Question

88. An inductor of inductance 10 H is connected in
series with a resistance R = 6ohm. A 12 V battery is
connected for a long time. When the circuit is
switched off, the induced emf in inductor, if current
reduces to zero in 10 ms, is
(a) 1000 V
(b) 2000 V
(c) 3000 V
(d) 4000 V​

Answer 1

the required answer is option b 2000 volt

GIVEN:

• Inductance of inductor ,L=10H

• resistance,R=6 ohm

• voltage of battery,v=12 volt

• time ,t=10 milli second

TO FIND:

emf induced in the Circuit ,e

FORMULAE:

• according to OHM'S law, V=IR

• According to faraday's law of electromagnetism, $e=-d \phi/ dt$

• $\phi = LI$

SOLUTION:

STEP1: TO CONVERT THE UNIT OF TIME:

• it is necessary to convert unit of time to standard unit.

• the SI unit of time is second

• 1 milli second= 10^-3 second

• 10 milli second =10 × 10^-3 second

• time ,dt = 10^-2 second

STEP 2: TO FIND THE CURRENT:

V=IR

I=V/R

=12/6

I=2 ampere

current ,I = 2A

STEP 3: TO COMBINE THE FARADAYS LAW:

$e=-d\phi/dt$......1

$\phi = LI$........2

substitute equation 2 in 1, we get,

e=-LI/dt

STEP 4: TO FIND INDUCED EMF:

$e = \frac{ - LI}{dt} \\ \\ = \frac{ - (10)(2)}{ {10}^{ - 2} } \\ \\ = \frac{ - 20}{ {10}^{ - 2} } \\ \\ = - 20 \times {10}^{2} \\ \\ e = - 2000 \: volt$

induced emf , e= 2000 volt

NOTE:

here the negative sign indicates that the emf is produced in the direction opposite to the flow of current.

Answer 2

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