Electrica Flashcards
Electric charge can be distributed from one place to another under the influence of an electric field but the algebraic sum of positive and negative charges in the system cannot change
Conservation of charge
The total electric flux generated by a point charge is proportional to the charge. Formula
electric flux = Q/ε
Que determina el flux que pasa a traves de un medium?
Permittivity of a medium , ε
Columbus Law formula
F2 = Q_1Q_2/(4piεr^2) a_r12
___ Law states that the total electric flux passing out of an enclosing surface is proportional tot the total charge withing the surface.
Gauss’ Law
(T/F) The mathematical formulation of Gauss law states that the total enclosed charge can be determined by summing all of the electric fields on the Gaussian surface.
True
Formula - Work performed by a moving charge Q1 from point 1 to point 2
W = - Q* integral (E*dl)
Formula - Work performed in moving a point charge QB in the radial direction withing a field crated by point charge QA
W = - integral (F * dr) = - integral (Q_A&Q_B/(4piεr^2) =
(Q_A * Q_B/4piε)* (1/r_2 - 1/r_1)
Formula - Work done in moving an object of charge Q a distance l parallel to a uniform field
W = - F* l = -EQl = -V_plates Ql/d = -Q* delta V
___ Law states that an induced voltage, v, will be generated in a circuit when there is a change in the magnetic flux.
Faraday’s Law of induction
Propiedad del circuito para oponerse al flujo de corriente
Resistance
Resistors in serie
Rs = R1 + R2 + R3
Resistors in parallel
Rp = 1/(1/R1 + 1/R2 + 1/R3)
(T/F) Resistors in serie comparte la misma corriente
True
(T/F) Resistors in parallel comparten el mismo voltaje drop
True
Capacitors in serie
1/Cs = 1/C1 + 1/C2
Capacitors in parallel
Cp = C1 + C2 + C3
Inductores en paralelo
1/Lp = 1/L1 + 1/L2
Inductores en serie
Ls = L1 + L2 + L3
Resolver circuitos con Ley de Ohm, pasos para hallar los I, delta V por cada resistor
- Simplificar el circuito lo mas posible hasta dejarlo en 1 solo resistor
- Calcular la corriente ya que el Voltaje antes del resistor sera el de la bateria y el voltaje luego del resistor sera 0.
- Ir expandiendo al circuito mas complejo. Por ejemplo, si antes del mas simple habia un circuito con 3 resistores en serie - se sabe que la corriente no cambia en los resistores en serie, por lo tanto se puede calcular el voltaje drop en el resistor. Pero el voltaje real del nodo sera la resta del V antes del resistor menos el V drop durante el resistor.
- Seguir expandiendose al circuito mas complicada, por ejemplo si ahora lo que continua son resistores en paralelo, se sabe que el V no cambia entre ellos solo la corriente. por lo tanto se calcula el V drop con los valores que ya tenemos de V y se calcula la corriente por cada resistor
Resolver circuitos con Ley de Kirchhoff’s
- Establecer las corrientes por cada branch del circuito. Le dimos direcciones que no es necesario estar correcto en ellas.
- Creamos la ecuacion de corrientes en uno de los junctions que las una todas
- Se crean Loops
- En estos loops si la corriente va desde el lado - al + de la bateria entonces hay un rise en voltaje y es (+) la ecuacion
- Si pasamos un resistor y la corriente va en la misma direccion que nos estamos moviento entonces hay un voltaje drop y es negativo
- Todo se iguala a 0. - Se crean los loops necesarios para encontrar la cantidad de variables desconocidas.
Amplificator factor formula
Av = V_out/V_in
Ai = Beta = i_out/i_in
Que son los amplificadores
Pueden aumentar o disminuir la senal, cambiar de signo o de phase.
Ideal voltage amplifier has an infinite ____ impedence and zero ___ impedance
- input / v_in appears across the amplifier and no current or power is drawn from the source
- output / all of the output current flows thorugh the load resistor.
Class ____ amplifier has a quiescent point in the center of the active region of the operationg curve. Greatest Linearity and least distortion.
Load current flows throughout the full input signal cycle.
Efficiency 50%
Class A
Class __ amplifiers has the quiescent point established at the cutoff point.
A load current flows only if the signal drives the amplifier into its active region, and the circuit acts like an amplifying half-wave rectifier. In pairs/transistors/push-pull operation. Efficiency 78%
Class B
Class __ amplifier has a quiescent point somewhat above cutoff but where a portion of the input signal still produces no load current. The outpout current flows dor more than half of the input cycle. Each output transistors conducts for slightliy less than 180 degrees of the cycle.
Class AB
Class ___ amplifier have a quiescent points wll into the cutoff region. Load curent flows during less than one-half of the input cycle. Efficiency 100%
Class C
Para resolver KCL con nodos se tiene que igualar las I de entradas al nodo = con las I de salidas al nodo
Cierto
Op Amp - The outpout voltage should remain withing 3V from either value to avoid distortion of the output signal at rated values.
|v1 - v2| < (V_dc - 3V) / A
Ideal Operational Amplifier Characteristics / Rules
- Zin = infinite
- Zout = 0
- A = infinite
- BW = 0
- Vout = 0 when v+ =v-
- Current to each input is zero
- Voltage between the two inputs terminals is zero
- op amp operates in the linear range
n
If a potential difference is generated by a single conductor passing through a magnetic field the potential difference depends on-
-Speed with which the conductor cuts the magnetic field
- Lenght of the conductor
- Magnetic field density
- No depende del diametro del conductor!
Como saber si los resistores estan paralell or series
Paralelo - tienen junction y dos cables con los que comparten voltajes diferentes (comparten los voltajes de los cables)
Serie - Es el mismo voltaje atraves de todo el loop
Which measurements are required to determine the phase angle of a single-phase circuit
Power, Voltage and Current
AC - Power Factor relation to real power P, reactive power W and complex power S
pf = P/S
Un circuito con un pf =1 quiere decir que es resonant, entonces f=60Hz
True
Para calcular Vout con amplificadores
Se utiliza KCL en nodos. La suma de los I que entran igual a los I que salen
Si el I entra en el nodo la resta de los voltages es el voltaje de afuera menos el voltaje del nodo.
Si el I sale del nodo la resta es el voltaje del nodo menos el voltaje de afuera.
