Topic: Electronics Engineering
For a particular toroidal inductor core, the inductance value obtained for 50 turns is 200 µH. If two such inductors are wired in parallel, the resulting inductance is
- A. 200 µH
- B. 50 µH
- C. 400 µH
- D. 100 µH
Answer: D. 100 µH
Explanation: If two identical inductors of 200 µH are connected in parallel, assuming no mutual coupling, the equivalent inductance is $L_{eq} = \frac{L_{1} \times L_{2}}{L_{1} + L_{2}} = \frac{200 \times 200}{200 + 200} = 100~\mu H$.