How do I solve this??????? Engineers help me :(?

Question

The no load voltage in the voltage divider circuit shown below is 20 V. The smallest load resistor that is ever connected to the divider is 48 kilo ohms. When the divider is loaded, V0 is not to drop below 16 V.
a) Design the divider circuit to meet the specifications just mentioned. Specify the numerical value of R1 and R2
b) Assume the power ratings of commercially available resistors are 1/16, 1/ 8, 1/4, 1, and 2 W. What power rating would you specify?

http://img466.imageshack.us/img466/236/gtuw0.jpg
please help me, i'd be grateful forever!!!!

Answer 1

First look at the no-load condition. Since we know the output voltage will be 20 volts and the source voltage is 100 volts, we can use a voltage divider to show that R1 is 4 times the value of R2.

Now look at the worst case load. We are told that at RL = 48Kohm is the lowest value. This should pull the voltage down to 16 volts, assuming that R2 is much less than 48Kohm. This part takes some algebra. Set up a voltage divider using R1 = 4*R2 (from above) and the parallel combination of R2 and RL with RL = 48Kohm. You can solve for R2 if you keep you wits about you.

I ended up with R2 = 15 Kohm. Then R1 = 60 Kohm. Note R2 is much less than 48 Kohm, so the assumption we made earlier is OK.

To find the power ratings, you need to know the max current through the voltage source. This will occur at no-load, as any specified loading will lower the current through the voltage source. So the max current is simply the 100 volts divided by 60 plus 15 Kohm, or 100/75 or 1.33 mA.

Use that current to find the power required in each resister. For R1 and R2, I think 1/8 and 1/16, respectively, should do.

I hope that helps.

Answer 2

I like you already :-) be gratefull (it's cool)

with 48K you are allowed a 4 volt drop (20 to 16)

so with 16 volts on 48k that's 1/3 mA

the output impedance maximum of the voltage divider must be given by 4 volts and 1/3 mA ie 12 K

So R1//R2 = 12k = R1*R2/(R1+R2)

100 * R2/(R1+R2) =20 (th open circuit volts

one can substitute R2/(R1+R2) = 12/R1

From which we get R1 = 60K R2 = 15K

These give exactly 20 volts open circuit
and 16 volts with 48k load

Power 100-16 = 84 volts across the 60K
= 0.1176 Watts max

power in the 15K is 0.0266667 max (at 20 volts)

so use 1/8th watt for the 60K
(not that 60k is a E24 value)

Answer 3

joe_ska answer is correct although I would quibble that although the max power in R2 is at no-load the maximum power in R1 is when full-load and is (84^2)/60K =.117 Watt. Still an 1/8 watt resistor but closer than you would want to use in real engineering practice.

Answer 4

I'd love to help, but sadly I cant. Oh yeah, and you're pretty cute.

Answer 5

all man were back on again lol