InterviewSolution
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InterviewSolution
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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`?A. `95 Omega`B. `995 Omega`C. `195 Omega`D. `1995 Omega` |
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Answer» Correct Answer - B 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` |
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