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A sandard cell or 1080V gives a balance length of 512 mm on a potentiometer. If the p.d across a standard 10.0 (ohm) resistor in a circuir gives a balance length of 784 mm, calculate the current in the circuit with the 10.0 (ohm) resistor. Draw a circuit sketch og the two circuit arrangements required in the above case.
I am stuck here..am i going right?.please help someone..
E = k * 512*10^-3 (512 times ten to the power of minus 3)
V = E - Ir
V = k * 784*10^-3
V = IR I = V/R = V/10
E = V + Ir
K* 784*10^-3 = k * 512*10^-3 - K*784*10^-3(r)/10
K's cancel
784*10^-3(r)/10 = 784*10^-3 - 512*10^-3
= 0.0784r = 0.272
= 3.46(ohms)
...what else do i do from here?
Homework Statement
A sandard cell or 1080V gives a balance length of 512 mm on a potentiometer. If the p.d across a standard 10.0 (ohm) resistor in a circuir gives a balance length of 784 mm, calculate the current in the circuit with the 10.0 (ohm) resistor. Draw a circuit sketch og the two circuit arrangements required in the above case.
I am stuck here..am i going right?.please help someone..
The Attempt at a Solution
E = k * 512*10^-3 (512 times ten to the power of minus 3)
V = E - Ir
V = k * 784*10^-3
V = IR I = V/R = V/10
E = V + Ir
K* 784*10^-3 = k * 512*10^-3 - K*784*10^-3(r)/10
K's cancel
784*10^-3(r)/10 = 784*10^-3 - 512*10^-3
= 0.0784r = 0.272
= 3.46(ohms)
...what else do i do from here?