Measuring unknown emf by potentiometer

AI Thread Summary
In measuring unknown emf with a potentiometer, the stable current source must have a voltage greater than the voltage being measured to ensure that there is a point along the resistance where the voltage can equal the unknown emf. The voltage drop across the entire length of the potentiometer wire must exceed the emf to allow for a measurable balance point. If the source voltage is insufficient, it won't be possible to find a corresponding voltage drop that matches the unknown emf. This principle is crucial for accurate measurements in physics experiments. Understanding this relationship is essential for conducting effective lab work with potentiometers.
tin llenaresas
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in measuring unknown emf by potentiometer, why must the stable current source(working battery) have a voltage greater than any voltage to be measured?

i can't understand why..
in our exercise in physics, we used a slide-wire potentiometer with the assumption that 1 meter of the slide-wire gives a voltage drop of 1V. so, the battery we used has an emf slightly greater than 2 volts.

is there a connection between the length of the slide wire used and the use of a battery with an emf greater than 2V?

i am currently doing our physics lab report and honestly, its part of the discussion. I am asking for help for i can't understand.

tanx to anybody who dares to reply.. :smile:
 
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tin llenaresas said:
in measuring unknown emf by potentiometer, why must the stable current source(working battery) have a voltage greater than any voltage to be measured?

i can't understand why.
Because you are finding the point along a length of an electrical resistance where the voltage is equal (and opposite) to the voltage you are trying to measure. The potential difference goes from 0 to maximum along that resistance, maximum being the voltage across the entire length.

If the voltage across that entire length is not greater than the voltage you are trying to measure, there will be no point where it is equal to the measured voltage.

AM
 
tanx buddy.ü
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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