# Attenuator's decibels question

#### BlackMelon

Joined Mar 19, 2015
143
Hello,

According to this link, http://www.allaboutcircuits.com/textbook/semiconductors/chpt-1/attenuators/, I'm curious in the equation PI = VI II = (VI )^2 / R
Since, after plugging an attenuator in, the VI does not lie across an input resistance directly, so the II is not VI/R, which will contradict the equation above.

If I don't make this clear, please look at the attached picture or inform me on this thread.

Thank you
BlackMelon

#### crutschow

Joined Mar 14, 2008
31,112
Sorry, but I don't really understand your question. #### AnalogKid

Joined Aug 1, 2013
10,173
Your statement is correct. So ... ?

#### WBahn

Joined Mar 31, 2012
27,866
Hello,

According to this link, http://www.allaboutcircuits.com/textbook/semiconductors/chpt-1/attenuators/, I'm curious in the equation PI = VI II = (VI )^2 / R
Since, after plugging an attenuator in, the VI does not lie across an input resistance directly, so the II is not VI/R, which will contradict the equation above.

If I don't make this clear, please look at the attached picture or inform me on this thread.

Thank you
BlackMelon
To use P = V·I to find the power delivered to a device (or subcircuit) you must use the voltage across THAT device and the current through THAT device. The same for Ohm's Law. You can't just pick and V, any I, and any R, throw them at some equation, and expect to get valid results.

If PI is the power delivered by the source in your diagram, then you are correct that PI = VI II since those are the voltage across the current through that source. But trying to write the current as II = VI/R is simply incorrect because VI does not appear across R. So there's no contradiction, you simply are doing something that is wrong.

#### BlackMelon

Joined Mar 19, 2015
143
To use P = V·I to find the power delivered to a device (or subcircuit) you must use the voltage across THAT device and the current through THAT device. The same for Ohm's Law. You can't just pick and V, any I, and any R, throw them at some equation, and expect to get valid results.

If PI is the power delivered by the source in your diagram, then you are correct that PI = VI II since those are the voltage across the current through that source. But trying to write the current as II = VI/R is simply incorrect because VI does not appear across R. So there's no contradiction, you simply are doing something that is wrong.

I just wanted to ask whether the equation encircled with red box is wrong or not. It's about an attenuator. The equation is from the following link:

I'm sorry that I didn't make a question clear enough in the first place.

#### dl324

Joined Mar 30, 2015
15,439

#### BlackMelon

Joined Mar 19, 2015
143
My intention is want to ask whether the equation is correct or not. Or, is it an "approximate" form?

Equation is correct.
Look at the first and the second figure on the link, the Vin is not equal to Ii/R because Vi does not lie across R directly in the circuit. So, I'm afraid we can't say Pin=Vi*(Vi/R)=(Vi^2)/R.
So who is this "R?". In the article, It says that Ro=Ri. And the equation Po= Io^2 * R implies that Ro=R also. After we plugging in the T or pi attenuator, the Vi does not lie across the input resistance as I said.

So is the equation an approximate form?

#### WBahn

Joined Mar 31, 2012
27,866
My intention is want to ask whether the equation is correct or not. Or, is it an "approximate" form?

Look at the first and the second figure on the link, the Vin is not equal to Ii/R because Vi does not lie across R directly in the circuit. So, I'm afraid we can't say Pin=Vi*(Vi/R)=(Vi^2)/R.
So who is this "R?". In the article, It says that Ro=Ri. And the equation Po= Io^2 * R implies that Ro=R also. After we plugging in the T or pi attenuator, the Vi does not lie across the input resistance as I said.

So is the equation an approximate form?
The article doesn't imply anything, it explicitly states: "Assuming that the load RI at PI is the same as the load resistor RO at PO (RI = RO), the decibels may be derived from the voltage ratio (VI / VO) or current ratio (II / IO):"

I don't think this article makes things too clear. When we look at voltage/current attenuation/gain in dB we almost always do it using a "what if" scenario. That is to say, we ask what the power ratios would be IF both voltages (currents) were applied across (or running through) the same resistance (and we usually ignore the effect of source resistance). This doesn't have to be the case, but it is often an approximation that is "good enough".