Two Questions (MCQ) on A.C. Circuits

Questions on alternating current circuits were discussed on 31^st October, 2006 and on 20^th March 2007. You will find those questions by clicking on the label ‘AC circuit’ below this post. Here are two more questions on AC circuits:

(1) In a simple AC generator, a plane rectangular coil PQRS rotates in a magnetic field B. During the rotation, the emf induced in the coil will be maximum when the plane of the coil is

(a) parallel to the magnetic field

(b) perpendicular to the magnetic field

(c) inclined at 30º with the magnetic field

(d) inclined at 45º with the magnetic field

(e) inclined at 60º with the magnetic field

It is the magnetic (Lorentz) force on the electrons in the sides QR ans PS of the coil, which is responsible for the shifting of charges and the consequent motional emf in these sides. The force (and therefore, the induced emf) is maximum when these sides move perpendicular to the magnetic field and this happens when the plane of the coil is parallel to the magnetic field. To put this in a different manner, the sides QR and PS will cut the magnetic field lines normally when the plane of the coil is parallel to the magnetic field (fig).

The correct option therefore is (a).

(2) In the AC circuit shown, the inductive reactance is X_L and the capacitive reactance is X_C. The readings of the voltmeter and ammeter are respectively

(a) 110 V, 0.25 A

(b) 55 V, 0.25 A

(c) 55 V, 0.5 A

(d) 110 V, 0.5 A

(e) 0 V, 0.5 A

The voltage across the inductance leads the current in the series circuit by 90º and the voltage across the capacitor lags behind the current by 90º. The voltages across the inductance and capacitance therefore are in opposition (phase difference of 180º). Since the inductive and capacitive reactances have the same value, the voltages across these elements have the same value, but they are in opposition. They get canceled and the voltmeter reading is zero. The entire supply voltage appears across the resistance which alone controls the current. The current is 110/220 = 0.5 A, which is indicated by the ammeter.

[If you remember that the inductive reactance and the capacitive reactance have equal magnitudes at resonance and the circuit behaves as a pure resistance, you will be able to answer this question in no time].