Question

Two coils has mutual inductance of 1.5 henry if the current in the primary circuit is raised by 5a in one millisecond after closing the circuit ,then the induced emf in secondary coil is ?

Answer 1

Answer:

7500

Explanation:

emf = mdI/dt

here

m = 1.5

dl/ dt = 5 / .001

Answer 2

Given,

The mutual inductance of two coils (M) = 1.5 Henry

The change in current in the primary circuit (dI) = 5 A

The time in which the current changes (dt) = 10⁻³ sec

Induced E.M.F can be calculated by the following formula:

$E=M\frac{dI}{dt}$

Hence,

$E=1.5\times \frac{5}{10^{-3} }$

$E= 7.5\times 10^{3} V$

The induced E.M.F. in the secondary coil is 7.5×10³ Volts.

Hence, the correct answer is 7.5×10³ Volts.