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A circuit has a 10 kΩ potentiometer with a 5 kΩ load. Determine the position of the slider on the ‘pot’ when the voltage across points ‘XX' is 3 V.

Think I'm there but not sure of my working out is excessive or I've taken the long route

R1 in series with 2 resistors in parallel (R2 and 5k ohm)

R1=10x, R2=10(1-x)

R2 + 5k ohm = RC

1/RC = 1/(10(1-x)) + 1/5.

1/RC = (5+10(1-x))/(50(1-x))

RC = 50(1-x) / (5+10(1-x))

= 10(1-x) / (1+2(1-x)

Voltage drop will be 9-3=6v

I = V/R

I = 6/R1 = 3/RC

R1 = 2RC

10x= 2x10(1-x)/ (1+2(1-x))

x = 2(1-x)/(1+2(1-x))

x(1+2(1-x))=2(1-x)

x+2x-2x^2 =2- 2x

2x^2-5x+2 = 0

Quadratic gives

x= 2 or x= .5

x=0.5 (halfway point)

Think I'm there but not sure of my working out is excessive or I've taken the long route

R1 in series with 2 resistors in parallel (R2 and 5k ohm)

R1=10x, R2=10(1-x)

R2 + 5k ohm = RC

1/RC = 1/(10(1-x)) + 1/5.

1/RC = (5+10(1-x))/(50(1-x))

RC = 50(1-x) / (5+10(1-x))

= 10(1-x) / (1+2(1-x)

Voltage drop will be 9-3=6v

I = V/R

I = 6/R1 = 3/RC

R1 = 2RC

10x= 2x10(1-x)/ (1+2(1-x))

x = 2(1-x)/(1+2(1-x))

x(1+2(1-x))=2(1-x)

x+2x-2x^2 =2- 2x

2x^2-5x+2 = 0

Quadratic gives

x= 2 or x= .5

x=0.5 (halfway point)

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