In an induction coil, the coefficient of mutual induction is 4 henry. If a current of 5 ampere in 1 the primary coil is cut off in $\frac{1}{1500}$ s, the e.m.f at the terminals of the secondary coil will be:

15 kV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

60 kV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 kV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30 kV

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D 30 kV
$ϕ_{21} = M . i,$
$\Rightarrow v = \frac{d ϕ_{21}}{d t} = M \frac{d i}{d t}$
$= 4 \times \frac{5}{(1 / 1500)}$
$= 20 \times 1500$
$= 30 k v$
Hence,
option $(D)$ is correct answer.

Suggest Corrections

Similar questions

Q. In an induction coil, the coefficient of mutual induction is

4 H

. If a current of

5 A

in the primary coil is cut-off in

1 / 1500 s

, the emf at the terminals of secondary coil will be:

In a transformer, the coefficient of mutual inductance between the primary and the secondary coil is 0.2 henry. When the current changes by 5 ampere/second in the primary coil, the induced e.m.f. in the secondary coil will be

Q. Alternating current of peak value

(\frac{2}{π})

ampere flows through the primary coil of the transformer. The coefficient of mutual inductance between primary and secondary coil is 1 henry. The peak e.m.f. induced in secondary coil is
(Frequency of a.c. = 50 Hz)

Q. In an induction coil, the coefficient of mutual inductance is 4 H. If a current of 5 A in the primary circuit is cut off in

\frac{1}{500}

s, the emf induced in the secondary circuit will be

The mutual inductance between two coils is 1.25 henry. If the current in the primary coil changes at a rate of 80 ampere/second, then the induced e.m.f. in the secondary coil is

Join BYJU'S Learning Program

In an induction coil, the coefficient of mutual induction is 4 henry. If a current of 5 ampere in 1 the primary coil is cut off in 11500s, the e.m.f at the terminals of the secondary coil will be:

The correct option is D 30 kVϕ21=M.i,⇒v=dϕ21dt=Mdidt=4×5(1/1500)=20×1500=30kvHence,option (D) is correct answer.

In an induction coil, the coefficient of mutual induction is 4 henry. If a current of 5 ampere in 1 the primary coil is cut off in $\frac{1}{1500}$ s, the e.m.f at the terminals of the secondary coil will be:

The correct option is D 30 kV
$ϕ_{21} = M . i,$
$\Rightarrow v = \frac{d ϕ_{21}}{d t} = M \frac{d i}{d t}$
$= 4 \times \frac{5}{(1 / 1500)}$
$= 20 \times 1500$
$= 30 k v$
Hence,
option $(D)$ is correct answer.