Home
Class 12
PHYSICS
A resistance of R draws current from a p...

A resistance of `R` draws current from a potentiometer. The potentiometer has a total resistance `R _(0)`. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potenttometer.

Text Solution

Verified by Experts

While the silde is in the middle of the potentiometer only half of its resistance `(R _(0) //2)` will be between the points A and B. Hence, the toal resictance between A and B, say, `R _(1),` will be given by the following expression:
` (1)/(R _(1)) = (1)/(R) + (1)/((R _(0) //2))`
`R _(1) = (R_(0) R )/( R _(0) + 2R)`
The total resistance between A and C will be sum of resistance between A and B and B and C, i.e., `R _(1) + R _(0) //2`
`therefore` The current following through the potentiometer will be
`I = (V)/(R _(1) + R _(0) //2) = (2V)/( 2 R_(1) + R _(0))`
The voltage `V _(1)` taken from the potentiometer will be the product of current I and resistance `R _(1),`
`V _(1) = I R _(1) ((2V)/(2R_(1) + R _(0))) xx R_(1)`
`V _(1) = (2 V)/(2 ((R _(0) xx R)/( R _(0) + 2R )) + R) xx (R_(0) xx R)/( R _(0) +2R)`
`V _(1) = (2VR)/( 2R+ R_(0) + 2R)`
or ` V _(1) = (2VR)/( R _(0) + 4R)`
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    NCERT TAMIL|Exercise EXERCISES|13 Videos
  • CURRENT ELECTRICITY

    NCERT TAMIL|Exercise ADDITIONAL EXERCISES|9 Videos
  • COMMUNICATION SYSTEMS

    NCERT TAMIL|Exercise Evaluation (I. MULTIPLE CHOICE QUESTIONS) |5 Videos
  • DUAL NATURE OF RADIATION AND MATTER

    NCERT TAMIL|Exercise EXERCISES|41 Videos

Similar Questions

Explore conceptually related problems

A potentiometer wire has a length 400 cm and a resistance of 20 Omega . It is connected in series with an external resistance R and a cell of e.m.f. 1.5 V and internal resistance 1 Omega . A source of e.m.f.6 mV is balanced against a length of 30 cm of the potentiometer wire. Find R. According to the principle of potentiometer Vpropl . i.e., V = ilp where l'is the balancing length & p the resistance per cm of the potentiometer wire.

A. Potentiometer measures the potential difference more accurately than a voltmeter, because the potentiometer a. does not draw current from external circuit. b. has a wire of high resistance. c. draws a heavy current from external circuit. d. has a wire of low resistance. B. With the help of a diagram explain the principle of a potentiometer.

A cell of emf 2V and internal resistance 0.1 Omega is connected with a resistance of 3.9Omega . The voltage across the cell terminals will be

Shown an arrangement to measure the emf (epsilon) and internal resistance r of a battery. The voltmeter has a very high resistance and the ammeter also has some resistance.The voltmeter reads 1.52 V when the switch S is open.When the switch is closed the voltmeter reading drops to 1.45 V and the ammeter read 1.0A Find the emf and the internal resistance of the battery.

A potentiometer wiere has a length of 4 and resistance of 20 Omega . It is connected in series with resistance of 2980 Omega and a cell of emf 4 V. Calcaulte the potential along the wire.

The potentiometer wire AB shown in figure is 50cm long.When AD=30cm, no deflection occurs in the galvanmeter.Find R.

Shown a circular coil of N turns and radius a, connected to a battery of emf epsilon through a rheostat. The rheostat has a total length L and resistance R. The resistance of the coil is r. A small circular loop of radius a' and resistance r' is placed coaxially with the coil. The centre of the loop is at a distance x from the centre of hte coil. In the beginning, the sliding contact of the rehostat is at the left end and then on wards it is moved towards right at a constant speed v. Find the emf induced int he small circular loop at the instant (a) the contact begins to slide and (b) it has slid thrugh half the length of the rheostat.

A potentiometer circuit has been set up for finding the internal resistance of a given cell.The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance.The potentiomter were itself is 4m long.When the resistance, R, connected across the given cell, has values of (i) Infinity (ii)9.5 Omega the'balancing lengths', on the potentiometerwire are found to be 3m and 2.85m, respectively. The value of internal resistance of cell is