Respuesta :
Answer:
Check attachment for circuit diagram
And better understanding
Explanation:
Let find the equivalent inductance.
The 6mH in parallel to 6mH
Parallel connection will give
1 / Leq = 1 / L1 + 1 / L2
L1 = L2 = L = 6mH
1 / Leq = 1 / 6 + 1 / 6
1 / Leq = 2 / 6
1 / Leq = 1 / 3
Then, take reciprocal
Leq = 3 mH
Now the combInation of this parallel connection is connect in series with an 8mH inductor
Series connection is given as
Leq = L1 + L2
So, the equivalent of the parallel connection is 3mH and this will be connect in series with 8mH
Then, final inductance is
Leq = 3 + 8
Leq = 11 mH
Therefore, the repairman need to replace the element with one inductor of inductance 11 mH.

This question involves the concepts of inductance, series combination, and parallel combination.
The inductance to be used is found to be "11 mH".
RESULTANT INDUCTANCE
First, we will find out the resultant inductance of the two inductors connected in a parallel combination. Since it is given that inductors behave like resistors in combination. Therefore, resultant of two inductances connected in parallel combination will be:
[tex]\frac{1}{L}=\frac{1}{L_1}+\frac{1}{L_2}[/tex]
where,
- L = resultant inductance of the two inductors in parallel combination = ?
- L₁ = L₂ = inductance of the two inductors = 6 mH = 6 x 10⁻³ H
Therefore,
[tex]\frac{1}{L}=\frac{1}{6\ x\ 10^{-3}\ H}+\frac{1}{6\ x\ 10^{-3}\ H}\\\\[/tex]
L = 3 x 10⁻³ H = 3 mH
Now, we will calculate the total inductance of this combination in series combination with the third inductor as follows:
[tex]L_T=L+L_3=3\ mH+8\ mH\\L_T=11\ mH[/tex]
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