Question on the test - Electrician Talk - Professional Electrical Contractors Forum

Carmexx · 05-22-2009, 10:20 PM

A 3.0 Ohm resistor bank has a rating of 450.0 W. What is the maximum permissible volt drop across this bank?

The equation I know of to solve this problem are:

P= E x I

I = E/R

But wouldn't you need two values from one of those equation to solve the problem?

jbfan · 05-22-2009, 10:31 PM

Here is the pdf of the ohm's law wheel.

http://www.leonaudio.com.au/pwheel.pdf

waco · 05-22-2009, 10:47 PM

Quote:

Originally Posted by Carmexx

A 3.0 Ohm resistor bank has a rating of 450.0 W. What is the maximum permissible volt drop across this bank?

The equation I know of to solve this problem are:

P= E x I

I = E/R

But wouldn't you need two values from one of those equation to solve the problem?

Yes and you have them, 3 ohms is the value of R and 450 is the value of P. It is a "gotcha" type of question since the main things are really I (current) because it creates P (power, heat) since "voltage drop" is really not very important in the real world.

Carmexx · 05-22-2009, 11:06 PM

Thanks that really helps...

But heres another question, I know the right answer but I can't seem to figure out how they got the answer.

Q: Five loads are to be energized from a 240 V supply. Three of the loads require 12.6 V and two loads require 35 V each. All the loads require 225mA. Find the resistance of a suitable dropping resistor if the five loads operate in series.

The right answer is: 587.556 Ohms... But how did they get that answer?

InPhase277 · 05-22-2009, 11:45 PM

Quote:

Originally Posted by Carmexx

Thanks that really helps...

But heres another question, I know the right answer but I can't seem to figure out how they got the answer.

Q: Five loads are to be energized from a 240 V supply. Three of the loads require 12.6 V and two loads require 35 V each. All the loads require 225mA. Find the resistance of a suitable dropping resistor if the five loads operate in series.

The right answer is: 587.556 Ohms... But how did they get that answer?

I haven't worked the math, but I'll let you in on how it's done. First, total circuit resistance is 240/0.225 = ?

The current through each component of a series circuit is the same, in this case 0.225 A. You can figure the resistance of each load by using
R = E/I. So... 3(12.6/0.225) = ? And... 2(35/0.225) = ? Add them together, then subtract that from the total resistance to get the missing resistance.

wildleg · 05-23-2009, 06:21 AM

loads are in series, so voltage drops add and current is same
35 + 35 + 12.6 +12.6 +12.6 + V(R) = 240
solve for V(R)
solve for R using ohms law
V/I = R (I = .225)

waco · 05-23-2009, 08:49 PM

add up the known voltage drops, subtract them from the total (240) and divide that by the current, .225 amps.

The idea is that the current is constant in a series circuit, so what you need to find is the value of one more resistor in that circuit which will result in the values given. The "gotcha" is that adding another resistor will change the current and the other voltage drops.

I hate questions like that.