An inductor of inductive reactance is connected in series with a bulb B and an AC source. How would the brightness of the bulb change when
(i) the number of turns in the inductor is reduced,
(ii) an iron rod is inserted in the inductor, and
(iii) a capacitor of reactance is inserted in series in the circuit? Justify your answer in each case.
(i) the number of turns in the inductor is reduced,
(ii) an iron rod is inserted in the inductor, and
(iii) a capacitor of reactance is inserted in series in the circuit? Justify your answer in each case.
Answer: (i) Brighter;
(ii) dimmer;
(iii) brightest (resonance).
Solution
The current, and hence the brightness, is set by the impedance .
(i) Fewer turns lower and so ; decreases, the current rises, and the bulb glows brighter.
(ii) An iron rod raises and so ; increases, the current falls, and the bulb grows dimmer.
(iii) With the reactances cancel (, resonance); the current is maximum, so the bulb is brightest.
(i) Fewer turns lower and so ; decreases, the current rises, and the bulb glows brighter.
(ii) An iron rod raises and so ; increases, the current falls, and the bulb grows dimmer.
(iii) With the reactances cancel (, resonance); the current is maximum, so the bulb is brightest.
CBSE Class 12 Board · 2015 · Delhi · Subjective