CBSE Class 12-science Answered
Please solve this numerical?
Asked by Anshul | 06 Aug, 2017, 10:01: AM
Expert Answer
A) The two circuits provided have an inductor and a capacitor, respectively connected to the bulb.
B) The resistance of the devices are XL = 2πfL and XC = 1/(2πfC), respectively.
1) When the frequency of the source decreases, the inductive reactance decreases. Therefore, the current in the circuit having L increases.
Hence, the brightness of the bulb increases. In the second circuit, when the frequency of the source decreases, the capacitive reactance increases. Therefore, the current in the circuit having C decreases. Eventually, this leads to reduced brightness of the bulb.
2) If the number of turns in circuit I is increased, then the inductance L increases. This increases the inductive reactance thereby reducing the current through the coil. This causes a reduction in the brightness of the bulb.
3) When a capacitor is connected in series, the total impedance of the circuit increases. This reduces the current through the bulb, thereby reducing its brightness.
Answered by Romal Bhansali | 07 Aug, 2017, 07:10: PM
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