255 Final Flashcards

Question

Charge

Answer 1

What a person feels walking on a carpet then receiving a shock from a metal object. It is from electron exchange with the surroundings. Measured in coulombs

Answer 2

1 electron = 1.602x10^-19C

Answer 3

Time rate of change of charge measured in amperes. (C/s)

Answer 4

Q= integral from to to t of (i dt)

Answer 5

A current that remains constant in time and does not change direction

Answer 6

A current that varies sinusoidally in time(does change direction)

Answer 7

Energy required to move a unit charge through an element

Answer 8

Vab= dw/dq (V= J/C = Nm/C)

Answer 9

Vab = -Vba

Answer 10

Time rate if change of expending/absorbing energy (W)

Answer 11

P = dw/dt = dw/dq*dq/dt

Answer 12

P= vi (VA=W)

Answer 13

If current enters through the positive terminal power is positive. If entering through the negative terminal, negative sign is used

Answer 14

+ power absorbed = - power supplied Or Sum = 0

Answer 15

W = integral from to to t of (p dt) or W = integral from to to t of (vi dt)

Answer 16

Voltage is set, current determined by circuit equations

Answer 17

Current is set voltage dictated by circuit equations

Answer 18

Voltage is function of current/voltage elsewhere in circuit Current determined by circuit equations

Answer 19

Current is functin of current/voltage elsewhere in circuit. Voltage determined by circuit equations

Answer 20

1) define problem 2) present everything known 3) establish potential solution paths 4) attempt a problem solution 5) evaluate solution, is it accurate 6) check to make sure problem has been solved satisfactory

Answer 21

p , the tendency if a materials to resist the flow of current. Unit is ohms

Answer 22

R= pl/A (ohms) p-material resistivity l- length A- cross sectional area

Answer 23

Resistance is 0

Answer 24

Resistance is infinite

Answer 25

G = 1/R = i/V (Units is S of 1/ohm or A/V)

Answer 26

P = iv = i^2R = V^2/R = V^2G = i^2/G

Answer 27

Power is always positive

Answer 28

Black (0), Brown (1), R(2), O(3), Y(4), G(5), B(6), V(7), Grey (8), white(9)

Answer 29

A single element such as a voltage source or a resistor

Answer 30

Point of connection between two or more branches

Answer 31

Any closed path in a circuit

Answer 32

The current going into a node has to be equal to the current leaving the node.

Answer 33

All voltages over a loop must sum to zero

Answer 34

Req = R1+R2…+Rn

Answer 35

Vn = Rn/ (R1+R2+…+Rn) *v Or Vn = Rn/Req *v

Answer 36

1/Req = 1/R1 + 1/R2 +…+1/Rn

Answer 37

in = Req/Rn *i

Answer 38

1/Geq = 1/G1 + 1/G2 + …+ 1/Gn

Answer 39

Geq = G1+G2

Answer 40

1) select node as reference node (often ground) and assign voltages to other nodes 2)Apply RCL to each non reference node. Use ohms law to express currents as node voltages 3) solve equations to get unknown node voltages

Answer 41

A dependent or independent voltage source between two non reference nodes.

Answer 42

Loop not containing another loop

Answer 43

Using kvl to solve for currents in a circuit with multiple meshes

Answer 44

1) Assign mesh currents (typically clockwise) 2) Apply KVL to each mesh. Use ohms law to express voltages in terms of mesh currents. 3) solve the system of equations

Answer 45

Two meshes have a dependent or independent current source in common

Answer 46

I = GV = V/R

Answer 47

Only applicable when circuit contains independent current sources. -diagonal terms are aim of conductance’s connected directly to node 1 or 2 - off diagonal terms are the conductance’s connected between the nodes but negative. -right hand side is the sum of the currents entering through the node

Answer 48

Only apply when circuit only contains independent voltage sources. -diagonal terms are sum of resistance in mesh -off diagonal terms are negative if resistance common to meshes Right hand side is sum (taken clockwise) of independent voltage sources

Answer 49

If v1=Ri1 and v2 = Ri2 then v = (i1+i2)R = i1R +i2R = v1+v2

Answer 50

p=i^2R and p=R/v^2. Are not linear relationships

Answer 51

Voltage/current for an element is the sum of the voltage/current from each independent source

Answer 52

1) turn of independent sources except one. Find the output v or i. (Voltage= short circuit, current = open) 2) repeat step 1 for each independent source 3) find total contribution by adding all the contributions from independent sources

Answer 53

Vs = isR or is= Vs/R Can be used for dependent or independent sources

Answer 54

A linear two part circuit can be replaced with a voltage source and resistor in series

Answer 55

1) put Vab = 1volt (test) find io into terminal a 2) put Iab = 1A (test) find Vo into a

Answer 56

A linear circuit can be replaced with a current source and resistor in parallel.

Answer 57

In = Isc = Vth/Rth

Answer 58

p= vi = i^2Rl = (Vth/Rth+Rl)^2*Rl = Vth^2 Rl/(Rth +Rl)^2

Answer 59

Pmax = Vth^2/ 4Rth

Answer 60

Real world sources will not provide a consistent output and will vary depending on the load resistance.

Answer 61

A type of ammeter

Answer 62

Designed to accurately measure resistance. The variable resistor is adjusted until there is no current through the galvanometer. Forms two voltage divider circuits

Answer 63

Rx = R2R3/R1

Answer 64

They are passive elements that dissipate energy

Answer 65

Is a circuit element that is passive and stores energy. A capacitor has two parallel conducting plates that are separated by an insulator. Capacitors apply voltage v over the plates creating a positive charge q on the positive side and a negative charge q on the negative side. It is a linear relationship between charge and voltage

Answer 66

Ratio between charge (on one plate) and voltage on capacitor.

Answer 67

C = q/v (farads)

Answer 68

C = EA/d = ErEoA/d

Answer 69

Current can be found by taking the derivative of this charge with respect to time. i=dq/dt = Cdv/dt

Answer 70

Integrate the current relationship. Or V = 1/C integral from to to t of (i(z)dz) +vo(t)

Answer 71

W = 1/2 Cv^2 = 1/2 q^2/C

Answer 72

Gives open circuit. I = 0

Answer 73

Gives short circuit. I = infinity

Answer 74

Ceq = C1 + C2 +…+ Cn

Answer 75

1/Ceq = 1/C1 + 1/C2 + …. 1/Cn

Answer 76

Tendency to oppose change in current

Answer 77

L = NuA/l N- number of coils u-permeability A- cross section area l- length

Answer 78

Proportional to the inductance and the derivative of the current with respect to time V = L di/dt

Answer 79

i = 1/L integral from to to t of v(z)dz + i(to)

Answer 80

W = 1/2 Li^2

Answer 81

Leq = L1 + L2 +… +Ln Like resistors

Answer 82

1/Leq = 1/L1 + 1/L2 + … + 1/Ln

Answer 83

V(t) = Vm sin(wt) Here Vm is the amplitude of the sinusoid w is the angular frequency and wt is the argument. The period is T = 2pi/w.

Answer 84

V(t) = (t+nT) It repeats every nT where n is an integer

Answer 85

V(t) = Vm sin(wt + theta) Theta is an offset

Answer 86

A complex number representing amplitude and phase of a sinusoid.

Answer 87

Z = re^(jtheta)

Answer 88

r = sqrt(x^2 + y^2) Theta = arctan(y/x) X = rcos(theta) y = rsin(theta) re^(jtheta) = rcos(theta) + j rsin(theta)

Answer 89

e^(jtheta) = cos(theta) +- j sin(theta) -> Re{e^jtheta} = cos(theta)

Answer 90

It is always for cosine. For sine a phase shift of -90 is included Vmsin(wt+theta) = Vm

Answer 91

dv/dt (time domain) = jwV (phasor domain) Or Integral vdt = V/jw

Answer 92

v(t) is the instantaneous time domain representation and V is the phasor domain representation. v(t) is the time dependent and V is not v(t) is real while V is complex

Answer 93

The AC version of resistnce. It is the ratio of phasor, voltage, V to phasor current I.

Answer 94

Is admittance

Answer 95

Y = 1/Z = I/V (S) And Y = G +jB

Answer 96

Vth = ZnIn Zn = Zth

Answer 97

ZL should be Zth* = Rtb -jXth P max = |Vth|^2/8Rth

Answer 98

p(t) = v(t) i(t) (W)

Answer 99

i(t) = Im cos(wt +thetai)

Answer 100

v(t) = Vmcos(wt +theta v)

Answer 101

p(t) = 1/2 VmImcos(thetav -thetai) + 1/2 VmIm cos(2wt + theta v +theta i)

Answer 102

P = 1/T integral T to 0 p(t) dt Or P = 1/2 VmIm cos(thetav -theta i) Or P = 1/2 Real {VI*} Or P = 1/2 VI*

Answer 103

A resistive load (R) absorbs power at all times while a reactive load( L kr C) absorbs 0 average power

Answer 104

First set the impedance of the Thevenin equivalent and the load. Zth = Rth +JXth Zl = Rl + jXl Need to find where Rl = Rth and Xth =-Xl

Answer 105

Zl = Rl + jXL = Rth -jXth = Zth*

Answer 106

Pave = |Vth|^2/8Rth If the load is only real Rl = |Zth|

Answer 107

The effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current.

Answer 108

Ieff = sqrt(1/T integral to to t i^2 dt)

Answer 109

Veff = sqrt (1/T integral from t to to v^2 dt)

Answer 110

Ieff = Irms Veff = Vrms

Answer 111

Xrms = sqrt(1/T integral to to t x^2 dt)

Answer 112

The effective value of a periodic signal os its root mean square rms value

Answer 113

The product of the rms value of current and rms value of voltage. The units are VA. S = VrmsIrms

Answer 114

The ratio between the power and apparent power pf = P/S = cos(thetav -thetai)

Answer 115

Thetav -theta i

Answer 116

Means current leads ahead of voltage (like capacitor). Power factor angle is negative

Answer 117

Means current is behind voltage( like inductor). Power factor angle is positive

Answer 118

S is complex power |S| is apparent power

Answer 119

S = Irms*Z = Irms^2 |Z|

Answer 120

Q, imaginary part

Answer 121

Q= 0 for resistive load Q < 0 for capacitive loads Q> 0 for inductive loads

Answer 122

S is hypotenuse P and Q are legs and angle is Thetav-Theta i

Answer 123

|Z| is hypotenuse and X and R are legs. The angle is thetav -theta i

Answer 124

Units are VARS. S = S1+S2+…+Sn

Answer 125

C = Qc/wVrms^2

255 Final Flashcards

(162 cards)