Potentiometer - calculating internal resistance of a cell | Electricity | Physics | Khan Academy

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me too

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Best explanation of potentiometer on KZbin dhanyawad (in English='thankyou')

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[@cypherfps3695]

6:06 how to measure voltage without drawing current. Does the measuring device work without current. Please help

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[@elizabethelson1124]

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[@anshik.k.t]

When I am measuring emf without bulb ... There is still a internal resistance... so isn't that gonna draw some current and hence disturb reading?

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From where do you study all these things?

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Clarity of this level is only attained by studying not for marks but for curiosity At that point it doesnt really matter where they study from.

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[@vivekgupta3692]

*_Summary Of The Video :-_* *Potentiometer* * We use a voltmeter to measure the potential difference (Voltage) but it does not give us accurate value. Because it draw some ammount of current from the circuit. Which can change the value of Voltage as well. * We can measure the accurate value of Voltage by using a _Potentiometer_. *Principle Of Potentiometer -* The principle of a potentiometer is that the potential dropped across a segment of a wire of uniform cross-section carrying a constant current is directly proportional to its length. *Wire Used In Potentiometer -* Generally alloys like constantan or manganin are used as potentiometer wire because they posses high specific resistance and low temperature coefficient *Null Point or Balancing Length Of The Potentiometer -* The balancing point or null point of the potentiometer is the point on the slide wire when the galvanometer shows zero deflection. *Primary Circuit & Secondary Circuit -* primary circuit consists of a secondary cell( driver cell). and a secondary circuit is the one in which the primary cell whose emf is to be determined is connected. *Potential Gradient -* Potential drop per unit length of the wire is known as potential gradient. *Sensitivity Of Potentiometer -* Sensitivity of Potentiometer means the smallest potential difference that can be measured by using it and this can be achieved by decreasing the potential gradient by increasing the length of the wire.

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​@@Nishantsingh-bq1mllts actually OK, cause for someone somewhere this might help in understanding certain definitions

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[@niranjan7457]

Sir at 10:46, why didn't the Galvanometer stay at 0? Why did the Voltage drop when Current was passed through it? Please clarify sir

[@KhanAcademyIndiaEnglish]

This is clarified in our previous videos on EMF and cells. In short, if you have a 9V battery, it means a coulomb of charge gains 9J of energy as it gets pushed from lower to higher terminal. But, due to internal resistance, some of that energy is converted to heat, as a result the net energy gained will be less than 9. This net energy gained is the potential difference across the cell. The more current we draw, the faster the coulomb moves, the more energy is lost to the resistance, the smaller net energy gained, and hence smaller the p.d. across the cell!

[@niranjan7457]

@@KhanAcademyIndiaEnglish Thank you so much sir!!! 🙏🙏🙏

[@sashwatyo1293]

Sir one doubt , shouldn’t the voltage of battery be E-ir?

[@KhanAcademyIndiaEnglish]

That's true as well, but we wanted a relation connecting the EMF and the external resistor, since its value is directly known.

[@sashwatyo1293]

@@KhanAcademyIndiaEnglish Oh okay sir got it. But sir I got another doubt- why is I=E/(R+r) ,E isn’t the terminal voltage,sir.

[@KhanAcademyIndiaEnglish]

@@sashwatyo1293 Maybe go through our videos on cells and EMF?

[@sashwatyo1293]

@@KhanAcademyIndiaEnglish yes sir sure sir...actually I have done that Concept once , but I have forgotten most of them over the year......Thanks sir ...;))

[@subramanianloganathan8696]

Sir why do we have to get rid of the bulb while we are calculating the emf of the battery?

[@tiredcat6653]

Because with bulb the voltage that potentiometer measure is terminal voltage and without bulb the voltage that potentiometer measures is emf

[@kiranbhat4902]

Sir is the magnitude of charge of an electron -1.6 * 10^-19C or 1.6*10^-19 C ? How can 1C of charge contain 6.25 times 10^18 electrons? If 6.25 *10^18C Move to right can we say 1C of charge moves to left? Please please answer my doubts I have these doubts from a few years and they aren't clarified....🙏🙏🙏

[@manojmaheshwari4055]

Charge on an electron is -1.6 *10^-¹⁹ C but the magnitude willl be positive of that value . And through milikans oil drop experiment we found that value , and if there are n electrons the the charge would be n times that value (e=1.6*10-¹⁹) . So q=ne . So if there is 1C of charge n=q/e so 1C/ 1.6*10-¹⁹ which will be equal to 6.25 *10¹⁸ electrons. If ther was 6.25 times 10^18 C of charge divide that by "e" and u will get no of electrons in that amount of charge. We generally use conventional current where positive charges flow but really not , electrons ( negative charges ) are actually flowing from negative terminal to positive terminal) but between 80-90's they used to follow that positive charges flow . So we r till now following that convention that's y it is called as conventional current . So mathematically , a positive charge flowing in a direction is equal to negative charge flowing in the opposite direction. Hope u get that

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[@krishnaniti6208]

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[@krishnaniti6208]

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[@krishnaniti6208]

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[@krishnaniti6208]

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Just practice the questions bro from question bank....