The figure given below shows an LCR series circuit with two switches S1 and S2. When switch S1 is closed keeping S2 open, the phase difference ( ) between the current and source voltage is 30° and phase difference is 60° when S2 is closed keeping S1 open. The value of (3L1-L2) is H. [figure]
- 92
- 29
- 13
- 3
Answer
(2) 29
Solution
From the figure, S1 is across L2 and S2 is across L1, and a closed switch shorts that inductor.
Here = 300 rad/s and XC = 1 C = 1300 100 10-6 = 1003 Ω.
With S1 closed only L1 is in the circuit, so 30° = L1 - XCR … (i)
With S2 closed only L2 is in the circuit, so 60° = L2 - XCR … (ii)
Dividing (ii) by (i): L2 - XC L1 - XC = 31/ 3 = 3
L2 - XC = 3 L1 - 3XC
(3L1 - L2) = 2XC
3L1 - L2 = 2XC = 2 100/3300 = 29 H
Answer: (2)