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A varying current in a coil changes from 10A tozero in 0.5 sec. If the average e.m.f induced inthe coil is 220V, the self-inductance of the coil is [1995]
- a)5 H
- b)6 H
- c)11 H
- d)12 H
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
A varying current in a coil changes from 10A tozero in 0.5 sec. If the...
Initial current (I1) = 10 A; Final current
(I2)= 0; Time (t) = 0.5 sec and induced e.m.f.
(ε) = 220 V.
Induced e.m.f. (ε)
(I2)= 0; Time (t) = 0.5 sec and induced e.m.f.
(ε) = 220 V.
Induced e.m.f. (ε)
[where L = Self inductance of coil]
Most Upvoted Answer
A varying current in a coil changes from 10A tozero in 0.5 sec. If the...
To find the self-inductance of the coil, we can use the formula:
emf = -L * (change in current / change in time)
Given:
emf = 220V
change in current = 10A - 0A = 10A
change in time = 0.5s
Let's substitute the given values into the formula and solve for L:
220V = -L * (10A / 0.5s)
Simplifying the equation:
220V = -L * 20A/s
Dividing both sides by -20A/s:
-11V/s = L
Therefore, the self-inductance of the coil is 11H (Henry).
Explanation:
- The average emf induced in a coil is given by the formula: emf = -L * (change in current / change in time).
- In this case, the change in current is from 10A to 0A, and the change in time is 0.5s.
- Substituting these values into the formula, we get the equation: 220V = -L * (10A / 0.5s).
- Simplifying the equation, we find that the self-inductance of the coil is -11V/s.
- The negative sign indicates that the induced emf opposes the change in current.
- The unit of self-inductance is Henry (H).
Therefore, the self-inductance of the coil is 11H (Henry).
emf = -L * (change in current / change in time)
Given:
emf = 220V
change in current = 10A - 0A = 10A
change in time = 0.5s
Let's substitute the given values into the formula and solve for L:
220V = -L * (10A / 0.5s)
Simplifying the equation:
220V = -L * 20A/s
Dividing both sides by -20A/s:
-11V/s = L
Therefore, the self-inductance of the coil is 11H (Henry).
Explanation:
- The average emf induced in a coil is given by the formula: emf = -L * (change in current / change in time).
- In this case, the change in current is from 10A to 0A, and the change in time is 0.5s.
- Substituting these values into the formula, we get the equation: 220V = -L * (10A / 0.5s).
- Simplifying the equation, we find that the self-inductance of the coil is -11V/s.
- The negative sign indicates that the induced emf opposes the change in current.
- The unit of self-inductance is Henry (H).
Therefore, the self-inductance of the coil is 11H (Henry).
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A varying current in a coil changes from 10A tozero in 0.5 sec. If the average e.m.f induced inthe coil is 220V, the self-inductance of the coil is [1995]a)5 Hb)6 Hc)11 Hd)12 HCorrect answer is option 'C'. Can you explain this answer?
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A varying current in a coil changes from 10A tozero in 0.5 sec. If the average e.m.f induced inthe coil is 220V, the self-inductance of the coil is [1995]a)5 Hb)6 Hc)11 Hd)12 HCorrect answer is option 'C'. Can you explain this answer? 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 varying current in a coil changes from 10A tozero in 0.5 sec. If the average e.m.f induced inthe coil is 220V, the self-inductance of the coil is [1995]a)5 Hb)6 Hc)11 Hd)12 HCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A varying current in a coil changes from 10A tozero in 0.5 sec. If the average e.m.f induced inthe coil is 220V, the self-inductance of the coil is [1995]a)5 Hb)6 Hc)11 Hd)12 HCorrect answer is option 'C'. Can you explain this answer?.
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