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A 10 Henry inductor carries a Steady Current of 2 ampere. How can a 100 volt self induced EMF be made to appear in the inductor?
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A 10 Henry inductor carries a Steady Current of 2 ampere. How can a 10...
Induced emf in inductor is given by

e = -LdI/dt

L = 10 H, e = 100 V,

we can calculate the rate of change of current (dI/dt)

100 = -10 x dI/dt

dI/dt = - 10

(0 - 2)/dt = - 10

dt = 2/10 = 0.2 second.

Thus if we change the current to zero in 0.2 s, we can induce 100 V emf across the inductor.
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A 10 Henry inductor carries a Steady Current of 2 ampere. How can a 10...
Introduction:
To generate a self-induced electromotive force (EMF) of 100 volts in a 10 Henry inductor carrying a steady current of 2 amperes, we can make use of Faraday's law of electromagnetic induction. According to this law, a change in the magnetic field through a circuit induces an EMF in the circuit.

Steps to create a 100 volt self-induced EMF:

1. Change in Current:
- The self-induced EMF can be generated by changing the current flowing through the inductor.
- To achieve this, we can connect a switch in series with the inductor.
- Initially, the switch is open, and the inductor carries a steady current of 2 amperes.

2. Closing the Switch:
- When the switch is closed, the circuit is completed, allowing the current to flow through the inductor.
- As the current starts to flow, the magnetic field around the inductor begins to change.

3. Change in Magnetic Field:
- As the current changes, the magnetic field through the inductor also changes.
- This change in magnetic field induces an electromotive force (EMF) in the inductor according to Faraday's law.

4. Calculation of Self-Induced EMF:
- The self-induced EMF can be calculated using the formula: EMF = -L * (di/dt), where L is the inductance and (di/dt) is the rate of change of current.
- Given that the inductance is 10 Henry and the initial current is 2 amperes, we need to calculate the rate of change of current.

5. Rate of Change of Current:
- Let's assume that the current changes linearly with time.
- If the current changes from 2 amperes to 0 amperes in t seconds, then the rate of change of current is (0 - 2)/t = -2/t amperes per second.

6. Substituting values:
- Substituting the values into the formula, we get: EMF = -10 * (-2/t).
- Simplifying further, EMF = 20/t volts.

7. Choosing the Time Interval:
- To generate a self-induced EMF of 100 volts, we need to choose an appropriate time interval.
- Let's assume we want the self-induced EMF to reach 100 volts within 5 seconds.

8. Final Calculation:
- Substituting the chosen time interval into the formula, we get: EMF = 20/5 = 4 volts per second.
- Therefore, to generate a 100 volt self-induced EMF, the current should decrease at a rate of 4 amperes per second.

Conclusion:
By closing the switch and allowing the current to decrease at a rate of 4 amperes per second, a self-induced EMF of 100 volts can be generated in the 10 Henry inductor, as per Faraday's law of electromagnetic induction.
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A 10 Henry inductor carries a Steady Current of 2 ampere. How can a 100 volt self induced EMF be made to appear in the inductor?
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A 10 Henry inductor carries a Steady Current of 2 ampere. How can a 100 volt self induced EMF be made to appear in the inductor? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A 10 Henry inductor carries a Steady Current of 2 ampere. How can a 100 volt self induced EMF be made to appear in the inductor? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 10 Henry inductor carries a Steady Current of 2 ampere. How can a 100 volt self induced EMF be made to appear in the inductor?.
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