Circuit Theory/Phasors/Examples/Example 10

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Parallel RL circuit for example 10

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

Label Loops Junctions[edit]

Parallel RL circuit marked up for analysis

In a parallel circuit, all devices have the exact same voltage across them.

Knowns, Unknowns and Equations[edit]

Knowns: I_s, R, L
Unknowns: V_s, i_R, i_L
Equations: V_s(t) = R * i_R(t), V_s = L * {d \over dt}i_L(t), i_R(t) + i_L(t) - I_s(t) = 0

Phasor Symbolic[edit]

mupad attempt to do some of phasor algebra ... need to find a way to solve symbolically for the real and imaginary parts in rectangular form and polar form, plus find a way to numerically evaluate .. m-file

time domain[edit]

Evaluate the terminal relations in this order:

i_R(t) = \frac{V_s}{R}
i_L(t) = \int\frac{V_s}{L} dt

Substitute into this equation:

I_s(t)= i_R(t) + i_L(t)
I_s(t)= \frac{V_s}{R} + \int\frac{V_s}{L} dt

Need to formulate as a differential equation (to keep mathematicians happy) so differentiate all sides (don't let them see you doing this):

{dI_s(t) \over dt}= \frac{1}{R}{dV_s \over dt} + \frac{V_s}{L}

So have a differential equation that can be solved for V_s.

Phasor domain[edit]

now transform both operations into the phasor domain ..

{I_s}(t) \rightarrow \mathbb{I}_s \rightarrow I_m\angle \phi
V_s(t) \rightarrow \mathbb{V}_s
\int V_s(t) dt \rightarrow  \frac{1}{j\omega} \mathbb{V}_s

So:

\mathbb{I}_s = \frac{\mathbb{V}_s}{R} + \frac{\mathbb{V}_s}{j\omega L} or j\omega \mathbb{I}_s = j\omega \frac{\mathbb{V}_s}{R} + \frac{\mathbb{V}_s}{L}

Solving for \mathbb{V}_s

\mathbb{V}_s = \frac{\mathbb{I}_s j\omega L R}{R + j\omega L}

Putting in Rectangular form:

\mathbb{V}_s = \mathbb{I}_s * \frac{\omega^2L^2R + j\omega LR^2}{R^2+\omega^2L^2}

Putting in Polar form:

\mathbb{V}_s = \mathbb{I}_s \frac{\sqrt{(\omega^2L^2R)^2 + (\omega L R^2)^2}}{R^2+\omega^2L^2}\angle \arctan(\frac{R}{\omega L})

back to time domain[edit]

V_s(t) = I_m \frac{\sqrt{(\omega^2L^2R)^2 + (\omega L R^2)^2}}{R^2+\omega^2L^2}\cos(\phi + \arctan(\frac{R}{\omega L}))
There will be an integration constant, but it is impossible to calculate now.
This is from a differential equation.
Will calculate the integration constant after adding the homogeneous solution to the above particular solution.

Phasor Numeric[edit]

phasor numeric solution combines all phasors in rectangular form then goes straight to the polar answer form. Angle is in the third quadrant... m-file

time domain[edit]

Evaluate the terminal relations in this order:

i_R(t) = \frac{V_s}{10}
i_L(t) = \int\frac{V_s}{.01} dt

Substitute into this equation:

I_s(t)= i_R(t) + i_L(t)

So have a differential equation that can be solved for V_s:

I_s(t)= \frac{V_s}{10} + \int\frac{V_s}{.01} dt

Phasor domain[edit]

plugging into the symbolic time domain solution created from phasor analysis ... same angle in the third quadrant ... positive this time m-file
Choosing to do calculation in time domain

back to time domain[edit]

V_s(t) = 120*\sqrt{2} \frac{\sqrt{(377^2 * .01^2 * 10)^2 + (377 *.01*10^2)^2}}{10^2+377^2*.01^2}\cos(377 t + \frac{2*\pi}{3} + \arctan(\frac{10}{377*.01}))
V_s(t) = 599 \cos(377t + 3.30)
i_R(t) = 59.9 \cos(377t + 3.30)
i_L(t) = 159 \sin(377t + 3.30) = 159 \cos(377t + 1.74)

Laplace Symbolic[edit]

time domain[edit]

Evaluate the terminal relations in this order:

i_R(t) = \frac{V_s}{R}
i_L(t) = \int\frac{V_s}{L} dt

Substitute into this equation:

I_s(t)= i_R(t) + i_L(t)

So have a differential equation that can be solved for V_s:

I_s(t)= \frac{V_s}{R} + \int\frac{V_s}{L} dt

laplace domain[edit]

now transform both into the laplace domain ..

{I_s}(t) \rightarrow \mathcal{L} \left\{{I_s}(t)\right\}
V_s(t) \rightarrow \mathcal{L} \left\{V_s(t)\right\}
\int V_s(t) dt \rightarrow  \frac{1}{s} \mathcal{L} \left\{V_s(t)\right\}\

So:

\mathcal{L} \left\{{I_s}(t)\right\} = \frac{\mathcal{L} \left\{V_s(t)\right\}}{R} + \frac{\mathcal{L} \left\{V_s(t)\right\}}{sL}

Solving for \mathcal{L} \left\{V_s(t)\right\}

\mathcal{L} \left\{V_s(t)\right\} = \frac{{\mathcal{L} \left\{{I_s }(t)\right\}}LsR}{R + sL}

At this point the Laplace symbolic solution has to stop. The next steps depend upon the form of {\mathcal{L} \left\{{I_s}(t)\right\}}.

back to time domain[edit]

This can not be done without {\mathcal{L} \left\{{I_s}(t)\right\}}. The form of the function {\mathcal{L} \left\{{I_s}(t)\right\}} determines the inverse mapping.

Laplace Numeric[edit]

laplace solution has the same problems as before ... constants ... sin term with wrong sign m-file

Time Domain[edit]

 I_s(t) = 120 \sqrt{2} cos(377t+ 2.09)
 R = 10
 L = .01

Laplace Domain[edit]

{I_s}(t) \rightarrow \mathcal{L} \left\{{I_s}(t)\right\} = \mathcal{L} \left\{120 \sqrt{2} cos(377t+120^\circ)\right\} = -\frac{(170.0*(0.5*s + 326.0))}{(s^2 + 142129)}
\mathcal{L} \left\{V_s(t)\right\} = \frac{\frac{-17.0*s*(0.5*s + 326.0)}{(s^2 + 142129)}}{10 + .01 *s}

Time Domain[edit]

V_s(t)=\mathcal{L}^{-1} \left\{i(t)\right\} = (-2.58)*e^{-\frac{t}{.001}} + 97.2 sin(377t) - 591 cos(377*t)

Comparison of answers[edit]

V_s(t) = 599 \cos(377t - 2.98) ... from numeric solution out of phasor domain
V_s(t) = 599 \cos(377t + 3.30) ... solution from substituting into symbolic time domain

Phase angles are exactly the same place in the third quadrant.

Simulation[edit]

of the simulation.

i_r is in phase with V_s as it should. V_s is leading i_L through the inductor by \frac{\pi}{2} as it should. Everything appears centered around 0 volts, so no constants are in play.

period check[edit]

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

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

magnitude check[edit]

The magnitude of V_s is above 500 volts .. although perhaps not 600 volts.

transient response check[edit]

Everything starts out at zero. This means that only the middle to right side of the simulation windows are going to display steady state information that we can compare to calculated values.

phase check[edit]

of the simulation.

I_s brown, i_R blue (with same phase angle as source voltage), and i_L (orange). i_R V_s (blue) leads i_L (orange) by \frac{\pi}{2}. The brown current (the starting point) is the starting point at 2.09 radians. The blue current should lead the brown by 3.30-2.09=1.21 radians which is 69.3 degrees which is about 3.21 ms or 1.5 squares above. This can not be seen at the beginning where the simulation software is dealing with the initial energy embalance, but by the middle of the simulation, the blue is leading the brown by about 1.5 squares. So all is well.

Power Analysis[edit]

The constants are going to add some DC power or real power to this analysis. Without knowing what they are, we can not compute their impact. So for now we stick with phasor domain power analysis:

\mathbb{I}_s =  120 \sqrt{2}\angle 2.09
\mathbb{V}_s = 599\angle3.30
\mathbb{I}_s^* = 120\sqrt{2}\angle -2.09

if :\mathbb{V} = M_v\angle\phi_v\quad, and \quad\mathbb{I} = M_i\angle\phi_i then

\mathbb{S} = \mathbb{V}\mathbb{I}^* = \frac{M_vM_i}{2}\angle(\phi_v - \phi_i) = \frac{120 \sqrt{2} * 599}{2}\angle(3.30-2.09)= 10,164\angle 1.21=3,586 + 9,510j

and

cos(1.21) = .357

This is a bad power factor. Power factors below .9 will cause you a visit from the utility company or the transformer on the power pole might blow. The utility's apparent power is much larger than what the customer is willing to pay.

Value Units Description
10,164 volt-ampere va apparent power what utility companies manage: peak power they design for, peak power they have to deliver
.357 unitless power factor, ratio of real power to apparent power, ideally 1
3,590 watt W real, average, active power ... what consumers want to pay for (watt-hours)
9,510 volt-amp-reactive var reactive power ... why not all outlets in a room are on the same circuit breaker

Intuition[edit]

Should be way to do phasor math in symbol form in mupad ... probably with matrices.