A `50 Hz` alternating current of peak value `1` ampere flows through the primary coil of a transformer. If the mutual inductance between the primary secondary be `1.5` henry, then the peak value of the induced voltage is

Options

A. `30pi V`
B. `60 pi V`
C. `15pi V`
D. `300 pi V`

Answer

A

Explanation

The current flows through the coil 1 is `I_1 = I_0 sin omega t `
where `I_0` is the peak value of current. Magnetic flux linked with the coil 2 is `phi_2 = MI_1 = MI_0 sin omega t ` where M is the mutual inductance between the two coils. The magnitude of induced emf in coil 2 is
`|epsi_2| = (dphi_2)/(dt) = (d)/(dt) (MI_0 sin omega t) = MI_0 omega cos omega t `
`therefore `Peak value of voltage induced in the coil 2 is ` = MI_0 omega `
` = 150 xx 10^(-3) xx 2 xx 2pi xx 50 = 30pi V`