The number of turs of primary and secondary coils of a transformer are...
Given:
Number of turns in primary coil, N1 = 5
Number of turns in secondary coil, N2 = 10
Mutual inductance, M = 25 H
New Number of turns in primary coil, N1’ = 10
New Number of turns in secondary coil, N2’ = 5
To find:
Mutual inductance of the transformer, M’
Solution:
According to the formula for mutual inductance,
M = (N1*N2)*√(L1*L2)
where L1 and L2 are the self-inductances of primary and secondary coils respectively.
So,
M = (5*10)*√(L1*L2)
Squaring both sides,
M^2 = 500L1L2 …(i)
Now, when the number of turns in primary and secondary coils are changed to N1’ and N2’ respectively, the new self-inductances will be L1’ and L2’ respectively.
According to the formula for self-inductance,
L = (N^2 * μ * A) / l
where N is the number of turns, A is the cross-sectional area of the coil, l is the length of the coil and μ is the permeability of the medium.
Since the geometry of the coil and the medium remain the same, we can say that L1/L1’ = (N1’/N1)^2 and L2/L2’ = (N2’/N2)^2
Therefore, L1L2 = L1’L2’ * (N1N2 / N1’N2’) …(ii)
Substituting equation (ii) in equation (i),
M^2 = 500L1’L2’ * (N1N2 / N1’N2’)
Or, M’^2 = 500L1’L2’ * (N1’N2’ / N1N2)
Substituting the given values we get,
M’^2 = 500 * 10 * 5
M’^2 = 25000
Taking the square root of both sides,
M’ = 25 H
Therefore, the mutual inductance of the transformer when the number of turns in primary and secondary coils are changed to 10 and 5 respectively, is 25 H.
The number of turs of primary and secondary coils of a transformer are...
Mutual induction inductance has a factor of multiplication of number of coils of both secondary and primary .
since we have been Knowing that multiplication is commutative by law ,so n1*n2=n2*n1
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.