Question

Two inductors of self inductance $L$ each and mutual inductance $M$ are connected in series as shown in the figure.
Here ${\epsilon }_0=100V$, $R=30\Omega ,L=2mH$
$\omega ={10}^{4}rad/s$.

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

Answer 1

Answer: (A), (C), (D)

The correct options are
A Maximum current in the circuit may be 3.33 A.
C Maximum current in the circuit may be less than 2A.
D Source voltage is ahead of current in circuit.

With the combination of two inductors along with their mutual inductances,
$L_{eq}=2L-2M$ or $L_{eq}=2L+2M$
If $L=M$, then $L_{eq}$ can be zero. In that case,
$i_0=\frac{{\epsilon }_0}{R}=\frac{100}{30}=3.33A$
Depending on the value of $M$, the value of $i_0$ will change but it cannot take a value above $3.33A$
In an L-R circuit, voltage is always ahead of current.