Home
Class 12
PHYSICS
A sinusoidal voltage V (t) = 100 sin (50...

A sinusoidal voltage `V (t) = 100 sin (500 t)` is applied across a pure inductance of `L = 0.02 H`. The current through the coil is :

A

`10 cos (500 t)`

B

`- 10 cos (500 t)`

C

`10 sin (500 t)`

D

`- 10 sin (500 t)`

Text Solution

Verified by Experts

The correct Answer is:
2
`i = i_(0) sin (omega t - (pi)/(2)), i_(0) = (V_(0))/(X_(L))`
Promotional Banner

Topper's Solved these Questions

  • ALTERNATING CURRENT

    NARAYNA|Exercise NCERT Based Question|1 Videos

  • ALTERNATING CURRENT

    NARAYNA|Exercise LEVEL - IV NCERT Based Questions|15 Videos

  • ALTERNATING CURRENT

    NARAYNA|Exercise LEVEL - II (C.W)|11 Videos

  • ATOMIC PHYSICS

    NARAYNA|Exercise LEVEL-II (H.W)|14 Videos

Similar Questions

Explore conceptually related problems

Alternating voltage V = 400 sin ( 500 pi t ) is applied across a resistance of 0.2 kOmega . The r.m.s. value of current will be equal to

Alternating voltage V = 400 sin ( 500 pi t ) is applied across a resistance of 0.2 kOmega . The r.m.s. value of current will be equal to

A small signal voltage V(t)=V_(0)sin omegat is applied across an ideal capacitor C :

An alternating voltage E=E_0 sin omega t , is applied across a coil of inductor L. The current flowing through the circuit at any instant is

An alternative voltage V = 10 sin omega t (in volts) is applied across a parallel arrangement as shown current (in A ) through the source is best described by

A sinusoidal voltage Vsin(at) is applied across a series combination of resistance R and inductor L. The amplitude of the current in the circuit is

NARAYNA-ALTERNATING CURRENT-LEVEL - III (C.W)
  1. A circuit contanining resistance R(1), Inductance L(1) and capacitance...

    Text Solution

    |

  2. An AC source of variable frequency is applied across a series L-C-R ci...

    Text Solution

    |

  3. A 750 Hz, 20 V source is connected to as resistance of 100 Omega an in...

    Text Solution

    |

  4. An ac source of angular frequency omega is fed across a resistor R and...

    Text Solution

    |

  5. An LCR circuit has L = 10 mL, R = 3 Omega, and C = 1 mu F connected in...

    Text Solution

    |

  6. In the circuit shown, R is a pure resistor, L is an inductor of neglig...

    Text Solution

    |

  7. A capacitor has a resistance of 1200 M Omega and capacitance of 22 mu ...

    Text Solution

    |

  8. A 120 V, 60 Hz a.c. power is connected 800 Omega non-inductive resista...

    Text Solution

    |

  9. A 100 V a.c. source of frequency 50 Hz is connected to a LCR circuit w...

    Text Solution

    |

  10. A coil has an inductance of 0.7 H and is joined in series with a resis...

    Text Solution

    |

  11. Two alternating voltage generators produce emfs of the same amplitude(...

    Text Solution

    |

  12. The potential difference across a 2 H inductor as a function of time i...

    Text Solution

    |

  13. For the circuit shown in the figure the rms value of voltage across R ...

    Text Solution

    |

  14. A bulb is rated at 100V, 100W. It can be treated as a resistor. Find o...

    Text Solution

    |

  15. In the circuit shown in fig. X(C ) = 100 Omega,(XL)=200 Omega and R=10...

    Text Solution

    |

  16. When the rms voltages V(L), V(C ) and V(R ) are measured respectively ...

    Text Solution

    |

  17. A sinusoidal voltage V (t) = 100 sin (500 t) is applied across a pure ...

    Text Solution

    |

  18. For the LCR circuit, shown here, the current is observed to lead the a...

    Text Solution

    |

  19. An LCR curcuit is equivalent to a damped pendulum. In an LCR circuit t...

    Text Solution

    |

  20. An are lamp requires a direct current of 10A at 80V to function. If it...

    Text Solution

    |