# Steady state sinusodial analysis - equivalent impedance

#### Sebastian Eriksson

Joined Aug 15, 2015
7
I have the circuit below and have found the equivalent impedance to be Z_EQ = Z_C || Z_R = (100 * - j 200)(100 - j 200) = 80 - j 40.

To find the voltage v over the resistance and capacitance I use the fact that the entire current will go through the equivalent resistance and the voltage (v) is then V=I_s/(80-j 40) and since I_s = 0.1 => V = 0.001 + j 0.0005 or in phasor notation: V = 0.001118033 ∠26.57°

The answer in the book is V = 8.944 ∠-26.56° but I don't see what I'm doing wrong...

#### shteii01

Joined Feb 19, 2010
4,647
What is Zr?
What is Zc?

#### Sebastian Eriksson

Joined Aug 15, 2015
7
What is Zr?
What is Zc?
Zr is the resistors impedance Zr = 100, and Zc is the impedance of the capacitance Zc = - j 1 / (ωC) = - j * 1 / (10^4 * 0.5 * 10^(-6)) = -j 200

#### shteii01

Joined Feb 19, 2010
4,647
Ok. I just finished it, I got book answer.

#### shteii01

Joined Feb 19, 2010
4,647
Zr is the resistors impedance Zr = 100, and Zc is the impedance of the capacitance Zc = - j 1 / (ωC) = - j * 1 / (10^4 * 0.5 * 10^(-6)) = -j 200
We match there.

What did you get for the Zr || Zc?

#### Sebastian Eriksson

Joined Aug 15, 2015
7
We match there.

What did you get for the Zr || Zc?
Z_C || Z_R = (100 * - j 200)/(100 - j 200) = 80 - j 40

#### shteii01

Joined Feb 19, 2010
4,647
Um... one more thing...
Ohm's Law is V=I*R.
So you should do V= 0.1*Zeq

#### shteii01

Joined Feb 19, 2010
4,647
Z_C || Z_R = (100 * - j 200)/(100 - j 200) = 80 - j 40
I saw that once I looked second time.

It looks like your error is in the V calculation. You are dividing the current by the resistance. That is wrong. Ohm's Law says that you find voltage you must multiply.

My work:
Zeq=Zr||Zc=100*(-200j)/(100-200j)=-20000j/(100-200j)=-200j/(1-2j)
V=Is*Zeq=0.1*(-200j)/(1-2j)=-20j/(1-2j)=8-4j=8.944@-26.56°

#### Sebastian Eriksson

Joined Aug 15, 2015
7
Um... one more thing...
Ohm's Law is V=I*R.
So you should do V= 0.1*Zeq
Haha, thank you, that's the problem

#### Sebastian Eriksson

Joined Aug 15, 2015
7

#### shteii01

Joined Feb 19, 2010
4,647
Nothing to be sorry about. This is homework forum.

#### WBahn

Joined Mar 31, 2012
25,100
This is a shining example of why you should always, always, ALWAYS do two things:

1) Track your units throughout your work.
2) Ask if the answer makes sense.

Let's start with the second one. You got an answer with a magnitude of just over 1 mV. Now, you give now indication of whether this is the voltage amplitude, the RMS voltage, or the peak-to-peak voltage. But since those are all within a factor of about three, it doesn't matter for this sanity check. If the circuit consisted of only the resistance, the amplitude of the voltage would be 10V. In order for it to be close to 1mV, that would require that only 0.01% of the current go through the resistor, which in turn would require that the impedance of the capacitor be about a factor of 10,000 smaller than the resistor. Yet you got an impedance that is actually LARGER in magnitude than that of the resistor, which would mean that most of the current will go through the resistor. If the impedances were the same (in magnitude), the voltage would be 7.07 V in amplitude, so you KNOW the answer if going to be between 7V and 10V in amplitude. Your answer is no where close to that, so you know you have a problem. The book answer is nicely within this range, so while this doesn't guarantee that it is correct, it is at least in the ballpark.

Now let's consider the units.

From your work, you state V=I_s/(80-j 40).

This is guaranteed wrong because (80-j40) is just a number and a current divided by a number is a current, not a voltage. If you had tracked your units to this point, you would have had
This should be

$$V \; = \; \frac{I_s}{\(80-j40$$\Omega}
\)

Since 1Ω = 1V / 1A, your next step should have been

$$V \; = \; \frac{0.1 A_p}{\(89.44 \angle -26.57^o$$\Omega} \cdot \frac{1 \Omega}{$$\frac{1V_p}{1A_p}$$}
\;
V \; = \; 0.00118 \angle 26.57^o \; \frac{A_p^2}{V_p}
\)

At this point you have an answer that clearly has the wrong units. So you KNOW the answer is wrong. But you also have a shining beacon pointing you to WHY the units didn't work out and you can scratch your head for a minute and spot that your equation SHOULD have been

$$V \; = \; I_s \cdot \(80-j40$$\Omega
\;
V \; = \; 0.1 A_p \cdot $$89.44 \angle -26.57^o$$\Omega \cdot \frac{$$\frac{1V_p}{1A_p}$$}{1 \Omega}
\;
V \; = \; 8.944 \angle -26.57^o \; V_p
\)

Another warning flag, but a much more subtle one and easy to miss, is that since this is a capacitive circuit, you know that the voltage must lag the current, which means that if the current has zero phase angle, that the phase angle of the voltage MUST be negative.

So before you even started plugging numbers into formulas, you should have determined that the angle on the answer would be negative. Then as soon as you determined that the capacitive impedance was more than the resistance (in magnitude), you should have determined that the amplitude of the voltage would be between 7V and 10V. Those should have been written off to the side for use as checks on whatever answer you got.

The bottom line is that we all make mistakes, and screwing up Ohm's Law is a pretty common one (heck, I still make it occasionally after more than 35 years doing this stuff). But most (not all) mistakes that you make will screw up the units allowing you to catch them almost immediately -- but ONLY if they are actually THERE to get screwed up. And note that this required that you actually TRACK the units and what happens to them from one line to the next -- you can't just tack on whatever units you WANT or EXPECT to see.

Does this take time. Yes. But consider how much time it would have saved you on this problem alone.

If you will always, always, ALWAYS do these two things, you will see your overall time spent on working problems go down and your homework and exam scores go up.

And this doesn't even touch on the value that being able to catch most of your mistakes quickly has in the real world where an uncaught mistake could literally cost someone their life -- and land you in prison for gross professional negligence. While both of those are unlikely, they have both happened, though the usual outcome for the engineer in something that results in major injury/death/damage is to be fired and blackballed from the industry.

#### Sebastian Eriksson

Joined Aug 15, 2015
7
This is a shining example of why you should always, always, ALWAYS do two things:

1) Track your units throughout your work.
2) Ask if the answer makes sense.

Let's start with the second one. You got an answer with a magnitude of just over 1 mV. Now, you give now indication of whether this is the voltage amplitude, the RMS voltage, or the peak-to-peak voltage. But since those are all within a factor of about three, it doesn't matter for this sanity check. If the circuit consisted of only the resistance, the amplitude of the voltage would be 10V. In order for it to be close to 1mV, that would require that only 0.01% of the current go through the resistor, which in turn would require that the impedance of the capacitor be about a factor of 10,000 smaller than the resistor. Yet you got an impedance that is actually LARGER in magnitude than that of the resistor, which would mean that most of the current will go through the resistor. If the impedances were the same (in magnitude), the voltage would be 7.07 V in amplitude, so you KNOW the answer if going to be between 7V and 10V in amplitude. Your answer is no where close to that, so you know you have a problem. The book answer is nicely within this range, so while this doesn't guarantee that it is correct, it is at least in the ballpark.

Now let's consider the units.

From your work, you state V=I_s/(80-j 40).

This is guaranteed wrong because (80-j40) is just a number and a current divided by a number is a current, not a voltage. If you had tracked your units to this point, you would have had
This should be

$$V \; = \; \frac{I_s}{\(80-j40$$\Omega}
\)

Since 1Ω = 1V / 1A, your next step should have been

$$V \; = \; \frac{0.1 A_p}{\(89.44 \angle -26.57^o$$\Omega} \cdot \frac{1 \Omega}{$$\frac{1V_p}{1A_p}$$}
\;
V \; = \; 0.00118 \angle 26.57^o \; \frac{A_p^2}{V_p}
\)

At this point you have an answer that clearly has the wrong units. So you KNOW the answer is wrong. But you also have a shining beacon pointing you to WHY the units didn't work out and you can scratch your head for a minute and spot that your equation SHOULD have been

$$V \; = \; I_s \cdot \(80-j40$$\Omega
\;
V \; = \; 0.1 A_p \cdot $$89.44 \angle -26.57^o$$\Omega \cdot \frac{$$\frac{1V_p}{1A_p}$$}{1 \Omega}
\;
V \; = \; 8.944 \angle -26.57^o \; V_p
\)

Another warning flag, but a much more subtle one and easy to miss, is that since this is a capacitive circuit, you know that the voltage must lag the current, which means that if the current has zero phase angle, that the phase angle of the voltage MUST be negative.

So before you even started plugging numbers into formulas, you should have determined that the angle on the answer would be negative. Then as soon as you determined that the capacitive impedance was more than the resistance (in magnitude), you should have determined that the amplitude of the voltage would be between 7V and 10V. Those should have been written off to the side for use as checks on whatever answer you got.

The bottom line is that we all make mistakes, and screwing up Ohm's Law is a pretty common one (heck, I still make it occasionally after more than 35 years doing this stuff). But most (not all) mistakes that you make will screw up the units allowing you to catch them almost immediately -- but ONLY if they are actually THERE to get screwed up. And note that this required that you actually TRACK the units and what happens to them from one line to the next -- you can't just tack on whatever units you WANT or EXPECT to see.

Does this take time. Yes. But consider how much time it would have saved you on this problem alone.

If you will always, always, ALWAYS do these two things, you will see your overall time spent on working problems go down and your homework and exam scores go up.

And this doesn't even touch on the value that being able to catch most of your mistakes quickly has in the real world where an uncaught mistake could literally cost someone their life -- and land you in prison for gross professional negligence. While both of those are unlikely, they have both happened, though the usual outcome for the engineer in something that results in major injury/death/damage is to be fired and blackballed from the industry.
Thank you, very good information in your response. From now I will definitely keep track of units!