Home
Class 12
PHYSICS
In Fig. key K is closed at t = 0. After ...

In Fig. key `K` is closed at `t = 0`. After a long time, the potential difference between `A` and `B` is zero, the value of `R` will be `[r_(1) = r_(2) = 1 Omega, E_(1) = 3 V` and `E_(2) = 7 V`, `C = 2 mu F`, `L = 4 mH`, where `r_(1)` and `r_(2)` are the internal resistances of cells `E_(1)` and `E_(2)`, respectively].

A

`(4)/(3) Omega`

B

`(4)/(9) Omega`

C

`(2)/(3) Omega`

D

`4 Omega`

Text Solution

Verified by Experts

The correct Answer is:
B

`E_(i) - ir_(1) = 0 rArr I = (E_(1))/(r_(1)) = (3)/(1) = 3 Omega`
`I = (10)/(3R + 2),(10)/(3R + 2) = 3, R = (4)/(9) Omega`
Promotional Banner

Topper's Solved these Questions

  • INDUCTANCE

    CENGAGE PHYSICS|Exercise Exercises (multiple Correct )|7 Videos
  • INDUCTANCE

    CENGAGE PHYSICS|Exercise Exercises (assertion-reasoning)|2 Videos
  • INDUCTANCE

    CENGAGE PHYSICS|Exercise Exercises (subjective)|7 Videos
  • HEATING EFFECT OF CURRENT

    CENGAGE PHYSICS|Exercise Thermal Power in Resistance Connected in Circuit|28 Videos
  • KINETIC THEORY

    CENGAGE PHYSICS|Exercise Question Bank|31 Videos

Similar Questions

Explore conceptually related problems

In the network shown the potential difference between A and B is (R = r_(1) = r_(2) = r_(3) = 1 Omega, E_(1) = 3V, E_(2) = 2V, E_(3) = 1V )

In the network shown the potential difference between A and B is R(=r_(1)=r_(2)=r_(3)=1Omega,E_(1)=3V,E_(2)=2V,E_(3)=1V)

In the circuit shown below E_(1) = 4.0 V, R_(1) = 2 Omega, E_(2) = 6.0 V, R_(2) = 4 Omega and R_(3) = 2 Omega . The current I_(1) is

Find a potentail difference varphi_(1) - varphi_(2) between points 1 and 2 of the circuit shown in Fig if R_(1) 1= 10 Omega, R_(2) = 20 Omega, E_(1) = 5.0 V , and E_(2) = 2.0 V . The internal resisances of the current sources are negligible.

Internal resistance of battery of emf 2E is R_1 and battery of emf E is R_2 . If the potential difference across the cell of emf 2E is zero, then value of R is

In the circuit R_(1) = 4 Omega , R_(2) = R_(3) = 15 Omega R_(4) = 30 Omega and E = 10 V Calculate the equivalent resistance of the circuit and current in each resistor

In the circuit shown, E_(1)= 7 V, E_(2) = 7 V, R_(1) = R_(2) = 1Omega and R_(3) = 3 Omega respectively. The current through the resistance R_(3)

For the circuit shown E=50 V R_(1)=10 Omega , R_(2)= 20 Omega, R_(3) = 30 Omega L = 2.0 mH . The current through R_(1) and R_(2) are ________ . And _______ immediate after the switch is closed.

In the following circuit , E_1 = 4V, R_1 = 2 Omega, E_2 = 6 V, R_2 = 2Oemga, and R_3 = 4Omega. Find the currents i_1 and i_2

CENGAGE PHYSICS-INDUCTANCE-Exercises (single Correct )
  1. In the given circuit (Fig), key K is witched on the at t = 0. The rati...

    Text Solution

    |

  2. A closed loop of cross-sectional area 10^(-2) m^(2) which has inductan...

    Text Solution

    |

  3. In Fig. key K is closed at t = 0. After a long time, the potential dif...

    Text Solution

    |

  4. A solenoid of inductance L with resistance r is connected in parallel ...

    Text Solution

    |

  5. The cell in the circuit shows in Fig is ideal. The coil has an inducta...

    Text Solution

    |

  6. Switch S shown in Fig. is closed for t lt 0 and is opened at t = 0. Wh...

    Text Solution

    |

  7. In Fig, the mutual inductance of a coil and a very long straight wire ...

    Text Solution

    |

  8. In the circuit shows in Fig switch S is shifed to position 2 from posi...

    Text Solution

    |

  9. The capacitance in an oscilatory LC circuit is increased by 1%. The ch...

    Text Solution

    |

  10. In the circuit of Fig. (1) and (2) are ammeters. Just after circuit K ...

    Text Solution

    |

  11. A wire of fixed length is wound on a solenoid of lengthl and redius r....

    Text Solution

    |

  12. The frequency of oscillation of current in the inductance is

    Text Solution

    |

  13. In Fig. switch S is closed for a long time. At t = 0, if it is opened,...

    Text Solution

    |

  14. A closed circuit of a resistor R, inductor of inductance L and a sourc...

    Text Solution

    |

  15. Given L(1) = 1 mH, R(1) = 1 Omega, L(2) = 2 mH, R(2) = 2 Omega ...

    Text Solution

    |

  16. A horizontal ring of radius r = (1)/(2) m is kept in a vertical consta...

    Text Solution

    |

  17. In the circuit shows in Fig. kay K is closed at t = 0, the current thr...

    Text Solution

    |

  18. In the circuit shows in Fig. switch k(2) is open and switch k(1) is cl...

    Text Solution

    |

  19. In an LC circuit shows in Fig. C = 1 F, L = 4 H. At time t = 0, charg...

    Text Solution

    |

  20. An aluminium ring hangs vertically from a thread with its axis pointin...

    Text Solution

    |