Question

A $50Hz$ ac current of peak value $2$ A flows through one of the pair of coils. If the mutual inductance between the pair of coils is $150mH$. then the peak value of voltage induced in the second coil is

• A. $30\pi V$
• B. $60\pi V$
• C. $15\pi V$
• D. $300\pi V$
• E. $3\pi V$

Answer 1

The correct option is A $30\pi V$

Let the current flows through coil 1 is, $I_1=I_0\sin \omega t$ where, $I_0$is peak value of current.

Magnetic flux linked with coil 2 is ${\varphi }_2=M{I}_1=M{I}_0\sin \omega t$

Emf in coil 2 is

${\epsilon }_2=\frac{d{\varphi }_2}{dt}=\frac{d\left(M{I}_0\sin \omega t\right)}{dt}=M\omega I_0\cos \omega t$

So, peak value of voltage induced in coil 2 is $=M{I}_o\omega ....\left(1\right)$

Given that,

$\nu =50Hz$

$I_0=2A$

$L=150mH$

$\omega =2\pi \nu =2\pi \times 50$

Put all values in equation (1)

$current=150\times {10}^{-3}\times 2\times 100\pi$

$I=30\pi V$