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Two coils of wire are placed close together. Initially, a current of 3.04 A exists in one of the coils, but there is no current in the other. The current is then switched off in a time of 1.75 x 10-2 s. During this time, the average emf induced in the other coil is 4.29 V. What is the mutual inductance of the two-coil system

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Khoso123

Answer:

[tex]M=0.0247H[/tex]

Explanation:

Given data

[tex]V_{voltage}=4.29V\\I_{current}=3.04A\\t_{time}=1.75*10^{-2}s[/tex]

To find

Mutual inductance of the two-coil system

Solution

The mutual inductance given as:

M= (-VΔt)/ΔI

Substitute the given values

So

[tex]M=-\frac{4.29V*1.75*10^{-2}s}{(0-3.04A)}\\ M=0.0247H[/tex]

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