Circuit Theory/Phasors/Examples/Example 9

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Parallel RL circuit for example 9

Given that the voltage source is defined by  V_s(t) = 120 \sqrt{2} cos(377t+120^\circ), find all other voltages, 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: V_s, R, L
Unknowns: I_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

Initial conditions[edit]

Saying the initial conditions are zero is impossible in this problem.

Look at the terminal relation:

V_s = V_L = L * {d \over dt}i_L(t)

VS has a non-zero value at t=0+. This means {d \over dt}i_L(t) = V_s/L. This one initial condition. The other initial condition is that i_L(0+) = 0.

Calculus Symbolic[edit]

Evaluate the terminal relations in this order:

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

Calculus Numeric[edit]

phasor matlab solution of RL circuit .... m file
 V_s(t) = 120 \sqrt{2} \cos(377t+ 2.09)
 R = 10
 L = .01
 i_R = 12 \sqrt{2} \cos(377t+ 2.09)
 i_L = \int \frac{120 \sqrt{2} \cos(377t+ 2.09)}{.01} dt = 45 \sin(377t+2.09) + C_1
 i_L(0) = 0 = 45 \sin(2.09) + C_1
 C_1 = -39
 i_L = 45 \sin(377t+2.09) - 39
I_s(t)=12 \sqrt{2} \cos(377t+ 2.09) + 45 \sin(377t+2.09) - 39
phasor math avoids trig! m-file

The next step is to combine the answer into a single cos term with a phase shift so it can be compared to i_L and i_R. The -39 turns into a DC bias.

The easiest method to combine the sin and cos terms is phasors.

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

12*\sqrt{2}*(377t+2.09) \Leftrightarrow 12*\sqrt{2}cos(2.09) + j12*\sqrt{2}sin(2.09)
45sin(377t +2.09) \Leftrightarrow 45*sin(2.09) - j45*cos(2.09)

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

\mathbb{I}_s = 48.1\angle0.884

Now put back into the time domain:

I_s = 48.1 \cos(377t + 0.884) - 39

Phasor Symbolic[edit]

time domain[edit]

i_R = V_s /R
i_L = \int \frac{V_s}{L}dt
i_S = i_R + i_L

phasor domain[edit]

V_s(t) \rightarrow \mathbb{V}
\mathbb{V} = V_m\angle\phi
\mathbb{I}_R = \frac{\mathbb{V}}{R} = \frac{V_m}{R}\angle\phi
\int V_s(t) dt \rightarrow \frac{\mathbb{V}}{j\omega} =\frac{\mathbb{V}}{\omega}\angle\frac{-\pi}{2} = \frac{V_m}{\omega} \angle (\phi - \frac{\pi}{2})
\mathbb{I}_L = \frac{V_m}{\omega L} \angle (\phi - \frac{\pi}{2})
\mathbb{I}_S = \frac{V_m}{R}\angle\phi + \frac{V_m}{\omega L} \angle (\phi - \frac{\pi}{2})
\mathbb{I}_S=V_m*\sqrt{\frac{1}{R^2} + \frac{1}{(L\omega)^2}} \angle (\phi-arctan(\frac{R}{L\omega}))

back to time domain[edit]

I_L(t) = \operatorname{Re}(\mathbb{I}_L e^{j\omega t}) = \frac{V_m}{\omega L}*cos(\omega t + \phi - \frac{\pi}{2}) + C_1
I_s(t) = \operatorname{Re}(\mathbb{I}_s e^{j\omega t})= V_m \sqrt{\frac{1}{R^2} + \frac{1}{(L\omega)^2}} cos(\omega t + \phi - arctan(\frac{R}{L\omega})) + C_1
 I_R(t) = \operatorname{Re}(\mathbb{I}_R e^{j\omega t}) = \frac{V_m}{R} cos(\omega t + \phi)

I_L involved an integral in the phasor domain, thus a constant as been added to their time domain equation.

doing problem using phasors rather than calculus m-filec

Phasor Numeric[edit]

there is no point in doing the math in the phasor domain if can solve in symbol form as above

time domain[edit]

i_r(t) = \frac{120*\sqrt{2}}{10}\cos(377t + 2.09)= 17.0 cos(377t + 2.09)
i_L(t) = \frac{120*\sqrt{2}}{.01*377}\cos(377t + 2.09 - \frac{\pi}{2}) + C_2 = 45 \cos(377t + 0.524) + C_1
0 =45 \cos(0.524) + C_1
C_1 = - 39.0
i_s(t) =  120*\sqrt{2} * \sqrt{\frac{1}{10^2} + \frac{1}{(.01*377)^2}} \cos(377 t + \frac{2*\phi}{3} - \arctan(\frac{10}{.01*377})) - 39
i_s(t) = 48.1 \cos(377t + 0.844) - 39

These are the same numbers as calculated with Calculus above ...


Vs, I_s, i_L, and i_R from the simulation simulation are graphed below. simulation with the current into a device which produces 180° phase shift
Symbol color Equation
I_s Brown 48.1\cos(377t + 50.6^\circ) - 39
i_L Orange 45\cos(377t + 30^\circ) - 39
i_R Example 17\cos(377t + 120^\circ)
Vs</math> Blue 170\cos(377t + 120^\circ)


showing what happens when circuitlab skip initial=no ..

Simulation matches the math. Circuitlab simulation requires taking current into the source in order to get the phase right. This can be seen by testing circuitlab with a sinusoidal source across one resistor.

The simulation also shows that the integration constant is dependent upon the phase angle of the source.

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

magnitude check[edit]

The simulation magnitudes, if measured peak to peak match the doubled magnitudes of the numeric answers. The integration constant adds a DC bias which is negative in this case.

phase check[edit]

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

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{V}_s =  120 \sqrt{2}\angle 2.09
\mathbb{I}_s = 48.1\angle0.884
\mathbb{I}_s^* = 48.1\sqrt{2}\angle -0.884

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} * 48.1}{2}\angle(2.09 - 0.884)= 4081\angle 1.206=1456 + 3813j


cos(1.206) = .357

This is a bad power factor. The utility's apparent power is much larger than what the customer is willing to pay. Pure inductive loads as described are found in large motors in industrial settings where engineers working for industry are talking to engineers working for the power company. Power conditioners may be needed to bring this power factor back up closer to 1. The utility will charge based upon apparent power ... which is real $$$ for them.

Value Units Description
4081 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
1456 watt W real, average, active power ... what consumers want to pay for (watt-hours)
3813 volt-amp-reactive var reactive power ... why not all outlets in a room are on the same circuit breaker


ELI ... voltage leads the current through an inductor

Integration constants matter. Initial conditions matter. Phasors don't deal with them.

A purely inductive load connected to a power source is going to create a bad power factor.

If integration constants show up, and there is no differential equation, must compute the integration constants using initial conditions.

Phase angle of the source influences the integration constant.