Review Part 3 of Course

Passive Circuit Elements i i i + -

Energy stored in the capacitor The instantaneous power delivered to the capacitor is The energy stored in the capacitor is thus

Energy stored in the capacitor Assuming the capacitor was uncharged at t = - , and knowing that represents the energy stored in the electric field established between the two plates of the capacitor. This energy can be retrieved. And, in fact, the word capacitor is derived from this element’s ability (or capacity) to store energy.

Parallel Capacitors Thus, the equivalent capacitance of N capacitors in parallel is the sum of the individual capacitances. Capacitors in parallel act like resistors in series.

Series Capacitors The equivalent capacitance of N series connected capacitors is the reciprocal of the sum of the reciprocals of the individual capacitors. Capacitors in series act like resistors in parallel.

Energy stored in an inductor The instantaneous power delivered to an inductor is The energy stored in the magnetic field is thus

Series Inductors The equivalent inductance of series connected inductors is the sum of the individual inductances. Thus, inductances in series combine in the same way as resistors in series.

Parallel Inductors The equivalent inductance of parallel connected inductors is the reciprocal of the sum of the reciprocals of the individual inductances. Thus inductances in parallel combine like resistors in parallel.

Complex Numbers real imag A  Euler's equation: Complex Plane  measured positive counter-clockwise Note: A

Comparing Sinusoids Note: positive angles are counter-clockwise

AC Circuits and Phasors  XMXM Recall that when the assumed form of the current cancelled out. We were then left with just the phasors A phasor is a complex number that represents the amplitude and phase of a sinusoid. was substituted into the differential equations, the

Impedance Note that impedance is a complex number containing a real, or resistive component, and an imaginary, or reactive, component. Units = ohms VRVR VLVL

Admittance conductance Units = siemens susceptance VRVR VLVL

Re Im Re Im Re Im V V V V V V I I I I I I I in phase with V I lags V I leads V

We see that if we replace Z by R the impedances add like resistances. Impedances in series add like resistors in series Impedances in parallel add like resistors in parallel

Voltage Division But Therefore

Instantaneous Power Note twice the frequency

Average Power Purely resistive circuit Purely reactive circuit

Effective or RMS Values We define the effective or rms value of a periodic current (voltage) source to be the dc current (voltage) that delivers the same average power to a resistor. root-mean-square

Effective or RMS Values Using and

Ideal Transformer - Voltage  The input AC voltage, v 1, produces a flux This changing flux through coil 2 induces a voltage, v 2 across coil 2

Ideal Transformer - Current  The total mmf applied to core is Magnetomotive force, mmf For ideal transformer, the reluctance R is zero.

Ideal Transformer - Impedance Input impedance Load impedance Turns ratio

Ideal Transformer - Power Power delivered to primary Power delivered to load Power delivered to an ideal transformer by the source is transferred to the load.

Force on current in a magnetic field Force on moving charge q -- Lorentz force Current density, j, is the amount of charge passing per unit area per unit time. N = number of charges, q, per unit volume moving with mean velocity, v. j Force per unit length on a wire is

Rotating Machine B i i X Force in Force out + - brushes commutator

B i i X Force in Force out + -  X r B l a b Back emf

B i i X Force in Force out + - brushes commutator

Armature with four coil loops X X X X N S

Motor Circuit Power and Torque

AC Nodal Analysis

For steady-state AC circuits we can use same the method of writing nodal equations by inspection that we learned for resistive circuits. To do so, we must replace resistances with impedances.

We solved Problem 4.31 in Class: Change impedances to admittances

Nodal Analysis for Circuits Containing Voltage Sources That Can’t be Transformed to Current Sources 1.Assume temporarily that the current across each voltage source is known and write the nodal equations in the same way we did for circuits with only independent voltage sources. 2.Express the voltage across each independent voltage source in terms of the node voltages and replace known node voltages in the equations. 3.Rewrite the equations with all unknown node voltages and source currents on the l.h.s. of the equality and all known currents and voltages on the r.h.s of the equality.

We solved #4.33 in text in Class: Note: V 2 = 10 assume I 2

AC Mesh Analysis

For steady-state AC circuits we can use same the method of writing mesh equations by inspection that we learned for resistive circuits. To do so, we must replace conductances with admittances.

We solved Problem 4.38 in text in Class: Find I 1 and I 2 :

What happens if we have independent current sources that can’t be transformed in the circuit? 1.Assume temporarily that the voltage across each current source is known and write the mesh equations in the same way we did for circuits with only independent voltage sources. 2.Express the current of each independent current source in terms of the mesh currents and replace known mesh currents in the equations. 3.Rewrite the equations with all unknown mesh currents and voltages on the left hand side of the equality and all known voltages and currents on the r.h.s of the equality.

We solved Problem 4.40 in text in Class: Find I 0 : Assume you know V 2 Note I 2 = -2

Matlab Solution: Know how to program

AC Thevenin's Theorem

Thevenin’s theorem states that the two circuits given below are equivalent as seen from the load Z L that is the same in both cases. V Th = Thevenin’s voltage = V ab with Z L disconnected (=  ) = the open-circuit voltage = V OC

AC Thevenin's Theorem Z Th = Thevenin’s impedance = the input impedance with all independent sources turned off (voltage sources replaced by short circuits and current sources replaced by open circuits). This is the impedance seen at the terminals ab when all independent sources are turned off.

We solved Problem 4.40 using Thevenin's Thm. in Class:

AC Superposition

Superposition Principle The superposition principle states that the voltage across (or the current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.

Steps in Applying the Superposition Principle 1.Turn off all independent sources except one. Find the output (voltage or current) due to the active source. 2.Repeat step 1 for each of the other independent sources. 3.Find the total output by adding algebraically all of the results found in steps 1 & 2 above.

Example Done in Class: Note that the voltage source and the current source have two different frequencies. Thus, if we want to use phasors, the only way we've solved sinusoidal steady-state problems, we MUST use superposition to solve this problem. We considered each source acting alone, and then found v 0 (t) by superposition. Remember that

By superposition: Example v 0 (t) = Response due to voltage source + Response due to voltage source