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What is potentiometer ? Explain principl...

What is potentiometer ? Explain principle of potentiometer.

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`rArr` Potentiometer is a device in which continously varying potential difference can be obtained and measured also.
`rArr` Device measuring potential difference is called potentiometer.
Principle:

`rArr` As shown in figure, a battery having emf t and internal resistance r is connected with resistance box R and wire with uniform cross- section and uniform resistance per unit length (AB) are connected in series.
`rArr` Resistance box (R) is always not needed.
`rArr` length of AB wire = L
`rArr` Let resistance of wire per unit length is `rho` .
`therefore ` Resistance of wire AB = L `rho`
`rArr` Let value of resistance obtained (used) from resistance box is R.
`thereforre ` Current flowing in the wire,
`I = (epsilon)/(R + L rho + r) " " `.... (1)
`rArr ` If length of AC wire is l then, resistance of AC wire = `rho ` l
`rArr` Potential difference between A and C = current `xx ` (resistance of AC )
`therefore V_(l) = epsilon (l) = I rho l " "`....(2)
`rArr therefore epsilon (l) = ((epsilon rho )/(L rho + R + r)) . l `
`therefore epsilon (l) = V = phi `l
Where `phi ` or `sigma ` of K is called potential gradient.
`rArr` Principle : Potential difference between two point of wire is directly proportional to distance between two points.
`rArr` Its unit is `Vm^(-1) ` and
dimensional formula `[M^(1) L^(1) T^(-3) A^(-1) ] `
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An experimental setup of verification of photoelectric effect is shown in the diagram. The voltage across the electrode is measured with the help of an ideal voltmetar, and which can be varied by moving jockey 'J' on the potentiometer wire. The battery used in potentiometer circuit is of 20 V and its internal resistance is 2omega . The resistance of 100 cm long potentiometer wire is 8 omega . The photo current is measured with the help of an ideal ammeter. Two plates of potassium oxide of area 50 cm^(2) at separation 0.5 mm are used in the vacuum tube. Photo current in the circuit is very small so we can treat potentiometer circuit an indepdent circuit. The wavelength of various colours is as follows : |{:("Light",underset("Violet")(1),underset("Blue")(2),underset("Green")(3),underset("Yellow")(4),underset("Orange")(5),underset("Red")(6)),(lambda "in" Årarr,4000-4500,4500-5000,5000-5500,5500-6000,6000-6500,6500-7000):}| It is found that ammeter current remians unchanged (2muA) even when the jockey is moved from the 'P' to the middle point of the potentiometer wire. Assuming all the incident photons eject electron and the power of the light incident is 4 xx 10^(-6) W . Then colour of the incident light is :

An experimental setup of verification of photoelectric effect is shown in the diagram. The voltage across the electrode is measured with the help of an ideal voltmetar, and which can be varied by moving jockey 'J' on the potentiometer wire. The battery used in potentiometer circuit is of 20 V and its internal resistance is 2omega . The resistance of 100 cm long potentiometer wire is 8 omega . The photo current is measured with the help of an ideal ammeter. Two plates of potassium oxide of area 50 cm^(2) at separation 0.5 mm are used in the vacuum tube. Photo current in the circuit is very small so we can treat potentiometer circuit an indepdent circuit. The wavelength of various colours is as follows : |{:("Light",underset("Violet")(1),underset("Blue")(2),underset("Green")(3),underset("Yellow")(4),underset("Orange")(5),underset("Red")(6)),(lambda "in" Årarr,4000-4500,4500-5000,5000-5500,5500-6000,6000-6500,6500-7000):}| When radiation falls on the cathode plate a current of 2muA is recorded in the ammeter. Assuming that the vecuum tube setup follows ohm's law, the equivalent resistance of vacuum tube operating in this case when jockey is at end P.

KUMAR PRAKASHAN-CURRENT ELECTRICITY-SECTION [D] MULTIPLE CHOICE QUESTIONS (MCQs) (MCQs ASKED IN BOARD EXAM AND GUJCET)
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