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Question

# Two inductors of self inductance L each and mutual inductance M are connected in series as shown in the figure. Here ε0=100V, R=30Ω,L=2mH ω=104 rad/s.

A
Maximum current in the circuit may be 3.33 A.
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B
Maximum current in circuit wil be 2A.
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C
Maximum current in the circuit may be less than 2A.
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D
Source voltage is ahead of current in circuit.
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Solution

## The correct option is D Source voltage is ahead of current in circuit.With the combination of two inductors along with their mutual inductances, Leq=2L−2M or Leq=2L+2M If L=M, then Leq can be zero. In that case, i0=ε0R=10030=3.33 A Depending on the value of M, the value of i0 will change but it cannot take a value above 3.33 A In an L-R circuit, voltage is always ahead of current.

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