Recent content by nmacholl

My equation for R2 is: R2=(VL*r + VL(375))/12 I made a calculator mistake: For VL= 4; r = 100 R2 = (4(100)+4(375))/12 R2 = 158.33 R1 = 375-R2 = 216.67 :) EEEEEEEEEEEeeeeeeeeeeeeeeeeeeeeeee!

Okay now I got these solution sets: VL = 5; r = 5 R1 = 216.67 ohms R2 = 158.33 ohms VL = 4; r = 100 R1 = 60.42 ohms R2 = 314.58 ohms I'm still uncomfortable with there being two values for each resistor.

NVM I solved for 12*R2 and pluged the first into the second to get R1+R2=375

VL=(12*R2)/(r+R1+R2) <algebra happens> R2=(VL*r+VL*R1)/(12-VL) But I can't substitute this back into VL=(12*R2)/(r+R1+R2) because then I would be plugging the equation back into itself.

Right well if r=100 ohms when VL = 5v R2=(500+5*R1)/7 and r=5 ohms when VL = 4v R2=(5+R1)/2 But I don't have an equation to use these in. We used my loop equation and ohms law to get these, don't I need another equation to plug these into?

I made a mistake in my first post but I have: 12-I(r+R1+R2)=0 VL=I*R2 Then you wrote this: VL = (R2*12v) / (r + R1 + R2) but I don't know where that came from. Using mathematica I can solve what you wrote for R2 in terms of R1 For VL = 4 :: R2=5(r+R1)/7 For VL = 5 :: R2=(r+R)/2

I'm still not getting anywhere. I tried doing all the algebra in Mathematica and ended up with worthless results.

Following your suggestions I've gotten two sets of solutions by plugging in r={5ohms, 100ohms} and VL={5v,4v} To get: R1={-4,-99ohms} R2={5/7,-1/2ohms} Now the resistors don't have to be "real" but I'm concerned because the resistors have different values for each load voltage and I don't...

Forgive me but I don't seem to have equations like that. My equation from Kirchoff's loop rule is: 12v - Ir - IR1 - IR2 = 0 This contains 3 unknowns: I, R1, and R2 Using ohm's law I can substitute; I = 12v / (r + R1 + R2) It seems to me that the only equation I have that relates R1 to R2...

I seem to be missing something. What relationship have I not used yet? I can't substitute the voltage equation back into itself and I've already used Ohm's law to replace "I" with 12v/Rtotal.

Okay. Is there any way to pull out another equation from the problem. If I am interpreting your suggestion correctly you want me to use r=5ohms then solve for the voltage across R2 and get 5 volts, then r=100ohms and get 4v. But even then I have one equation with two unknowns: R1 and R2, am I...

Homework Statement You want to build a voltage divider for a product that uses as it's source a 12v battery and a non-zero internal resistance. You want to have an output voltage between 4 and 5 volts. The internal resistance varies from 5 ohms when fully charged to 100 ohms when the battery...

Homework Statement Two point charges, -520x10-6C and -270x10-6C are 2 meters apart. At which point would a carbon nucleus experience zero net force? (positive test charge) Homework Equations k=9x109 F=k((q1*q2)/r2) Coulomb's Law E=k(q1/r2)=(F/q2) The Attempt at a Solution I really...

That certainly makes more sense than before! The worksheet must have a typo on the equation, I double checked and the handout and the -λt is the exponent of Ao and Euler's number is nowhere to be found. Thanks a lot.

If I substitute numbers in for a and b using my calculator I don't get a true statement. a = 2 b = 5 ln(e^(a*b)) = a + b ln(e^(10)) = 7 10 = 7 What am I doing wrong? :(