Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AIIMS2002PhysicsAlternating Current

Assertion : We use a thick wire in the secondary of a step down transformer to reduce the production heat. Reason : When the plane of the armature is parallel to the line of force of magnetic field, the magnitude of induced e.m.f. is maximum.

Options

  1. AIf both the assertion and reason are true and reason is a correct explanation of the assertion.
  2. BIf both assertion and reason are true but assertion is not a correct explanation of the assertion.
  3. CIf the assertion is true but the reason is false.
  4. DIf both assertion and reason are false.

Correct answer

B. If both assertion and reason are true but assertion is not a correct explanation of the assertion.

Step-by-step solution

A step-down transformer converts electrical energy from a high voltage to one at a low voltage. Accordingly the current in the secondary will be larger than that in the primary. In order to produce less heat in the secondary, we use a wire of lesser resistance i.e. thick wire. We also know that when the plane of the armature is parallel to the lines of force of magnetic field, the rate of change of magnetic flux linked with it is maximum. Therefore the e.m.f. induced in the armature in this orientation is maximum.

Practice Alternating Current on Quantrex Academy →

More from Alternating Current

A step down transformer connected to an a.c. mains of 220 V is made to operate at 5.5 V, 44 W lamp. The current in the primary circuit is (Ignore power losses) 2026Which phasor diagram represents LCR circuit at resonance? 2026For an R-L series circuit, the power factor is 3 2 , for R-L frequency f Hz. If the frequency doubles, the new power factor will be 2026In an AC circuit, the current is I = 100 (5t) A. The value of I_ rms is 2026The ratio of power factor of purely resistive circuit to purely reactive circuit is ( 0^ = 1 and 90^ = 0 ) 2026In an LCR series circuit, at resonance, 2026In an AC circuit, E and I are given by E = 150 (150t) V and I = 150 (150t + 3 ) A. The power dissipated in the circuit is (60)^ = 1/2 2026A series LCR circuit is connected across a source E of e.m.f. E=15 (50 t- 3 ) . The current from the supply is I=5 (50 t+ 6 ) . The impedance of the circuit and the phase differenc 2026 Full Alternating Current list All AIIMS PYQs