How about your text?I have no idea what impedance is, and my lecture notes does not show what impedance is either.
You tried Googling "impedance" and got no relevant hits?We don't use textbooks on this module, just the lecture notes. I tried Google but no luck.
It doesn't really do to just post your answers -- you need to show your work so that we can see what you are doing right and what you are doing wrong.well, I do not know if I can use that formulas since my tutor did not teach it nor it is in any of our lectures.
Anyways, here are my answers:
i. 36 ohms
ii. 0.05 Amps
iii. 36 ohms
iv. 0.17 Amps (should this be the current for the 50 ohm resistor?)
v. 0.04 Amps(again current for the middle resistors?)
I used principle of superposition to solve the problem, correct me if I am wrong.
That is correct.(i) it asks for impedance seen by the left supply, so I think I have to short the supply from the right and get the total resistance : (1/140 + 1/50)^-1 = 36 Ohms
So you are saying that the rightmost source has absolutely no effect on the current drawn from the other supply? Does that make sense.(ii) for this part, I think its just the total current if I short the rightmost voltage supply, so I = V/R = 2/36 = 0.05 Amps
If the leftmost supply is shorted, then how do you claim that the three resistances are in series. Does every electron that goes through the 50Ω have to go through the other two resistors? If not, then they are not in series.(iii) same as part i but I short the leftmost supply and work with just the rightmost supply, so total resistance: 50 + 60 + 80 = 190 Ohms
What is I(A) and (B). DEFINE YOUR TERMS!(iv) current in the right most branch, I think its the total current for the 50 Ohm resistor?
I(A) = middle resistor and right resistor is in parallel so voltage is the same, then to solve for current in 50 ohm resistor: I = V/R = 2/50 = 0.04 Amps
I(B) = if I short left most branch, resistors are in series, so current is the same: I = V/R = 5/190 = 0.02 Amps
I(total) = 0.02 + 0.04 = 0.06 Amps
Now ask if your answers are consistent. If (ii) says you have 50 mA flowing upward in the left branch and (iv) means you have 30 mA flowing upward in the right branch, wouldn't that require that you have 80mA flowing downward in the center branch? Or if the right branch was flowing downward, then you would have 20mA flowing downward in the center branch. Or, for the other two possibilities for your current polarities in the left and right branch, you could have 20 mA flowing upward or 80 mA flowing upward. Do you begin to see why current direction matters?(v) works the same as part iv: shorting right most branch: I(A)= 2/140 = 0.01 Amps
I(B) = shorting left most branch: 0.02 Amps
I(total) = 0.01 + 0.02 = 0.03 Amps
Yeah, because it only asks for the current drawn from the left supply voltage? Correct me if I am wrong.So you are saying that the rightmost source has absolutely no effect on the current drawn from the other supply? Does that make sense.
I don't know much about polarity, but how do you determine if its flowing down or up?You also need to indicate polarity, since 50mA flowing upward is very different than 50mA flowing downward. Don't make people, including yourself later, guess.
You also need to start tracking your units -- 2/36 is a number, not a current. 2V/36Ω is a current.
Ok, so the current will not flow through the middle resistors since it will prefer the shorted circuit from the left thus making the total resistance only 50 Ohms.If the leftmost supply is shorted, then how do you claim that the three resistances are in series. Does every electron that goes through the 50Ω have to go through the other two resistors? If not, then they are not in series.
I(A) means the current on the 50 ohm resistor when I short the right supply and I(B) means the current on the 50 ohm resistor when I short left supply.What is I(A) and (B). DEFINE YOUR TERMS!
To explore that line of thinking, let's use a simplified example:Yeah, because it only asks for the current drawn from the left supply voltage? Correct me if I am wrong.
The circuit you have been given is a bit ambiguous. The voltage sources are shown as being sinusoidal sources and so the voltages that each produce alternate between positive and negative over time, so the stated voltages, the 2V and the 5V, are the amplitudes of the sinusoidal waveforms. However, we don't know if the waveforms are sin(wt) or (cos(wt). But that really doesn't matter because we are only looking for the magnitude of the resulting currents so as long as we have the relative phase of the two sources correct we will get the correct answer. Normally one end of an AC source is labeled as positive to serve as a reference polarity, but that is not the case here. However, the orientation of both sources is the same, so it is reasonable to assume that the polarity of the top of each source is the same. So without loss of generality, we can simply declare the top of each source to be the positive terminal. We can declare the current to be positive in which ever direction we want, as long as we are then consistent with that choice.I don't know much about polarity, but how do you determine if its flowing down or up?
Correct.Ok, so the current will not flow through the middle resistors since it will prefer the shorted circuit from the left thus making the total resistance only 50 Ohms.
Again, what are the polarities of the two currents. When you calculate I(A), you get a current going downward in the 50Ω resistor and when you calculate I(B) you get a current that is going upward in the 50Ω resistor. So one of them is negative relative to the other.I(A) means the current on the 50 ohm resistor when I short the right supply and I(B) means the current on the 50 ohm resistor when I short left supply.
Going by my new answer for part iii. I(A) = V/R = 2V/50 ohms = 0.04 Amps
I(B) = 5v/50 omhs = 0.1 Amps
I just add them together = 0.14 Amps.
so if I assume downward is negative then: I(total) = -I(A) + I(B)Again, what are the polarities of the two currents. When you calculate I(A), you get a current going downward in the 50Ω resistor and when you calculate I(B) you get a current that is going upward in the 50Ω resistor. So one of them is negative relative to the other.
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by Jake Hertz