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Formula Packet Basic Definitions: dq(t ) , dt

i(t) =

(1)

where i(t) is the current (A), q(t) is charge (C) and t is time (sec).

Q(t) =

Z tf

i(t)dt,

(2)

t0

where i is the current (A), Q is net charge (C) and t is time (sec).

vab (t) =

dw(t) , dq(t)

(3)

where v is the voltage (V) between a and b, q is net charge (C) and w is energy in joules (J).

P=

v2 (t) dw = v(t)i(t) = = i2 (t)R, dt R

(4)

where P is the power in Watts (W), w is energy in joules (J), t is time (sec), v is the voltage (V), i is the current (A), and R is the resistance. Conservation Laws: Kirchhoff’s Current Law N

∑ in = 0,

(5)

n=1

where the algebraic sum of the currents at a node (or closed boundary) is zero. Kirchhoff’s Voltage Law M

∑ vm = 0,

(6)

m=1

where the algebraic sum of all the voltages around a closed path (or loop) is zero. DC Circuits: Ohm’s Law v = iR,

(7)

where v is the voltage (v), i is the current (A), and R is the resistance. Conductance G=

i 1 = , R v 1

(8)

where G is the conductance (S), v is the voltage (v), i is the current (A), and R is the resistance. N resistors in series Req = R1 + R2 + · · · + RN .

(9)

Voltage divider for two resistors in series v1 =

R1 ∗ vs , R1 + R2

v2 =

R2 ∗ vs , R1 + R2

(10)

where vs denotes the source voltage. N Parallel Resistors

Special case of two parallel resistors 1 1 1 1 , = + +···+ Req R1 R2 RN

Req =

R1 R2 . R1 + R2

(11)

Current Divider formula for two resistors in parallel i1 =

R2 ∗ is , R2 + R1

i2 =

R1 ∗ is , R1 + R2

(12)

where is is the current source. General current divider formula in =

Gn i, G1 + G2 + · · · GN s

(13)

where G is conductance, n is the desired branch, N is the total number of branches. A Wye Delta conversion is given as Rc a

b

R1

R2

Rb

R3

Ra

c

Converting Delta to Wye R1 =

Rb Rc , Ra + Rb + Rc

R2 =

Rc Ra , Ra + Rb + Rc

R3 =

Ra Rb . Ra + Rb + Rc

(14)

Converting Wye to Delta Ra =

R1 R2 + R2 R3 + R3 R1 , R1

Rb =

R1 R2 + R2 R3 + R3 R1 , R2 2

Rc =

R1 R2 + R2 R3 + R3 R1 . R3

(15)

Time-Domain: Capacitors: The voltage across a capacitor cannot change abruptly. q = CV,

(16)

where q is charge on one plate, C is capacitance, and V is voltage. dV (t ) , dt

(17)

1Zt i(τ)dτ + v(t0 ), C t0

(18)

i=C where i is current, C is capacitance, and V is voltage.

v(t) =

where i is current, C is capacitance, and V is voltage. 1 w = Cv2 , 2

(19)

where the energy w is in joules, C is capacitance, and V is voltage. For capacitors in parallel, Ceq = C1 +C2 + ... +CN . N Series Capacitors

(20)

Special case of two series capacitors 1 1 1 1 , = + +···+ Ceq C1 C2 CN

Ceq =

C1C2 . C1 + C2

(21)

Inductors: The current through an inductor cannot change abruptly.

V =L

di(t ) , dt

(22)

where i is current through the inductor, L is inductance, and V is voltage.

i(t) =

1Z t V (τ)dτ + i(t0 ), L t0

(23)

where i is current, C is capacitance, and V is voltage. 1 w = Li2 , 2 where the energy w is in joules, C is capacitance, and V is voltage.

3

(24)

For Inductors in series, Leq = L1 + L2 + · · · + LN . N Parallel Inductors

(25)

Special case of two parallel inductors 1 1 1 1 , = + +···+ Leq L1 L2 LN

Leq =

L1 L2 . L1 + L2

(26)

Advanced Analysis Techniques: Source transformation Vs = Is R

or

Is =

Vs . R

(27)

Phasor Analysis: Sinusoid is a signal that has the form of a sine or cosine function. v(t) = Vm sin(ωt + φ ) or Vm cos(ωt + φ ),

f=

1 , T

ω = 2π f ,

(28)

where Vm is the amplitude, ω is the angular freqency in rad/sec, φ is a phase shift, T is the period measured in seconds, and f is the frequency in Hz. Useful trig identities sin(ω t ± 180) = −sin(ω t),

(29)

cos(ω t ± 180) = −cos(ω t),

(30)

cos(ωt ± 90) = ∓sin(ωt),

(32)

sin(ωt ± 90) = ±cos(ωt),

(31)

Combing sinusoidal waveforms: Acos(ωt) + Bsin(ωt) = Ccos(ωt − θ ),

where C =

(33)

√ A2 + B2 and θ = atan(B/A).

Phasor is a complex number that represents the amplitude and phase of a sinusoid. Phasor, z, can be represented in equivalent rectangular, polar, exponential forms: z = x + jy = r φ = r(cosφ + jsinφ ) = re jφ . where

r=

p

x2 + y 2 ,

φ = atan(y/x),

x = rcos(φ ),

y = rsin(φ ),

(34) j=

√ −1.

Phasor Relationships for Circuit Elements: Impedance (Z = V/I): the ratio between the phasor voltage, V, to the phasor current, I, measured in Ohms (Ω). Admittance (Y = I/V): the ratio between the phasor current, I, to the phasor voltage, V, in Siemens (S). 4

Table 1: Phasor Operations Action

Method

+

z1 + z2 = (x1 + x2 ) + j(y 1 + y 2 )

-

z1 − z2 = (x1 − x2 ) + j(y 1 − y 2 )

*

z 1 z 2 = r 1 r 2 φ1 + φ2 z1 z2

/

1 z

Inverse

r1 r2

=

φ1 − φ2

= r1 −φ

Sqrt

√ √ z = r φ /2

Complex Conjugate

z∗ = x − jy = r −φ = re− jφ

Table 2: Equivalent forms based on Euler’s Identity: e± jφ = cos(φ ) + jsin(φ ) Time Domain

Phasor Domain

v(t) = Vm cos(ωt + φ )

V = Vm φ

dV dt

jωV

R

V jω

vdv

Ohm’s Law V = Z ∗ I.

(35)

Zeq = Z1 + Z2 + · · · + ZN .

(36)

N impedances in series

Voltage divider for two impedances in series v1 =

Z1 ∗ vs , Z1 + Z2

v2 =

Z2 ∗ vs , Z1 + Z2

(37)

where vs denotes the source voltage. N Parallel impedances

Special case of two parallel impedances

1 1 1 1 , = + +···+ Zeq Z1 Z2 ZN

Zeq =

5

Z1 Z2 . Z1 + Z2

(38)

Table 3: Impedances and admittances of passive elements. Element

Impedance

Admittance

R

Z=R

Y = R1

L

Z = jωL

Y=

C

Z=

1 jωC

1 jω L

Y = jωC

Current Divider formula for two impedances in parallel i1 =

Z2 ∗ is , Z2 + Z1

i2 =

Z1 ∗ is , Z1 + Z2

(39)

where is is the current source. General current divider formula in =

Yn i, Y1 +Y2 + · · ·YN s

(40)

where Y is admittance, n is the desired branch, N is the total number of branches. A Wye Delta using impedances is: a

b

Zc Z1

Z2

Zb

Z3

Za

c

Converting Delta to Wye: Z1 =

Zb Zc , Za + Zb + Zc

Z2 =

Zc Za , Za + Zb + Zc

Z3 =

Za Zb . Za + Zb + Zc

(41)

Converting Wye to Delta: Za =

Z1 Z2 + Z2 Z3 + Z3 Z1 , Z1

Zb =

Z1 Z2 + Z2 Z3 + Z3 Z1 , Z2

6

Zc =

Z1 Z2 + Z2 Z3 + Z3 Z1 . Z3

(42)

Maximum power transfer in DC When RL = RT H , pmax =

VT2H . 4RT H

(43)

When ZL = RT H , pmax =

|VT2H | . 8RT H

(44)

AC Power: Maximum power transfer in AC

The Rload for max power transfer is ZT∗H for a complex RL and | ZT H | if load can only be resistive. Instantaneous power 1 1 P(t) = Vs Is cos(φv − φi ) + Vs Is cos(2ωt + φv + φi ) 2 2

(45)

Where φv is the phase angle of the voltage and φi is the phase angle of the current. RMS of a Signal Vm Im Vrms = √ , Irms = √ 2 2 RMS of signal with multiple AC sources having different frequencies q 2 +V 22 + ... Vrms = V 1rms rms

(46)

(47)

Average power 1 P(t) = Vs Is cos(φv − φi ) = Vrms Irms cos(φv − φi ) 2

(48)

Where φv is the phase angle of the voltage and φi is the phase angle of the current. Power Triangle

S = Apparent Power (kVA) Q = Reactive Power (kVAR)  P = Real Power (kW)

where S =

p

2 Z and the power factor (PF) is given as: P2 + Q2 = VRMS IRMS = IRMS

PF = cos(θ ) = cos(φv − φi ) =

7

RealPower . Ap parentPower

(49)

Power factor can also be calculated from Z, impedance, with Z=

V φ I φ

(50)

Correcting power factor can be accomplished via a shunt capacitor: C=

P(tan(θ1 ) − tan(θ2 )) Qc = 2 2 ω ∗Vrms ω ∗Vrms

(51)

where θ1 and θ2 are the beginning and ending power factor respectively. If φv > φi then current lags voltage and the load is inductive. If φv < φi then current leads voltage and the load is capacitive.

3 Phase Power: 3 Phase for balanced Y-Y (A is the reference phase and E is magnitude of balance source voltage) PhaseVoltages : EAN = E φ , EBN = E φ − 120, ECN = E φ + 120

(52)

√ √ √ LineVoltages : EAB = E 3 φ + 30, EBC = E 3 φ − 90, ECA = E 3 φ + 150

(53)

Transformers: Let P be the primary and S be the secondary. Transformer does not pass DC power.

Efficiency η=

PS PP

(54)

Where PP is the power in the primary and PS is the power in the secondary winding respectively. Voltage ratio VP NP = NS VS Where N denotes the number of turns.

8

(55)

Current ratio IP NS = IS NP

(56)

N2 ZP = P2 ZS NS

(57)

I− = I+ = 0,

(58)

V− = V+ .

(59)

Rf Vs , R1

(60)

Where N denotes the number of turns. Reflected Impedance

Where N denotes the number of turns. Transistors:

The voltage drop across VBE is 0.7V . When the transistor is on, IC = BIB

Operational Amplifiers: Ideal Op-Amp Assumptions:

Inverting Amplifier: V0 = −

where R f is the feedback resistor, R1 is a resistor between the voltage source Vs and the inverting input, and V0 is the output voltage. Non-inverting Amplifier:   Rf Vs , V0 = 1 + R1 9

(61)

where R f is the feedback resistor, R1 is a resistor between the ground and the inverting input, and V0 is the output voltage. Summing Amplifier: V0 = −R f



 V1 V2 VN +···+ + , RN R1 R2

(62)

where R f is the feedback resistor, R1 is a resistor between the voltage sources and the inverting inputs, and V0 is the output voltage. Difference Amplifier: V0 =

R2 (1 + R1 /R2 ) R2 V2 − V1 R1 R1 (1 + R3 /R4 )

If R1 /R2 = R3 /R4 then V0 = (R2 /R1 ) ∗ (V2 −V1 ). If R2 = R1 and R3 = R4 then V0 = V2 −V1 .

10

(63)

First Order Systems: Steps to working with a source-free RC circuit: 1. Find the initial voltage, V0 , across the capacitor at time t = (-0).

+

2. KCL: IC + IR = 0. R

3. ODE: V dV + = 0. dt RC

C

V(t)

-

(64)

4. Find the time constant τ = RC. 5. Solution: Figure 1: The Source-free RC Circuit.

−t

V (t) = V0 e τ

(65)

Steps to working with a source-free RL circuit: I

1. Find the initial current, I0 , through the inductor at time t = (-0). 2. KVL: VL +VR = 0.

-

+

3. ODE:

VR

dI RI = 0. + L dt

(66)

-

R

L

VL

+

4. Find the time constant τ = L/R. 5. Solution: I(t) = I0 e

−t τ

(67)

11

Figure 2: The Source-free RL Circuit.

Steps to working with a forced RC circuit: 1. Find the initial voltage, V0 , across the capacitor at time t = (-0).

R

t=0

2. KCL: IC + IR = 0. 3. ODE:

Vs

Vs V dV = + . RC dt RC

+

+

-

C

V(t)

(68)

-

4. Find the time constant τ = RC. 5. Solution:

Figure 3: The Forced RC Circuit. V (t) = V0 , t < 0,

(69)

−t τ

V (t) = Vs + (V0 −Vs )e , t > 0.

(70)

Steps to working with a forced RL circuit: 1. Find the initial current, I0 , through the inductor at time t = (-0).

R

t=0

I

2. Find the final inductor current, I∞ .

Vs

3. Find the time constant τ = L/R.

+

+

-

V(t)

-

4. Solution: I(t) = I0 , t < 0, −t τ

I(t) = I∞ + (I0 − I∞ )e , t > 0,

(71) (72)

Figure 4: The Forced RL Circuit.

5. Obtain inductor voltage by VL = L ∗ di/dt . Second Order Systems: General Solution Steps: 1. Assume DC conditions for time = 0− (before the switch flips) and find the current through the inductor and the voltage across the capacitor. 2. Flip the switch, and calculate for time = 0+ . Recall the boxed text from Chapter 6, and then use equations (17) and (22). 3. Determine the configuration of the circuit, and calculate α and ω0 . 4. Find the appropriate dampening condition (under, critical, over) and write the equation. 5. Solve for the unknown coefficients and write the final solution. 6. Verify that you have solved for what the problem is asking.

12

The source-free series RLC circuit: Roots of the characteristic equation: q s1,2 = −α ± α 2 − ω02

ODE: d 2 i R di i = 0. + + 2 LC L dt dt

where

L

α=

R , 2L

1 ω0 = √ , LC

ωd =

q

ω 02 − α 2 .

The solutions are as follows:

+ C

R

1. If α > ω0 we have the overdamped case.

V

-

• i(t) = A1 es1 t + A2 es2 t . 2. If α = ω0 we have the critically damped case. • i(t) = (A1 + A2 t)e−αt . 3. If α < ω0 we have the underdamped case.

Figure 5: The Source-free RLC Series Circuit.

• i(t) = (A1 cos(ωd t) + A2 sin(ωd t))e−αt .

The source-free parallel RLC circuit: Roots of the characteristic equation: q s1,2 = −α ± α 2 − ω02

ODE: d2v v 1 dv = 0. + + LC dt 2 RC dt

where α=

+ R

L

C

V

1 , 2RC

1 ω0 = √ , LC

ωd =

q

ω02 − α 2 .

The solutions are as follows: 1. If α > ω0 we have the overdamped case.

-

• v(t) = A1 es1 t + A2 es2 t . 2. If α = ω0 we have the critically damped case. • v(t) = (A1 + A2 t)e−αt .

Figure 6: The Source-free RLC Series Circuit.

3. If α < ω0 we have the underdamped case. • v(t) = (A1 cos(ωd t) + A2 sin(ωd t))e−αt .

13

Step response of a series RLC circuit: Roots of the characteristic equation: q s1,2 = −α ± α 2 − ω02

ODE: d 2 v R dv Vs v . = + + LC L dt LC dt 2 t=0

where α=

L

R

R , 2L

1 ω0 = √ , LC

ωd =

q

ω 02 − α 2 .

The solutions are as follows: Vs

+

+

-

C

V

1. If α > ω0 we have the overdamped case.

-

• v(t) = Vs + A1 es1 t + A2 es2 t . 2. If α = ω0 we have the critically damped case. • v(t) = Vs + (A1 + A2 t)e−αt .

Figure 7: The Forced RLC Series Circuit.

3. If α < ω0 we have the underdamped case. • v(t) = Vs + (A1 cos(ωd t ) + A2 sin(ωd t ))e−αt .

Step response of a parallel RLC circuit: Roots of the characteristic equation: q s1,2 = −α ± α 2 − ω02

ODE:

where

d2i Is i 1 di = + + . dt 2 RC dt LC LC

α=

1 , 2RC

1 ω0 = √ , LC

ωd =

q

ω02 − α 2 .

The solutions are as follows: Is

+ t=0

R

L

C

V

1. If α > ω0 we have the overdamped case.

-

• i(t) = Is + A1 es1 t + A2 es2 t . 2. If α = ω0 we have the critically damped case. • i(t) = Is + (A1 + A2 t)e−αt .

Figure 8: The Forced RLC Parallel Circuit.

3. If α < ω0 we have the underdamped case. • i(t) = Is + (A1 cos(ωd t ) + A2 sin(ωd t ))e−αt .

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