A circuit contains two inductors of 6.0 mH inductance in parallel placed in series with an inductor of 8.0 mH inductance. After one of the 6.0 mH inductors burns out, the repairman wants to replace all three inductors with one inductor of equivalent inductance. Assuming inductors combine in series and parallel the same way resistors do, what inductance should he use

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.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west