Find V Max, V Min: Solutions & Equations

[nobrainer612]

Homework Statement

Homework Equations

I don't know the relevant equations for V max, V min and V rms.

But I found some equations say
Vpp ( voltage peak to peak ) = 2 * Vp ( V peak)
V rms ( RMS voltage of the AC signal) = Vp / square root(2)

The Attempt at a Solution

I calculated the Vpp for the output at 3 and 4, which I called channel 1 and channel 2. Vpp of channel 1 is 4.8, and Vpp for channel 2 is 2.4So if anyone knows how to find V max, V min and V rms, please tell me how to do it

Thanks a lot

 

[Zryn]

Can you re size this picture (much less white space) and show your calculations for each voltage?

I believe Vmax in this context is the maximum amplitude (magnitude) of the voltage waveform, and Vmin is correspondingly the minimum amplitude.

 

[nobrainer612]

Can you re size this picture (much less white space) and show your calculations for each voltage?

I believe Vmax in this context is the maximum amplitude (magnitude) of the voltage waveform, and Vmin is correspondingly the minimum amplitude.

I couldn't edited the original picture, so here is the resized one

The oscilloscope showed the V max and Vmin for channel 1 is: 3.938 V and -843.8mV
for channel 2 is: 2.031 V and -406.2

I got to compare my calculated result with the result shown by the oscilloscope

so here is how I calculated V max: 3 sin(2*pi*1000t) * (200/250) = 0.8208 for channel 1

The difference is too large so I think I did something wrong.

 

[Zryn]

The picture has changed from R1 = 1000R to R1 = 50R!

so here is how I calculated V max: 3 sin(2*pi*1000t) * (200/250) = 0.8208 for channel 1

I can't replicate this calculation. What did you use for t?

Since we're only using resistances there are no components in the circuit which will alter the angle and phase, so we can be content to use only the value of the sin wave.

i.e. 3 * 200/250 = 0.8208 ... which looks wrong to me. Additionally, the sin wave is not the only voltage, and the DC voltage is actually labeled 'offset' in the second picture. How does all this come together? Can you show a picture of the combination of the two voltages, and understand how the picture comes about?

How does the program that you use handle sin waves? In the picture you have 3 sin(wt), but is that 3 a Vrms or a Vmax?

What do you actually expect to get at Channel1 and Channel2?

Sorry for all the questions, I'm just trying to give you to the answer, instead of giving the answer to you.

 

[nobrainer612]

The picture has changed from R1 = 1000R to R1 = 50R!



I can't replicate this calculation. What did you use for t?

Since we're only using resistances there are no components in the circuit which will alter the angle and phase, so we can be content to use only the value of the sin wave.

i.e. 3 * 200/250 = 0.8208 ... which looks wrong to me. Additionally, the sin wave is not the only voltage, and the DC voltage is actually labeled 'offset' in the second picture. How does all this come together? Can you show a picture of the combination of the two voltages, and understand how the picture comes about?

How does the program that you use handle sin waves? In the picture you have 3 sin(wt), but is that 3 a Vrms or a Vmax?

What do you actually expect to get at Channel1 and Channel2?

Sorry for all the questions, I'm just trying to give you to the answer, instead of giving the answer to you.

ok I tried to calculate V max and V min and here is what I got:

I know max in sin wave is 1 and min is -1.

V max at channel 1 is: V max = V ac + V dc

V ac = 3(1) * ( 200/250) = 2.4
V dc = 1 * ( 200 / 250) = 0.8

V max = 2.4 + 0.8 = 3.2

V min at channel 1 is:

V ac = 3(-1) * (200/250) = -2.4
V dc = 1* (200/250) = 0.8

V min = -2.4 + 0.8 = -1.6

I don't know if this answer sounds right to you, thanks a lot

 

[Zryn]

Vmax = 3.2V and Vmin = -1.6V for Channel1 sounds right to me, since the sin wave would oscillate between 4V and -2V and Channel1 resistance to ground is 200R out of the circuits 250R, i.e. 4V * 200/250 = 3.2V and -2V * 200/250 = -1.6V

The oscilloscope showed the V max and Vmin for channel 1 is: 3.938 V and -843.8mV
for channel 2 is: 2.031 V and -406.2

If Channel1 represents 200/250 (or 4/5) of the circuits resistance, then the source Vmax would have to be 5V (~4V * 250/200 = 5V) and Vmin be -1V (~-0.8V * 250/200 = -1V) from that measurement, and then for Channel2 which represents 200/250 (or 2/5) of the circuits resistance, Vmax would have to be 5V and Vmin would have to be -1V.

So, either you have 3V Peak + 2V DC = Vmax of 5V and Vmin of -1V or you have 3V RMS * sqrt(2) = 4.24V Peak + 1V DC = Vmax of 5.24V and Vmin of -3.24V.

If your program is showing the sin wave with 3 as a RMS Voltage, which means Vmax is 5.24V, can you think of why you measure 3.938V instead of 5.24V * 200/250 = 4.192V as predicted?

 

[nobrainer612]

Vmax = 3.2V and Vmin = -1.6V for Channel1 sounds right to me, since the sin wave would oscillate between 4V and -2V and Channel1 resistance to ground is 200R out of the circuits 250R, i.e. 4V * 200/250 = 3.2V and -2V * 200/250 = -1.6V



If Channel1 represents 200/250 (or 4/5) of the circuits resistance, then the source Vmax would have to be 5V (~4V * 250/200 = 5V) and Vmin be -1V (~-0.8V * 250/200 = -1V) from that measurement, and then for Channel2 which represents 200/250 (or 2/5) of the circuits resistance, Vmax would have to be 5V and Vmin would have to be -1V.

So, either you have 3V Peak + 2V DC = Vmax of 5V and Vmin of -1V or you have 3V RMS * sqrt(2) = 4.24V Peak + 1V DC = Vmax of 5.24V and Vmin of -3.24V.

If your program is showing the sin wave with 3 as a RMS Voltage, which means Vmax is 5.24V, can you think of why you measure 3.938V instead of 5.24V * 200/250 = 4.192V as predicted?

what you just said is pretty complicated, so I need some time to digest

but the lab just asked me to find V max and V min for channel 1 and channel 2

so for channel 2:

V max :

V ac = 3(1) * (100/250) = 1.2
V dc = 1 * (100/250) = 0.4

V max = 1.2 + 0.4 = 1.6


V min:

V ac = 3(-1) * (100/250) = -1.2
V dc = 1 * (100/250) = 0.4

V min = -1.2 + 0.4 = -0.8



BUT what I just calculated is so much difference then what I got from oscilloscope:
The oscilloscope shown:

V max = 2.031 V, v min = -406.2

I don't know if this will be the problem. Part of the circuit is inside the function generator

according to the lab, it stated " The display on the function generator does not display the true output voltage of function generator because the internal voltage sources are always set to twice the value that appears on the display. The display only displays the actual output voltage of the generator when the function generator is loaded with a 50 ohms resistor. When the generator is loaded with a resistor much larger than 50 ohms the actual output of the generator is about twice as large the value on the display. If the function generator is loaded with a resistor much smaller than 50 ohms the displayed value will be much smaller than the output value."

Thanks a lot. Hope my answer sounds correct to you