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].

Power in A.C. Circuits

The power(P) dissipated in an alternating current circuit is given by
P = V_rms I_rms cosΦ where cosΦ is the power factor. [You should remember that Φ is the phase difference between current and voltage].
Power in an A.C. circuit is also given by P = (I_rms)²R. Note that power is dissipated in resistance only.
Another form of the power relation is P = (I_rms)²Z cosΦ, which follows from cosΦ = R/Z

Some of you may be wondering which relation is to be used for calculating power. Well, you may use any one of them which is convenient to you in the context of the problem. Now, consider the following MCQ:
A voltage v = V₀ cosωt is applied in series with an inductance L and a resistance R. The average power dissipated per cycle in the circuit is
(a) V₀² /R (b) V₀²/2R (c) V₀²/(R+L) (d) V₀²R/[(R²+ L²ω ²)] (e) V₀²R/[2(R²+ L²ω ²)]
We shall use the relation, P = (I_rms)²R. Since I_rms = V_rms/Z = (V₀/√2)/[√(R²+ L²ω ²)], we obtain P = V₀²R/[2(R² + L²ω ²)].
You could have used the other relations (for power) also to obtain the result.
Consider the folowing MCQ which appeared in C.E.T. J&K 2000 question paper:
In an A.C. circuit V and I are given by
V = 100 sin (1000t) volt
I = 1000 sin (1000t + π/3 ) mA.
The power dissipated in the circuit is
(a) 10⁴ W (b) 25 W (c) 10 W (d) 250 W
Here current, voltage and the phase difference (between current and voltage) are given so that it is convenient to use the relation P = V_rms I_rms cosΦ. [Note that the current has a phase lead of π/3 and is given in mA].
Therefore, P = (100/√2) × (1000×10 –3/√2) ×cos (π/3) = 25 W