Karnataka CET 2010 Questions on Alternating Currents

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Questions on alternating currents are generally interesting and often not tough. By clicking on the label ‘alternating current’ below this post, you can access all posts related to this section on this site.

Here are two questions on alternating currents which appeared in Karnataka CET 2010 question paper:

(1) In a series R-L-C circuit, the voltage across R is 100 V and the value of R = 1000 Ω The capacitance of the capacitor is 2×10^–6 F; angular frequency of A.C. is 200 rad s^–1. Then the P.D. across the inductance coil is

(a) 100 V

(b) 40 V

(c) 250 V

(d) 400 V

Since the voltage drop across the 1000 Ω resistor is 100 V, the current (I) in the R-L-C circuit is 100 V/1000 Ω = 0.1 A.

At resonance the magnitude of the voltage across the inductance (V_L) is the same as that across the capacitor (V_C).

Or, V_L = V_C = I/Cω where ω is the angular frequency of A.C.

Therefore, V_C = 0.1/(2×10^–6×200) = 250 V.

(2) A capacitor and an inductance coil are connected in separate AC circuits with a bulb glowing in both the circuits. The bulb glows more brightly when

(a) an iron rod is introduced into the inductance coil

(b) the number of turns in the inductance coil is increased

(c) separation between the plates of the capacitor is increased

(d) a dielectric is introduced into the gap between the plates of the capacitor

The reactance of the inductor is Lω where L is the inductance and ω is the angular frequency of AC. In options (a) and (b) the value of L is increased and hence the inductive reactance is increased. The current through the bulb is therefore decreased, thereby decreasing the brightness.

The capactive reactance is 1/Cω where C is the capacitance. When the separation between the plates is increased as given in option (c), the capacitance is decreased. The capacitive reactance is therefore increased, thereby decreasing the current and the brightness.

On introducing a dielectric slab between the plates, the capacitance is increased, thereby decreasing the capacitive reactance. The brightness of the bulb is then increased because of the increased current. So the correct option is (d).