Home
Class 12
PHYSICS
The equal resistances, 400 Omega each ar...

The equal resistances, `400 Omega` each are connected in series with a 8V battery. IF the resistance of first one increases by 0.5%, the charge required in the resistance of the second one in order to keep the potential difference across it unaltered is to

A

increase it by `1 Omega`

B

increase it by `2 Omega`

C

increase it by `4 Omega`

D

decrease it by `4 Omega`

Text Solution

Verified by Experts

The correct Answer is:
B

Increase in first resistance=`400 times 0.5/100=2 Omega`
Initially, the emf 8V will be divided equally between the two resistances.SO the voltage across each resistance will be 4V. When the first resistance is increased, the second resistance should also be inceased by `2 Omega` to keep the voltage across it unchanged.
Promotional Banner

Topper's Solved these Questions

  • CURRENT ELECTRICITY

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE WITH SOLUTIONS(JEE MAIN)|4 Videos
  • CURRENT ELECTRICITY

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE WITH SOLUTIONS(AIPMT)|2 Videos
  • CURRENT ELECTRICITY

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE WITH SOLUTIONS(WBCHSE)|18 Videos
  • COMMUNICATION STSTEM

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|28 Videos
  • DIFFRACTION AND POLARISATION OF LIGHT

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|27 Videos

Similar Questions

Explore conceptually related problems

Three resistance of magnitudes 20Omega,30Omega and 40Omega are connected in series.(i) What is the equivalent resistance? (ii) IF the potential difference across the resistance 20Omega is 1V, calculate the potential differences across the other two resistances and also the total potential difference across the combination.

A cell of emf 1.4V and internal resistance 2 Omega is connected in series with a resistance of 100 Omega and an ammeter. The resistance of the ammeter is 4/3 Omega .To measure the potential difference between the two ends of the resistance a voltmeter is connected IF the reading of the ammter is 0.02 A.what is the resistance of the voltmeter?

A cell of emf 1.4V and internal resistance 2 Omega is connected in series with a resistance of 100 Omega and an ammeter. The resistance of the ammeter is 4/3 Omega .To measure the potential difference between the two ends of the resistance a voltmeter is connected IF the reading of the voltmeter is 1.10V, what will be its error?

A cell of emf 1.4V and internal resistance 2 Omega is connected in series with a resistance of 100 Omega and an ammeter. The resistance of the ammeter is 4/3 Omega .To measure the potential difference between the two ends of the resistance a voltmeter is connected .Draw the circuit

Two resistances 2Omega and 6Omega are connected in series and the combination is then connected to a source of emf 12 V . How much power is consumed in each resistance ?

Two resistances 1 Omega and 2 Omega are connected in series and a potential difference of 6V is applied across the ends of this combination. What will be the terminal potential difference across the second resistance?

A battery of emf 1.4V and interna! resistance 2Omega is connected to a resistor of 100Omega resistance through an ammeter. The resistance of ammeter is 4/3Omega a voltmeter has also been connected to find the potential difference across the resistor. The ammeter reads 0.02 A. What is the resistance of voltmeter.