# Circuit Theory/Phasors/Examples/Example 8

Series RL circuit for example 8

Given that the current source is defined by ${\displaystyle I_{s}(t)=120{\sqrt {2}}cos(377t+120^{\circ })}$, find all other voltages, currents and check power.

### Label Loops Junctions

Series RL circuit marked up for analysis

The important point is that nothing changes even though the source is oscillating. + and - still have to capture the circuit topology and work their way into the equations.

## Knowns, Unknowns and Equations

Knowns: ${\displaystyle I_{s},R,L}$
Unknowns: ${\displaystyle V_{s},v_{R},v_{L}}$
Equations: ${\displaystyle v_{R}(t)=R*i(t),v_{L}=L*{d \over dt}i(t),v_{R}(t)+v_{L}(t)-V_{s}(t)=0}$

## Calculus Symbolic

Evaluate the terminal relations in this order:

${\displaystyle v_{R}(t)=R*i(t)}$
${\displaystyle v_{L}=L*{d \over dt}i(t)}$
${\displaystyle V_{s}(t)=v_{R}(t)+v_{L}(t)}$

## Calculus Numeric

phasor matlab solution of RL circuit .... m file
${\displaystyle v_{R}(t)=10*120{\sqrt {2}}cos(377t+{\frac {2\pi }{3}})=1700.0*cos(377.0*t+2.09)}$
${\displaystyle v_{L}=.01*{d \over dt}120{\sqrt {2}}cos(377t+{\frac {2\pi }{3}})=-640.0*sin(377.0*t+2.09)}$
${\displaystyle V_{s}(t)=v_{R}(t)+v_{L}(t)=1700.0*cos(377.0*t+2.09)-640.0*sin(377.0*t+2.09)}$
phasor math avoids trig! m-file

Now the trouble is how to wrestle the above answer into a form where it can be compared to the other answers and our intuition built up. There are three possible ways:

Euler
Trig
Phasors (which is derived from Euler).

Here is the phasor method. Notice the math is done in the phasor domain ... with imaginary numbers.

${\displaystyle 1700cos(377t+2.09)\Leftrightarrow 1700cos(2.09)+j1700sin(2.09)}$
${\displaystyle -640sin(377t+2.09)\Leftrightarrow -640*sin(2.09)+j640*cos(2.09)}$

Now the numbers are in the phasor world. The next step is to add them all up.

${\displaystyle \mathbb {V} _{s}=1814\angle 2.45}$

Now put back into the time domain:

${\displaystyle V_{s}=1814cos(377t+2.45)}$

Phasors are also an alternative to the calculus above.

Because no integrals were evaluated in this calculus solution, there is no integration constant!

## Phasor Symbolic

This problem could be done with Laplace transforms, but Calculus functions are less complex. It can also be done with phasors. The question is "Which mathmetical tool should be used?" Calculus above leaves the answer in a form that would require some trig to get back into the form:

${\displaystyle V_{s}=V_{m}cos(\omega t+\phi )}$

The phasor solution enables us to stay close to the above solution form.

The phasor solution provides a uniform way (one tool) for doing all inductor and capacitor problems. We will use it through out the rest of the course.

### time domain

${\displaystyle v_{R}(t)=R*i(t)}$
${\displaystyle v_{L}=L*{d \over dt}i(t)}$
${\displaystyle v_{R}(t)+v_{L}(t)-V_{s}(t)=0}$

### phasor domain

${\displaystyle i(t)\rightarrow \mathbb {I} }$
${\displaystyle \mathbb {I} =I_{m}\angle \phi }$
${\displaystyle \mathbb {V} _{R}=R*\mathbb {I} =R*I_{m}\angle \phi }$
${\displaystyle {d \over dt}i(t)\rightarrow j\omega \mathbb {I} =\omega \mathbb {I} \angle {\frac {\pi }{2}}=\omega I_{m}\angle ({\frac {\pi }{2}}+\phi )}$
${\displaystyle \mathbb {V} _{L}=jL\omega \mathbb {I} =L*\omega I_{m}\angle ({\frac {\pi }{2}}+\phi )}$
${\displaystyle \mathbb {V} _{S}=jL\omega \mathbb {I} +R*\mathbb {I} =(R+jL\omega )\mathbb {I} ={\sqrt {R^{2}+(L\omega )^{2}}}\angle arctan({\frac {L\omega }{R}})*I_{m}\angle \phi }$
${\displaystyle \mathbb {V} _{S}=I_{m}*{\sqrt {R^{2}+(L\omega )^{2}}}\angle (arctan({\frac {L\omega }{R}})+\phi )}$

### back to time domain

${\displaystyle V_{s}(t)=\operatorname {Re} (\mathbb {V} _{s}e^{j\omega t})=I_{m}{\sqrt {R^{2}+(L\omega )^{2}}}cos(\omega t+\phi +arctan({\frac {L\omega }{R}}))}$
${\displaystyle v_{L}(t)=\operatorname {Re} (\mathbb {V} _{L}e^{j\omega t})=I_{m}*L*\omega *cos(\omega t+\phi +{\frac {\pi }{2}})}$
${\displaystyle v_{R}(t)=\operatorname {Re} (\mathbb {V} _{R}e^{j\omega t})=R*I_{m}cos(\omega t+\phi )}$

## Phasor Numeric

### time domain

${\displaystyle v_{R}(t)=10*120{\sqrt {2}}cos(377t+{\frac {2*\pi }{3}})=1700cos(377t+2.09)}$

### phasor domain

Rather than substituting into the time domain as last time, showing phasor (imaginary math) by substituting into the phasor domain.

${\displaystyle i(t)\rightarrow {\mathbb {I} }=120{\sqrt {2}}\angle {\frac {2*\pi }{3}}}$
${\displaystyle R+jL\omega =10+.01*377j=10.6870\angle 0.3605}$
${\displaystyle {\mathbb {V} _{s}}=\mathbb {I} *(R+jL\omega )=120{\sqrt {2}}*10.6870\angle ({\frac {2*\pi }{3}}+0.3605)=1814\angle 2.45}$
${\displaystyle \mathbb {V} _{L}=I_{m}*\omega *L\angle (\phi +{\frac {\pi }{2}})=120{\sqrt {2}}*377*.01\angle ({\frac {2*\pi }{3}}+{\frac {\pi }{2}})=640\angle 3.67}$
doing problem using phasors rather than calculus m-filec

### back to time domain

${\displaystyle V_{s}(t)=1811cos(377t+2.45)}$
${\displaystyle v_{L}(t)=640cos(377t+3.67)}$
${\displaystyle v_{R}(t)=1700cos(377t+2.09)}$

### comparison

Only ${\displaystyle V_{s}}$ was calculated with both the phasor ${\displaystyle 1811}$ and calculus/phasor ${\displaystyle 1814}$ method. The ${\displaystyle 1814}$ number can be found in the phasor math of both the calculus method and the phasor method. But the phasor math in both sections had an intermediate calculation of R + jLw in the phasor domain. A matlab script was written for the phasor solution that merely plugged into the time domain symbolic solution. This did not allow errors to build up in matlab caused by reusing numbers .. and came up with ${\displaystyle 1811}$.

The goal is to type numbers into the calculator once, or into matLab once and calculate the answers directly from the given numbers with no intermediary calculations.

## Simulation

Vs,VL and I of the simulation graphed above.

### period check

The period above looks to be between 16ms and 17ms, closer to 17ms. This agrees with the formula:

${\displaystyle \omega =2\pi *f}$
${\displaystyle f={\frac {\omega }{2\pi }}}$
${\displaystyle T={\frac {1}{f}}={\frac {2\pi }{\omega }}={\frac {2\pi }{377}}=16.7ms}$

### magnitude check

The magnitude of Vs above appears to be close to 2000 volts. This is close to 1811/1814 of the math.

The magnitude of i(t) appears to be ${\displaystyle 120*{\sqrt {2}}}$.

The magnitude of VL appears to be above 500 volts which could be the 640 volts from the math.

Vr can not be graphed because of the choice of ground. Notice how Vr and VL share the same -. They can be measured with an oscilloscope or the above simulator. Vr can not.

### transient response check

The transient response is different. The straight lines and starting above 0 seems odd. Since this is a steady state analysis, going to save digging into this until later. Otherwise, everything is the same as when driven by a voltage source.

### phase check

Voltage will always lead the current through an inductor (think of the terminal relationship or ELI) by ${\displaystyle 90^{\circ },{\frac {\pi }{2}}}$ or ${\displaystyle {\frac {1}{4}}}$ of a period, whether a current source or voltage source is driving the circuit.

## Voltage Check

Could do the phasor math again to make sure they check, but this is the math that produced the values in the first place. Here the goal is to do a quick spot check. This check is not good, but it is quick. It can detect many mistakes. At the least it builds our confidence in the answer. Need to plug into:

${\displaystyle v_{L}=L*{d \over dt}i(t),v_{R}(t)+v_{L}(t)-V_{s}(t)=0}$

And make sure that they add to zero.

${\displaystyle V_{s}(t)=1811cos(377t+2.45)}$
${\displaystyle v_{L}(t)=640cos(377t+3.67)}$
${\displaystyle v_{R}(t)=1700cos(377t+2.09)}$

Pick t = 1:

${\displaystyle V_{s}(t)=1811cos(377+2.45)=-1410}$
${\displaystyle v_{L}(t)=640cos(377+3.67)=-560}$
${\displaystyle v_{R}(t)=1700cos(377+2.09)=-857}$

The equation above is true. The actual numbers above would need more accuracy (decimal places) to get closer to zero.

## Power Analysis

Power Analysis is rooted in the phasor domain!

${\displaystyle \mathbb {V} _{s}=1814\angle 2.45}$
${\displaystyle \mathbb {I} =120{\sqrt {2}}\angle 2.09}$
${\displaystyle \mathbb {I} ^{*}=120{\sqrt {2}}\angle -2.09}$

if :${\displaystyle \mathbb {V} =M_{v}\angle \phi _{v}\quad }$, and ${\displaystyle \quad \mathbb {I} =M_{i}\angle \phi _{i}}$ then

${\displaystyle \mathbb {S} =\mathbb {V} \mathbb {I} ^{*}={\frac {M_{v}M_{i}}{2}}\angle (\phi _{v}-\phi _{i})={\frac {120{\sqrt {2}}*1814}{2}}\angle (2.45-2.09)=15400\angle 0.36=14400+5420j}$

and

${\displaystyle cos(.36)=.936}$

The power is much larger ... by a factor of 100 because of the resistor, but the phase angle is still the same. This means the power factor is still the same.

Value Units Description
${\displaystyle 15400}$ volt-ampere va apparent power what utility companies manage: peak power they design for, peak power they have to deliver
${\displaystyle .936}$ unitless power factor, ratio of real power to apparent power, ideally 1
${\displaystyle 14400}$ watt W real, average, active power ... what consumers want to pay for (watt-hours)
${\displaystyle 5420}$ volt-amp-reactive var reactive power ... why not all outlets in a room are on the same circuit breaker

## Intuition

ELI ... voltage leads the current through an inductor

Derivatives cause a lag ... a delay in time ... which is a positive angle in sinusoidal.