Question

The number of turns in two coils A and B are 300 and 400 respectively. They are placed close to each other. Co-efficient of mutual induction between them is 24 mH. If the current passing through the coil A is 2 Amp then the flux linkage with coil B will be

• A. 24 mwb
• B. 12 × 10^–5wb
• C. 48 mwb
• D. 48 × 10^–5wb

Answer 1

Answer: (C)

48 mwb

Answer 2

Flux linkage $= N _ { 2 } \phi _ { 2 } = M i _ { 1 } = 24 \times 2 = 48 \mathrm { mwb }$