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The e.m.f induced is 2 V when current is changing at the rate of 0.5 A/s. The self-inductance of the coil is:
- a)4
- b)2
- c)1
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
The e.m.f induced is 2 V when current is changing at the rate of 0.5 A...
Given that, emf (E) = 2 V
Rate of change of current = 0.5 A/s
We know that, induced emf
⇒ 2 = L(0.5) ⇒ L = 4H
Rate of change of current = 0.5 A/s
We know that, induced emf
⇒ 2 = L(0.5) ⇒ L = 4H
Most Upvoted Answer
The e.m.f induced is 2 V when current is changing at the rate of 0.5 A...
The self-inductance of a coil is a measure of its ability to induce an electromotive force (emf) in itself when the current passing through it changes. It is denoted by the symbol L and is measured in henries (H).
Given:
emf induced (e) = 2 V
rate of change of current (di/dt) = 0.5 A/s
The formula to calculate the emf induced in a coil is given by:
e = -L * (di/dt)
Here, the negative sign indicates that the induced emf opposes the change in current.
To find the self-inductance of the coil, we rearrange the formula as:
L = -e / (di/dt)
Substituting the given values, we have:
L = -2 V / (0.5 A/s)
Simplifying the expression, we get:
L = -4 H
Since self-inductance cannot be negative, we take the magnitude of the value:
L = 4 H
Therefore, the self-inductance of the coil is 4 henries (H), which corresponds to option A.
Summary:
- emf induced (e) = 2 V
- rate of change of current (di/dt) = 0.5 A/s
- The formula to calculate self-inductance is L = -e / (di/dt)
- Substituting the given values, we find L = -4 H
- Taking the magnitude of the value, the self-inductance of the coil is 4 H, which corresponds to option A.
Given:
emf induced (e) = 2 V
rate of change of current (di/dt) = 0.5 A/s
The formula to calculate the emf induced in a coil is given by:
e = -L * (di/dt)
Here, the negative sign indicates that the induced emf opposes the change in current.
To find the self-inductance of the coil, we rearrange the formula as:
L = -e / (di/dt)
Substituting the given values, we have:
L = -2 V / (0.5 A/s)
Simplifying the expression, we get:
L = -4 H
Since self-inductance cannot be negative, we take the magnitude of the value:
L = 4 H
Therefore, the self-inductance of the coil is 4 henries (H), which corresponds to option A.
Summary:
- emf induced (e) = 2 V
- rate of change of current (di/dt) = 0.5 A/s
- The formula to calculate self-inductance is L = -e / (di/dt)
- Substituting the given values, we find L = -4 H
- Taking the magnitude of the value, the self-inductance of the coil is 4 H, which corresponds to option A.
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