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A circuit has inductance of 2 H. If the circuit current changes at the rate of 10 A/'sec, then self-induced emf is-
- a)5 V
- b)0.2 V
- c)20 V
- d)10 V
Correct answer is option 'C'. Can you explain this answer?
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A circuit has inductance of 2 H. If the circuit current changes at th...
= 2 X 10 = 20 V
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A circuit has inductance of 2 H. If the circuit current changes at th...
Calculation of Self-Induced EMF in a Circuit
Given:
- Inductance (L) = 2 H
- Rate of change of current (di/dt) = 10 A/sec
To find:
- Self-induced EMF (ε)
Solution:
- According to Faraday's Law of Electromagnetic Induction, the self-induced EMF (ε) in a circuit is given by:
ε = -L(di/dt)
- Substituting the given values, we get:
ε = -2 x 10 = -20 V
- The negative sign indicates that the self-induced EMF opposes the change in current.
- However, the question asks for the magnitude of the self-induced EMF, which is always positive. Therefore, we take the absolute value of ε:
|ε| = 20 V
- Hence, the correct option is (c) 20 V.
Conclusion:
The self-induced EMF in a circuit can be calculated using Faraday's Law of Electromagnetic Induction, which relates the EMF to the rate of change of current and the inductance of the circuit. In this question, we found that the self-induced EMF is 20 V when the current changes at a rate of 10 A/sec in a circuit with an inductance of 2 H.
Given:
- Inductance (L) = 2 H
- Rate of change of current (di/dt) = 10 A/sec
To find:
- Self-induced EMF (ε)
Solution:
- According to Faraday's Law of Electromagnetic Induction, the self-induced EMF (ε) in a circuit is given by:
ε = -L(di/dt)
- Substituting the given values, we get:
ε = -2 x 10 = -20 V
- The negative sign indicates that the self-induced EMF opposes the change in current.
- However, the question asks for the magnitude of the self-induced EMF, which is always positive. Therefore, we take the absolute value of ε:
|ε| = 20 V
- Hence, the correct option is (c) 20 V.
Conclusion:
The self-induced EMF in a circuit can be calculated using Faraday's Law of Electromagnetic Induction, which relates the EMF to the rate of change of current and the inductance of the circuit. In this question, we found that the self-induced EMF is 20 V when the current changes at a rate of 10 A/sec in a circuit with an inductance of 2 H.
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A circuit has inductance of 2 H. If the circuit current changes at th...
= 2 X 10 = 20 V
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A circuit has inductance of 2 H. If the circuit current changes at the rate of 10 A/'sec, then self-induced emf is-a)5 Vb)0.2 Vc)20 Vd)10 VCorrect answer is option 'C'. Can you explain this answer?
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