𝐼 𝑠 = 0 𝑡 −𝑑𝑦 𝑅𝐶 ln 𝑣−𝑅 𝐼 𝑠 − ln 𝑣 0 −𝑅 𝐼 𝑠 = −𝑡 𝑅𝐶 ln 𝑣−𝑅 𝐼 𝑠 𝑣 0 −𝑅 𝐼 𝑠 = −𝑡 𝑅𝐶 v=R 𝐼 𝑠 + 𝑣 0 −𝑅 𝐼 𝑠 𝑒 −𝑡 𝑅𝐶 , t≥0

30. Capacitor current iC 𝐼 𝑠 = 𝑣 𝐶 𝑅 + 𝑖 𝐶 ⇒ 𝑖 𝐶 = 𝐼 𝑠 − 𝑣 𝐶 𝑅
𝐼 𝑠 = 𝑣 𝐶 𝑅 + 𝑖 𝐶 ⇒ 𝑖 𝐶 = 𝐼 𝑠 − 𝑣 𝐶 𝑅
𝑣 𝑐 =R 𝐼 𝑠 + 𝑣 0 −𝑅 𝐼 𝑠 𝑒 −𝑡 𝑅𝐶
𝑖 𝐶 = 𝐼 𝑠 − R 𝐼 𝑠 + 𝑣 0 −𝑅 𝐼 𝑠 𝑒 −𝑡 𝑅𝐶 𝑅
𝑖 𝐶 = 𝐼 𝑠 − R 𝐼 𝑠 𝑅 + − 𝑣 0 𝑅 + 𝐼 𝑠 𝑒 −𝑡 𝑅𝐶
𝑖 𝐶 = 𝐼 𝑠 − 𝑣 0 𝑅 𝑒 −𝑡 𝑅𝐶 , t≥0

31. Assessment Problem 7.6 𝑣 𝑜 =−60+90 𝑒 −𝑡 𝜏 , 𝑡≥0
From example 7.6 vo and i0 are determined to be:
𝑣 𝑜 =−60+90 𝑒 −𝑡 𝜏 , 𝑡≥0
𝑖 𝑜 =−2.25 𝑒 −𝑡 𝜏 , 𝑡≥0
Both equations are valid after the switch is thrown to 2.

32. Assessment Problem 7.6
Using these values already determined find the Voltage across the160 resistor: note current through the 8K resistor causes the polarity to be as shown v1
v8K
𝑣0+8𝐾 𝑖 0 =𝑣1 −60+90 𝑒 −100𝑡 +8𝐾 −2.25 𝑒 −100𝑡 𝑚= 𝑣 1 𝑣 1 =−60+ 90−18 𝑒 −100𝑡 𝑣 1 =−60+72 𝑒 −100𝑡 𝑉

33. General Solution
You can see that both inductor and capacitor single order system exhibit the same behavior.
For natural the stored energy is released in a exponential decay from a stored value.
For step response the energy accumulates to the maximum value in an exponential
The response curves are similar and depend on the time constant of the circuit.
For inductor circuit: 𝜏= 𝐿 𝑅
For capacitor circuit: 𝜏=𝑅𝐶

34. Simulation and lab
In the lab we use a square wave to mimic switch we can also simulate the same thing in spice.
Voltage source can be implemented as a pulse generator
Pulse(V1 V2 Tdelay Trise Tfall Ton Tperiod Ncycles)
Pulse can be specified as 0 to 5 V with rise abd fall tiems of .001us ata value of 5V for 500u and repeating ever 1000u number of cycles =2
Which is ~ 5 time constants for the circuit
How would you extend it to 10 time constants?

35. Spice circuit

36. Spice results

37. Chapter 8
RLC circuits

38. RLC circuits
Two energy conserving components lead to second order systems.
Two configurations
Parallel
Series
Look at the second order systems in general:
The system has a solution of the form 𝑥=𝐴 𝑒 𝑠𝑡
Where x is a variable representing either v or I depending on the topology.
A represents the max stored value
S represents the time constant ( or rate of the decay)

39. Parallel RLC circuit
The common variable here is v so we write the sum of currents in terms of v
𝑣 𝑅 + 1 𝐿 0 𝑡 𝑣𝑑𝜏+ 𝐼 0 +𝐶 𝑑𝑣 𝑑𝑡 =0 1 𝑅 𝑑𝑣 𝑑𝑡 + 𝑣 𝐿 +𝐶 𝑑 2 𝑣 𝑑𝑡 2 =0 𝑑 2 𝑣 𝑑𝑡 𝑅𝐶 𝑑𝑣 𝑑𝑡 + 𝑣 𝐿𝐶 =0
Now we have a second order differential equation We can use the techniques of differential equations to solve.

40. Solution to second order equation
Assuming v has the form: v=𝐴 𝑒 𝑠𝑡 : we substitute v in the equation and after differentiating the appropriate terms we get:
𝐴 𝑠 2 𝑒 𝑠𝑡 + 𝐴𝑠 𝑅𝐶 𝑒 𝑠𝑡 + 𝐴 𝑒 𝑠𝑡 𝐿𝐶 =0
Factoring out the common exponential term:
𝑠 2 + 𝑠 𝑅𝐶 + 1 𝐿𝐶 𝐴𝑒 𝑠𝑡 =0
Of the three terms
𝐴𝑒 𝑠𝑡 if A=0 the solution would be trivial
𝑒 𝑠𝑡 cannot be 0
𝑠 2 + 𝑠 𝑅𝐶 + 1 𝐿𝐶 =0 This quadratic can take on 3 cases depending on the values

41. Characteristic equation
The form of: 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 is the general form of the quadratic equation.
The roots are:
−b ± 𝑏 2 −4𝑎𝑐 2𝑎
The discriminant : 𝑏 2 −4𝑎𝑐 suggests the type of solution
> 0 indicated 2 distinct roots
= 0 duplicate roots
< 0 complex conjugate roots

42. Using the discriminant to suggest the solution
a=1 b= 1 𝑅𝐶 ,c= 1 𝐿𝐶
1 𝑅𝐶 2 −2 1 𝐿𝐶 compared to 0
Now we introduce an alternative variables,  and ω0
The roots can be specified in terms of these:
𝑠 1 =−𝛼+ 𝛼 2 − 𝜔 0 2
𝑠 2 =−𝛼− 𝛼 2 − 𝜔 0 2

43. Alternative variables for the parallel circuit
Neper Frequency: 𝜶= 𝟏 𝟐𝑹𝑪
Resonant radian frequency: 𝝎 𝟎 = 𝟏 𝑳𝑪
𝛼 2 > 𝜔 overdamped: real distinct roots
𝛼 2 < 𝜔 underdamped: complex roots
𝛼 2 = 𝜔 critically damped: repeated real roots

44. Assessment 8.1 a.
If R= 100 and L= 20mH what value of C will make the system critically damped?
Criteria for critically damped: 𝛼 2 = 𝜔 0 2
𝜶= 𝟏 𝟐𝑹𝑪 𝝎 𝟎 = 𝟏 𝑳𝑪
Thus: 𝟏 𝟐𝑹𝑪 2 = 𝟏 𝑳𝑪
Rearranging lets us write 𝐶= 𝐿 4 𝑅 2 = 20𝑚 40𝐾Ω =500𝑛𝐹

45. Assessment 8.1 b.
If C is adjusted to give =5 K rad/sec what is the value of C and the roots of the equation (form of solution)
𝟏 𝟐𝟎𝟎𝑪 = 5K => C=1uF
𝑠 1 =−5𝐾+ 25𝑀𝑒𝑔−50𝑀𝑒𝑔 =−5𝐾+𝑗5𝐾 𝑠 2 =−5𝐾−𝑗5𝐾
𝑠 1,2 =−𝛼± 𝛼 2 − 𝜔 0 2

46. Form of solutions for the 3 cases
Real distinct roots: 𝑣= 𝐴 1 𝑒 𝑠 1 𝑡 + 𝐴 2 𝑒 𝑠 2 𝑡 𝑡≥0
To solve this the following steps are necessary:
Find the roots of the characteristic equation.
Find v(0+) and 𝑑𝑣(0+) 𝑑𝑡 using circuit analysis
Find the values of 𝐴 1 and 𝐴 2 solving :
𝑣 0+ = 𝐴 1 + 𝐴 2
𝑑𝑣(0) 𝑑𝑡 = 𝑖 𝐶 (0+) 𝐶 = 𝑠 1 𝐴 1 + 𝑠 2 𝐴 2
Substitute the values into the equation for the solution above.

47. Finding the solutions
Underdamped solution: 𝑣= 𝐵 1 𝑒 −𝛼𝑡 cos 𝜔 0 𝑡 + 𝐵 2 𝑒 −𝛼𝑡 sin 𝜔 0 𝑡
Put the roots of the characteristic equation in the form of:
−𝛼±𝑗 𝜔 0 2 − 𝛼 2
Where 𝜔 0 2 − 𝛼 2 = 𝜔 𝑑 the damped radian frequency.
𝑣 𝑡 = 𝐴 1 𝑒 −𝛼+𝑗 𝜔 𝑑 𝑡 + 𝐴 2 𝑒 −𝛼−𝑗 𝜔 𝑑 𝑡 𝑣 𝑡 = 𝐴 1 𝑒 −𝛼𝑡 𝑒 𝑗 𝜔 𝑑 𝑡 + 𝐴 2 𝑒 −𝛼𝑡 𝑒 −𝑗 𝜔 𝑑 𝑡 now factor out the 𝑒 −𝛼𝑡 term
𝑣 𝑡 = 𝑒 −𝛼𝑡 𝐴 1 𝑒 𝑗 𝜔 𝑑 𝑡 + 𝐴 2 𝑒 −𝑗 𝜔 𝑑 𝑡 and we have the exponential form of the sine expanding each ej using euler’s identity we have
𝑒 ±𝑗𝜃 =𝑐𝑜𝑠𝜃±𝑗𝑠𝑖𝑛𝜃
𝑣 𝑡 = 𝑒 −𝛼𝑡 𝐴 1 𝑐𝑜𝑠 𝜔 𝑑 +𝑗𝑠𝑖𝑛 𝜔 𝑑 + 𝐴 2 (𝑐𝑜𝑠 𝜔 𝑑 −𝑗𝑠𝑖𝑛 𝜔 𝑑 )

48. Underdamped solution
𝑣 𝑡 = 𝑒 −𝛼𝑡 𝐴 1 𝑐𝑜𝑠 𝜔 𝑑 +𝑗𝑠𝑖𝑛 𝜔 𝑑 + 𝐴 2 (𝑐𝑜𝑠 𝜔 𝑑 −𝑗𝑠𝑖𝑛 𝜔 𝑑 )
𝑣 𝑡 = 𝑒 −𝛼𝑡 ( 𝐴 1 + 𝐴 2 )_𝑐𝑜𝑠 𝜔 𝑑 +𝑗(𝐴 1 − 𝐴 2 ) 𝑠𝑖𝑛 𝜔 𝑑
𝑣 𝑡 = 𝑒 −𝛼𝑡 𝐵 1 𝑐𝑜𝑠 𝜔 𝑑 + 𝐵 2 𝑠𝑖𝑛 𝜔 𝑑
Determine the coefficients B1 and B2 by the initial energy stored in the system by evaluating the voltage and its derivative at t=0+
V(0+)=V0 = 𝐵 1
𝑑𝑣(0+) 𝑑𝑡 = 𝑖 𝑐 ( 0 + ) 𝐶 =−𝛼 𝐵 1 + 𝜔 𝑑 𝐵 2
The underdamped solution describes a decaying oscillation as shown also known as “ringing”.

49. Assessment problem 8.4
A 10mH inductor and 1uF capacitor are connected with a resistor R. The roots of the characteristic equation are:
-8000+j6000 and j6000
V(t=0)=10V iL(t=0)=80mA
a. What is R?
b. dv(0)/dt?

50. Assessment 8.4: 10mH inductor and 1uF
Roots of Characteristic equation are complex conjugates :
-8000+j6000 and j6000
−𝛼±𝑗 𝜔 𝑑
Has the form 𝑣 𝑡 = 𝑒 −𝛼𝑡 𝐵 1 𝑐𝑜𝑠 𝜔 𝑑 + 𝐵 2 𝑠𝑖𝑛 𝜔 𝑑
𝛼= 1 2𝑅𝐶 𝑎𝑛𝑑 𝜔 0 = 1 𝐿𝐶
Plug in 0 in the equation for v(t): V(0)=B1
Part a. = = 1 2𝑅𝐶 ⇒𝑅= 𝑢𝐹 =62.5Ω
iL(0+)=80m and VC (0+)=10V
V(0+)=V0 =𝟏𝟎V = 𝑩 𝟏

51. Assessment 8.4: b and c. dv(0+) dt = i c ( 0 + ) C
Since V(0+) = 10V and iC+ iL + iR = 0 = 10/ m + iC
b: iC =-240mA
𝑖 𝑐 ( 0 + ) 𝐶 =−𝛼 𝐵 1 + 𝜔 𝐷 𝐵 2
−240𝑚𝐴 1𝑢 =− 𝐵 2
-240K + 80K =6000*B2= -160K/6K= -80/3 = B2

52. Critically Damped solution
When 𝜔 0 =𝛼 then the roots are both: 𝛼= −1 2𝑅𝐶
The voltage must take another form: 𝑣 𝑡 = 𝐷 1 t 𝑒 −𝛼𝑡 + 𝐷 2 𝑒 −𝛼𝑡
To solve for D1 and D2 we use:
𝑣 = 𝑉 0 = 𝐷 2
𝑑𝑣( 0 + ) 𝑑𝑡 = 𝑖 𝑐 𝐶 = 𝐷 1 −𝛼 𝐷 2
You will rarely see this case with real components since a slight fluctuation in the component values will cause the response to go either over or under damped

53. Assessment Problem 8.5
L= 0.4H and C= 10 Find the value of R that makes the circuit critically damped.
For critical damping  = ω0 so we can find 𝜔 0 = 1 𝐿𝐶 𝑎𝑛𝑑 𝛼= 1 2𝑅𝐶
𝜔 0 =500 so 𝑅= 1 2𝛼𝐶 = 1 20(500)( 10 −6 ) =100Ω

54. Assessment Problem 8.5 b,c B: 𝐯(𝟎)= 𝟐𝟓𝟎𝟎 =𝟓𝟎𝑽
Given the initial energy stored is 25m and is distributed equally L and C:
𝑊 𝐿 = 1 2 L i 𝐿 2 and 𝑊 𝐶 = 1 2 C 𝑣 2
So each energy equation is equal to 25m/2
25𝑚 2 = 𝜇 𝑣 2 ⇒ 𝑣 2 = 25𝑚 10𝜇 = 25𝐾 10 =2.5𝑥 10 3 =2500
B: 𝐯(𝟎)= 𝟐𝟓𝟎𝟎 =𝟓𝟎𝑽
Similarly for initial inductor current
25𝑚 2 = 1 2 L 𝑖 𝐿 2 = 1 2 (0.4) 𝑖 𝐿 2
C: 𝟐𝟓𝒎= 𝟎.𝟒 𝒊 𝑳 𝟐 ⇒ 𝒊 𝑳 𝟐 = 𝟐𝟓𝐦 𝟎.𝟒 =.𝟎𝟔𝟐𝟓⇒ 𝒊 𝑳 (𝟎)=𝟎.𝟐𝟓𝑨=𝟐𝟓𝟎𝒎𝑨

55. Assessment Problem 8.5 d 𝑖 𝑐 𝐶 = 𝐷 1 −𝛼 𝐷 2 ⇒ 250𝑚 10𝜇 = 𝐷 1 −500 50
D2=v(0)=50V
𝑖 𝑐 𝐶 = 𝐷 1 −𝛼 𝐷 2 ⇒ 250𝑚 10𝜇 = 𝐷 1 −500 50
𝑫 𝟏 =𝟐𝟓,𝟎𝟎𝟎+𝟐𝟓𝟎𝟎𝟎⇒ 𝑫 𝟏 =𝟓𝟎,𝟎𝟎𝟎 𝑽 𝒔

56. Natural and step response of the Series RLC circuit
We use the same procedures to analyze the series RLC circuit.
Sum the voltages around the closed loop
𝑅𝑖+𝐿 𝑑𝑖 𝑑𝑡 + 1 𝐶 0 𝑡 𝑖𝑑𝜏+ 𝑉 0 =0 now differentiate with respect to t
𝑅 𝑑𝑖 𝑑𝑡 +𝐿 𝑑2𝑖 𝑑𝑡2 + 𝑖 𝐶 =0 rearrange to give:
𝑑2𝑖 𝑑𝑡2 + 𝑅 𝐿 𝑑𝑖 𝑑𝑡 + 𝑖 𝐿𝐶 =0

57. Characteristic equation
𝑠 2 + 𝑅 𝐿 𝑠+ 1 𝐿𝐶 =0
Now while similar the CE is slightly different
The roots are: 𝑠 1,2 = −𝑅 2𝐿 ± 𝑅 2𝐿 2 − 1 𝐿𝐶
We can convert to the notation we’ve user before:
𝛼= 𝑅 2𝐿 and 𝜔 0 = 1 𝐿𝐶 rad/S
The definition for 𝛼 is slightly different but the 𝜔 0 is the same.

58. Response types are the same type: Natural Response
Depending on the values of 𝛼 𝑎𝑛𝑑 𝜔 0 we have the same 3 situations :
𝜔2 0 < 𝛼2 Overdamped
𝜔2 0 > 𝛼2 Underdamped
𝜔2 0 = 𝛼2 Critically damped
Overdamped: 𝑖 𝑡 =𝐴 1 𝑒 𝑠 1 𝑡 + 𝐴 2 𝑒 𝑠 2 𝑡
Underdamped 𝑖 t = 𝐵 1 𝑒 −𝛼𝑡 𝑐𝑜𝑠 𝜔 𝑑 𝑡+ 𝐵 2 𝑒 −𝛼𝑡 𝑠𝑖𝑛 𝜔 𝑑 𝑡
Critically Damped 𝑖 𝑡 =𝐷 1 𝑡 𝑒 −𝛼𝑡 + 𝐷 2 𝑒 −𝛼𝑡

59. Step Response 𝑑 2 𝑣 𝑐 𝑑𝑡 2 + 𝑅 𝐿 𝑑 𝑣 𝑐 𝑑𝑡 + 𝑣 𝐶 𝐿𝐶 = 𝑉 𝐿𝐶
For the series RLC circuit step response we must find the capacitor voltage
Now substitute these into the first equation and we have the differential equation for the series circuit
𝑑 2 𝑣 𝑐 𝑑𝑡 2 + 𝑅 𝐿 𝑑 𝑣 𝑐 𝑑𝑡 + 𝑣 𝐶 𝐿𝐶 = 𝑉 𝐿𝐶
𝑉=𝑖𝑅+𝐿 𝑑𝑖 𝑑𝑡 + 𝑣 𝑐 𝑖=𝐶 𝑑 𝑣 𝑐 𝑑𝑡 𝑑𝑖 𝑑𝑡 =𝐶 𝑑 2 𝑣 𝑐 𝑑𝑡 2

60. Series RLC The initial conditions can be 0 or something else
Now this is the same form as the equation for parallel circuit only the coefficients differ.
So the solution forms will be the same:
Overdamped: 𝑣 𝑐 =𝑉 𝑓 + 𝐴 1 ′ 𝑒 𝑠 1 𝑡 + 𝐴 2 ′ 𝑒 𝑠 2 𝑡
Underdamped: 𝑣 𝑐 =𝑉 𝑓 + 𝐵 1 ′ 𝑒 −𝛼𝑡 𝑐𝑜𝑠 𝜔 𝑑 𝑡+ 𝐵 2 ′ 𝑒 −𝛼𝑡 𝑠𝑖𝑛 𝜔 𝑑 𝑡
Critically damped: 𝑣 𝑐 = 𝑉 𝑓 + 𝐷 1 ′ 𝑒 −𝛼𝑡 + 𝐷 2 ′ 𝑒 −𝛼𝑡
Where 𝑉f is the final value of Vc
The initial conditions can be 0 or something else

61. Assessment problem 8.6
The switch is in position a for a long time then 80=9𝐾𝑖+15𝐾𝑖
𝑖= 80𝑉 24𝐾 =3.33𝑚𝐴
𝑣= =50𝑉
At t=0 the switch is moved to position b.
Initially it=0 since it cannot change instantaneously
𝑣 𝐿 =𝐿 𝑑𝑖 𝑑𝑡
Since i cannot change instantaneously so it is 0 at the start. Since the inductor has NO energy stored.
100=5𝑚 𝑑𝑖 𝑑𝑡 + 𝑣 𝑐 ⇒ 50 5𝑚 =10,000 A/s

62. Skip Op amp Circuits for now

63. Chapter 9
Sinusoidal analysis

64. Sinusoidal Solid State Analysis
Sinusoidal signal properties:
Period T and frequency f=1/T
Peak amplitude Vm
Pk to pk amplitude = 2Vm
𝒗= 𝑽 𝒎 𝐜𝐨𝐬⁡(𝝎𝒕+ 𝝓)
ω=2πf = 2π/T
𝜙 is the phase angle

65. Phase is frequency equivalent of time
If  is positive the function shifts to the left
If  is negative the function shifts to the right
Time is increasing to the right so
a signal sifted to the right occurs later in time
Similarly a signal sifted to he left occurs earlier in time
Radians is the default in many texts but radians are easily converted to degrees : 𝑑𝑒𝑔= 180 𝜋 radians
Another common term is rms or root mean square:
𝑉𝑟𝑚𝑠= 1 𝑇 𝑡 0 𝑡 0 +𝑇 𝑉 𝑚 2 𝑐𝑜𝑠 2 (𝜔𝑡+𝜙 )𝑑𝑡

66. Prob 9.3 a-f V(t)=25cos(400t+60)
Max Amp = 25V
Freq in hz : 400 = 2f => f = 200Hz
Freq in rad /s : 2f =400(3.14)=
Phase angle in rad : 2  =1.047 rad
Phase angle in deg = 60
Period in ms : T=1/f=> 1/200=5ms

67. RMS voltage
It is computed just like it sounds: (in reverse)
Square the voltage
Find the mean
Take the square root
𝑉 𝑅𝑀𝑆 = 𝑉 𝑚 2

68. Circuit solution 𝑅𝑖+𝐿 𝑑𝑖 𝑑𝑡 = 𝑉 𝑚 cos⁡(𝜔𝑡)
As with any differential equation (with constant coefficients)
The solution is the sum of the transient solution and the steady state solution
𝑅𝑖+𝐿 𝑑𝑖 𝑑𝑡 = 𝑉 𝑚 cos⁡(𝜔𝑡)
We solved the first part (natural response)
𝑅𝑖+𝐿 𝑑𝑖 𝑑𝑡 =0 as we did earlier.
The forced response has the form of the forcing function:
If the circuit is linear:
The magnitude and phase angle are affected not the frequency.

69. Solution con’t Evaluating yields:
The solution to the forced response is of the form:
𝑖 𝑡 =𝐴𝑐𝑜𝑠(𝜔𝑡+𝜙)
Now we can write this function as: 𝑖 𝑡 =𝐴𝑐𝑜𝑠 𝜙 cos 𝜔𝑡 +𝐴𝑠𝑖𝑛 𝜙 sin⁡(𝜔𝑡)
No absorb the constant cos 𝜙 𝑎𝑛𝑑 sin⁡(𝜙) into the constant A, A1and A2
𝑖 𝑡 = 𝐴 1 cos 𝜔𝑡 + 𝐴 2 sin⁡(𝜔𝑡) and substitute it into the equation:
R( 𝐴 1 cos 𝜔𝑡 + 𝐴 2 sin⁡(𝜔𝑡)) +𝐿 𝑑( 𝐴 1 cos 𝜔𝑡 + 𝐴 2 sin 𝜔𝑡 ) 𝑑𝑡 = 𝑉 𝑚 cos⁡(𝜔𝑡)
Evaluating yields:
𝑅 𝐴 1 cos 𝜔𝑡 +𝑅 𝐴 2 sin 𝜔𝑡 − 𝐴 1 𝜔𝐿𝑠𝑖𝑛 𝜔𝑡 + 𝐴 2 𝜔𝐿cos 𝜔𝑡 = 𝑉 𝑚 cos⁡(𝜔𝑡)
Now we equate the coefficients of the sine and cosine terms to arrive at 2 equations:
𝑅 𝐴 2 − 𝐴 1 𝜔𝐿=0 and 𝑅 𝐴 1 + 𝐴 2 𝜔𝐿= 𝑉 𝑚
Now we solve for A1 and A2 using these 2 equations
𝐴 1 = 𝑅 𝑉 𝑚 𝑅 2 + 𝜔 2 𝐿 2
𝐴 2 = 𝜔𝐿 𝑉 𝑚 𝑅 2 + 𝜔 2 𝐿 2

70. For the alternate expression
𝑖 𝑡 =𝐴𝑐𝑜𝑠(𝜔𝑡+𝜙)
We find A and 𝜙 using more trig expressions:
Acos 𝜙 = 𝑅 𝑉 𝑚 𝑅 2 + 𝜔 2 𝐿 2
𝐴𝑠𝑖𝑛(𝜙)= 𝜔𝐿 𝑉 𝑚 𝑅 2 + 𝜔 2 𝐿 2
tan 𝜙 = sin⁡(𝜙) cos⁡(𝜙) =− 𝜔𝐿 𝑅 hence 𝜙=− 𝑡𝑎𝑛 −1 𝜔𝐿 𝑅
A 2 cos 2 𝜙 + A 2 sin 2 𝜙 = 𝐴 2 = 𝑅 𝑉 𝑚 𝑅 2 + 𝜔 2 𝐿 𝑅 𝑉 𝑚 𝑅 2 + 𝜔 2 𝐿 = 𝑉 𝑚 2 𝑅 2 + 𝜔 2 𝐿 2
Now we can write A= 𝑉 𝑚 𝑅 2 + 𝜔 2 𝐿 2
So 𝒊 𝒕 = 𝑽 𝒎 𝑹 𝟐 + 𝝎 𝟐 𝑳 𝟐 𝐜𝐨𝐬(𝝎𝒕− 𝒕𝒂𝒏 −𝟏 𝝎𝑳 𝑹 )

71. Some simplifying methods
The solution to even a simple RL equation using a sinusoidal forcing function is complicated
We can simplify the analysis by noting that the frequency of the parameters i(t) and various v(t) in the circuit is unaffected.
Now use Euler’s equation: 𝑒 ±𝑗𝜃 =𝑐𝑜𝑠𝜃±𝑗𝑠𝑖𝑛𝜃
This complex function has a real part and an imaginary part
Now let’s select as our forcing function the non-realizable voltage:
𝑣 𝑡 = 𝑉 𝑀 𝑒 𝑗𝜔𝑡
𝑣 𝑡 = 𝑉 𝑀 cos 𝜔𝑡 +𝑗 𝑉 𝑀 sin 𝜔𝑡

72. Solution 𝒊 𝒕 = 𝑰 𝑴 𝐜𝐨𝐬 𝝎𝒕+𝝓 +𝒋 𝑰 𝑴 𝐬𝐢𝐧 𝝎𝒕+𝝓 V=Vm  0 and I = Im  
As a consequence of linearity and superposition the solution is of the form:
𝒊 𝒕 = 𝑰 𝑴 𝐜𝐨𝐬 𝝎𝒕+𝝓 +𝒋 𝑰 𝑴 𝐬𝐢𝐧 𝝎𝒕+𝝓
This can also be written as: 𝒊 𝒕 = 𝑰 𝑴 𝒆 𝒋(𝝎𝒕+𝝓)
Using this technique allows us to covert a differential equation into an algebraic equation which is much simpler to solve.
Now we can drop the frequency and just use the magnitude and phase angle to describe the forcing function or other parameters.
V=Vm  0 and I = Im  
These are called phasors.

73. Phasor notation
Sinusoidal current or voltage at a given frequency is characterized by only 2 parameters: Magnitude and phase.
𝑉 𝑀 cos 𝜔𝑡 =𝑉𝑚 𝑒 𝑗𝜔𝑡 = 𝑉 𝑀 <0°
This is a Phasor
Phasor notation assumes the frequency is the same as it will be if the circuit is linear as all the circuits are that we talk about in ENA.
This is our first step int the “frequency domain”

74. Process to get to the phasor
Given a sinusoidal function , v(t) or i(t), in the time domain, write it as a cosine wave with phase angle.
If the sin() form is used, convert it to a cosine with: sin(ωt)=cos(ωt-90)
Express the cosine as the real part of euler’s identity
Drop the RE
Surpress: 𝑒 𝑗𝜔

75. Resistor and inductor Phasors
This change to the frequency domain effects how we view the components R, L and C
V(t) = Ri(t)
In phasor notation : VM< = R IM <
Note the is no change in the phase angle, only the magnitude
But for L and C there is a change in the phase angle.
𝑣 𝑡 =𝐿 𝑑𝑖 𝑑𝑡 = 𝑉 𝑀 𝑒 𝑗 𝜔𝑡+𝜃 =𝐿 𝑑 𝐼 𝑀 𝑒 𝑗 𝜔𝑡+𝜙 𝑑𝑡 =𝑗𝜔𝐿 𝐼 𝑀 𝑒 𝑗 𝜔𝑡+𝜙
Now suppressing the 𝑒 𝑗 𝜔𝑡 we have:
𝑉 𝑀 𝑒 𝑗 𝜃 = 𝑗𝜔𝐿 𝐼 𝑀 𝑒 𝑗 𝜙
𝑉 𝑀 <𝜃= 𝑗𝜔𝐿 𝐼 𝑀 <𝜙

76. Capacitor phasor 𝑖=𝐶 𝑑𝑣 𝑑𝑡 v t = V M e j(ωt+ θ v )
𝑖 𝑡 = 𝐼 𝑀 𝑒 𝑗 𝜗 𝐼 =𝐶 V M 𝑑 𝒆 𝒋(𝝎𝒕+ 𝜽 𝒗 ) 𝑑𝑡 =𝑗𝜔𝐶 V M 𝑒 𝒋𝜽 𝒗
𝐼 𝑀 𝑒 𝑗 𝜗 𝐼 = 𝑗𝜔𝐶 V M 𝑒 𝒋𝜽 𝒗
𝑰= 𝑗𝜔𝐶V
𝑉= I 𝑗𝜔𝐶 < 𝜽 𝒗 = 𝐼 𝜔𝐶 < 𝜽 𝒗 −𝟗𝟎°

77. 1 𝑗𝜔𝐶 Impedance 𝑗𝑤𝐶 𝑗𝜔𝐿 Resistor R 1 𝑅 Capacitor Inductor 1 𝑗𝜔𝐿
Impedance the ratio of the voltage to the current in a component.
Angular Freq ω
Impedenace Z
Admittance Y
Resistor R
1 𝑅
Capacitor
1 𝑗𝜔𝐶
𝑗𝑤𝐶
Inductor
𝑗𝜔𝐿
1 𝑗𝜔𝐿

78. Assessment 9.1 a, b, c 𝒊 𝒕 =𝟓𝒄𝒐𝒔 𝝎𝒕+𝟑𝟔.𝟖𝟕 +𝟏𝟎𝒄𝒐𝒔(𝝎𝒕−𝟓𝟑.𝟏𝟑)
Transformv into a phasor: 𝒗 𝒕 =𝟏𝟕𝟎 𝐜𝐨𝐬 𝟑𝟕𝟕𝒕−𝟒𝟎° 𝑽
170 < 40
Transformv into a phasor: i 𝒕 =𝟏𝟎 𝒔𝒊𝒏 𝟏𝟎𝟎𝟎𝒕+𝟐𝟎° 𝑽
10 < => 10 < -70
𝒊 𝒕 =𝟓𝒄𝒐𝒔 𝝎𝒕+𝟑𝟔.𝟖𝟕 +𝟏𝟎𝒄𝒐𝒔(𝝎𝒕−𝟓𝟑.𝟏𝟑)
Phasors add better in rectangular rather than polar form
5 < <
Convert the first: a+jb=5cos(36.87) + j 5sin(36.87) = 4 + j3
C+jd = 10cos(-53.13) +j10sin(-53.13) = 6 - j8.0
= 10 – j5 => < 26.57

79. Reactance Impedance is a complex quantity.
The imaginary part of impedance is called reactance
So the inductor and capacitor have “reactance values”
Now we can go back through all the circuit techniques we did for resistors and repeat the process with phasors.
Kirchhoff's Voltage and current laws
Source transformations
Thevenin and Norton equivalent circuits
Node voltage and mesh current method

80. Kirchhoff's laws in the frequency domain
Voltages around a loop:
Phasor values of V sum to zero:
𝑽 𝟏 + 𝑽 𝟐 + 𝑽 𝟑 +..+ 𝑽 𝑵 =0
Currents at a node sum to zero:
𝑰 𝟏 + 𝑰 𝟐 + 𝑰 𝟑 +..+ 𝑰 𝑵 =0
Adding phasors (complex numbers) is easier in rectangular rather than polar coordinates:
𝑅<𝜃=𝑎+𝑗𝑏
𝑅= 𝑎 2 + 𝑏 2
𝜃= 𝑡𝑎𝑛 −1 ( 𝑏 𝑎 )
𝑎=𝑅𝑐𝑜𝑠 𝜃
𝑏=𝑅𝑠𝑖𝑛(𝜃)

81. Adding complex values
Add the real parts and imaginary parts separately.
The covert the result back to form polar form i.e. Phasor
𝑖 1 =100 cos 𝜔𝑡+25
𝑖 2 =100 cos 𝜔𝑡+145
𝑖 3 =100 cos 𝜔𝑡−95
Convert to phasors:
𝑖 1 =100 <25
𝑖 2 =100 <145
𝑖 3 =100 <−95
Covert to rectangular j
 − j
−8.72 − 99.62j

82. Completed problem Add the real and imaginary parts separately:j
 − j
−8.72 − 99.62j
______________________
0 +j 0
Sum is ~0

83. Summing impedances in series and parallel
𝑉 𝑎𝑏 = 𝑍 1 𝐼+ 𝑍 2 𝐼+ 𝑍 3 𝐼 𝑉 𝑎𝑏 = (𝑍 1 + 𝑍 2 + 𝑍 3 )𝐼 𝑉 𝑎𝑏 𝐼 = 𝑍 𝑒𝑞 = 𝑍 1 + 𝑍 2 + 𝑍 3
𝐼= 𝐼 1 + 𝐼 2 + 𝐼 3 𝑉 𝑍 𝐸𝑄 = 𝑉 𝑍 1 + 𝑉 𝑍 2 + 𝑉 𝑍 3 𝑉 𝑍 𝐸𝑄 = 1 𝑍 𝑍 𝑍 3 𝑉 1 𝑍 𝐸𝑄 = 1 𝑍 𝑍 𝑍 3

84. Assessment 9.7 A) Calculate impedance if ω=2Krad/sec first
Compute the impedance of each component first:
L1: Z=(2K*6m)= j12 ohms
C1: Z = -j/(25u*2K) = -j20 ohms
The resistor in series with the cap is: 5 - j20
Parallel comb of resistor and inductance is:
20𝑜 20+𝑗10 = 200< <26.57 = <90−26.57
= 8.9 <63.43= j7.96j
Now add this to the 5 - j20
8.98 +j ~= 9 - j12

85. Thevenin’s Equivalent and Source Transformations
Thevenin's and Norton's equivalent for phasors:
V, Z and I are phasor quantities
I.e. complex in general

86. Source Transformations and Thevenin/Norton Equivalent circuits
All voltages and currents and impedances are phasor values
Complex quantities

87. Node Voltage method and Mesh current method
Node voltage and mesh current methods work the same way but the values for Voltage and current sources and impedances are phasors

88. Examples: use Source transformations to compute v0
Transform LHS to current source
v1= 240<53.13=144+j192
jωL = j 15mH*4000 = j 60
I =V1/j60=-2.4j+3.2
Transform the RHS to Current source
V2= -96 < 90
I2= -96/20 <90 = -j4.8
1 𝑗𝜔𝐶 = 1 𝑗 𝜇𝐹 = 1 j16.667m = - j60

89. Example
Devices in parallel:
1 𝑍𝑡 = 1 𝑗 − 1 𝑗 = =12
Switch the direction of the RH Current source - j4.8A -> j4.8A
Sum the two current sources: 3.2 – j2.4A + j4.8A = j 4 < 36.86
V = IR =12 ( j )= j 28.8 = 48 < 36.86
V1 =48 cos(4000t+36.86)

90. Assessment 9.12
Use node voltage method to find the steady state value of v(t):
Where 𝑖 𝑠 𝑡 =10 cos 𝜔𝑡 , 𝑣 𝑠 =100 sin 𝜔𝑡 𝑎𝑛𝑑 𝜔=50𝑘 𝑟𝑎𝑑/𝑠
First convert everything to phasors:
𝐼 𝑠 =100 < 0 and 𝑉 𝑠 =100<−90
C -> 1 𝑗𝜔𝐶 = 1 𝑗50𝐾∗9𝜇 = 1 𝑗450𝑚 = −𝑗 = -j2.22
L-> 𝑗𝜔L = j50K*100u = j5

91. Assessment 9.12 con’t − 𝑖 𝑠 + 𝑣 5 + 𝑣 −𝑗20 9 + 𝑣 𝑗5 + 𝑣− 𝑣 𝑠 20 =0
Write a node equation :
− 𝑖 𝑠 + 𝑣 5 + 𝑣 −𝑗 𝑣 𝑗5 + 𝑣− 𝑣 𝑠 20 =0
Now 𝐼 𝑠 =10 < 0 and 𝑣 𝑠 =100<−90
𝑗 20 + −𝑗 5 𝑣= 𝑖 𝑠 + 𝑣 𝑠 20 = 10 − 5j
( j) v =10 - 5j
𝑣= 10−5𝑗 𝑗 = 11.08<− <45 =31.33<49.76
V = 10 − j30 = 31.62/−71.57◦

92. Transformer Analysis 𝑉 𝑠 = 𝑍 𝑠 + 𝑅 1 +𝑗𝜔 𝐿 1 𝐼 1 −𝑗𝜔𝑀 𝐼 2
R1: Resistance of the primary winding
R2: Resistance of the secondary winding
L1: self inductance of the primary winding
L2: self inductance of the secondary winding
M: mutual inductance
All values are phasors
𝑉 𝑠 = 𝑍 𝑠 + 𝑅 1 +𝑗𝜔 𝐿 1 𝐼 1 −𝑗𝜔𝑀 𝐼 2
0=−𝑗𝜔𝑀 𝐼 1 + 𝑅 2 +𝑗𝜔 𝐿 2 + 𝑍 𝐿 𝐼 2

93. Chapter 10 Sinusoidal power
Instantaneous Power
Real Power
Average Power
Reactive power
Power factor
RMS power
Complex power
Apparent power
Maximum power transfer

94. Instantaneous Power p(t)= i(t)v(t) 𝑣= 𝑉 𝑚 cos⁡(𝑤𝑡+ 𝜃 𝑣 )
0 time is defined when current is passing the + maximum:
𝑖= 𝐼 𝑚 cos⁡(𝑤𝑡)
𝑣= 𝑉 𝑚 cos⁡(𝑤𝑡+ 𝜃 𝑣 − 𝜃 𝑖 )
𝑝=𝑣𝑖= 𝑉 𝑚 cos⁡(𝑤𝑡+ 𝜃 𝑣 − 𝜃 𝑖 ) 𝐼 𝑚 cos⁡(𝑤𝑡)
𝑝= 𝑉 𝑚 𝐼 𝑚 cos⁡(𝑤𝑡+ 𝜃 𝑣 − 𝜃 𝑖 ) cos⁡(𝑤𝑡) :use trig identity cos(a) cos(b)
𝑝= 𝑉 𝑚 𝐼 𝑚 2 cos 𝜃 𝑣 − 𝜃 𝑖 + 𝑉 𝑚 𝐼 𝑚 2 cos⁡(𝜔𝑡+ 𝜃 𝑣 − 𝜃 𝑖 ) :use trig identity for cos(a + b)
𝑝= 𝑉 𝑚 𝐼 𝑚 2 cos 𝜃 𝑣 − 𝜃 𝑖 + 𝑉 𝑚 𝐼 𝑚 2 cos 𝜃 𝑣 − 𝜃 𝑖 cos 2𝜔𝑡 − 𝑉 𝑚 𝐼 𝑚 2 sin 𝜃 𝑣 − 𝜃 𝑖 sin⁡(2𝜔𝑡)

95. Average and Reactive Power
Notice that power is at 2x the frequency of I(t) and v(t)
Now lets identify the 3 components of the instantaneous power
𝒑=𝑷+𝑷𝒄𝒐𝒔 𝟐𝝎𝒕 +𝑸𝒔𝒊𝒏(𝟐𝝎𝒕)
Average power is (real power): 𝐏= 𝑽 𝒎 𝑰 𝒎 𝟐 𝒄𝒐𝒔 𝜽 𝒗 − 𝜽 𝒊 𝒄𝒐𝒔 𝟐𝝎𝒕
Reactive power is: 𝐐= 𝑽 𝒎 𝑰 𝒎 𝟐 𝒔𝒊𝒏 𝜽 𝒗 − 𝜽 𝒊 𝒔𝒊𝒏⁡(𝟐𝝎𝒕)
Real power is associated with purely resistive circuits
Describes power that is transformed from electrical to non-electrical energy (like heat)