Figure `6.51` shows a simple a potentiometer circuit for measuring a small emf produced by a thermocouple.

The meter wire `PQ` has a resistance of `5 Omega`, and the driver cell has an emf of `2.00 V`. If a balance point is obtained `0.600 m` along `PQ` when measuring an emf of `6.00 mV`,
what is the value of resistance `R`?

Options

A. `95 Omega`
B. `995 Omega`
C. `195 Omega`
D. `1995 Omega`

Answer

B

Explanation

The voltage per unit length on the meter wire `PQ` is
`(6.00 mV)/(0.60 m)`or `10 mVm^(-1)`
Hence, potential across the meter wire `PQ` is `10 mVm^(-1) (1 m)`
` = 10 mV`. Current draw from the driver cell is
`I = (10mV)/(5 Omega) = 2mA`
Resistance of the resistor `R` is
`R = (2V - 10 mV)/(2mA) = (1990 mV)/(2mV) = 995 Omega`