Recent content by damon669

The voltage across the left-most branch is 1/11∙1+13=13 1/11 =13.0909 V The voltage across the right-most branch is -5/11∙2+14 =13.0909V The voltage across the centre is 4/11∙3+12=13.0909V

oh that was easy -5/11 x 2 +14 = 13.0909

left hand branch 1/11 x 1 +13 = 13.0909 right hand branch = 5/11 x 2 + 14 =14.0909 confused

so to check voltage is just V=I/R

For R1 P=I_1^2 R =0.00826 W For R2 P= I_2^2 R =0.454*2 =0.90990 W For R P=I^2 R =0.1322*3 =0.3967 W

Homework Statement a) Calculate the value of the 12V battery b) Calculate the power dissipated in R_1,R_2 and R. https://www.physicsforums.com/attachment.php?attachmentid=60819&stc=1&d=1376202905 Homework Equations I_1+I_2+I_R=0 The Attempt at a Solution For the left-hand...

OK this is what i got then Let p represent the position of the ‘pot’ slider where 0≤p≤1 (i.e. the proportion of the ‘pot’ resistance that is in parallel with the load resistor). For p=0 the slider is at the bottom (minimum resistance) and for p=1 the slider is at the top (maximum resistance)...

so you DO have to use a Quadratic to solve then

Can this be done without using a Quadratic equation?

If the slider was at 5k ohm would that not give 3 volts across the load?

10k ohm is the max position of the slider @ 9V

voltage division. is it correct then?

That formula came from my notes

Vl = 3 volts across the load Ro = what I'm trying to find Vi = voltage on the circuit R1 = 10k ohms R2 = 5k ohms

Vl=RoVi/R1+R2 3=Ro•9/(10•10^3)+(5•10^3) = 5k ohms