Question

Calculate the equivalent inductance of the following inductive circuit.



A. 36.8mH
B. 19.6mH
C. 15mH
D. 150mH

Answer 1

Hint: We will find out the equivalent inductances one by one of each circuit by applying the formula for inductances in series and inductances in parallel. Refer to the solution below for further doubts.

Formula used: ${L_{total}} = {L_1} + {L_2} + .....{L_n}$ and ${L_{total}} = \dfrac{1}{{\dfrac{1}{{{L_1}}} + \dfrac{1}{{{L_2}}} + ....\dfrac{1}{{{L_n}}}}}$.

Complete Step-by-Step solution:
For series inductances- ${L_{total}} = {L_1} + {L_2} + .....{L_n}$
For parallel inductances- ${L_{total}} = \dfrac{1}{{\dfrac{1}{{{L_1}}} + \dfrac{1}{{{L_2}}} + ....\dfrac{1}{{{L_n}}}}}$
As we can see in the given circuit that L7 and L6 are in series. So, its equivalent L will be-
$
   \Rightarrow L = {L_6} + {L_7} \\
    \\
   \Rightarrow L = 40 + 100 \\
    \\
   \Rightarrow L = 140mH \\
$
Now, L and L5 are in parallel. So, its equivalent L’ will be-
$
   \Rightarrow L' = \dfrac{1}{{\dfrac{1}{L} + \dfrac{1}{{{L_5}}}}} \\
    \\
   \Rightarrow L' = \dfrac{1}{{\dfrac{{{L_5} + L}}{{L \times {L_5}}}}} \\
    \\
   \Rightarrow L' = \dfrac{{L \times {L_5}}}{{L + {L_5}}} \\
    \\
   \Rightarrow L' = \dfrac{{50 \times 140}}{{50 + 140}} \\
    \\
   \Rightarrow L' = \dfrac{{700}}{{19}} \\
    \\
   \Rightarrow L' = 36.84mH \\
$
L’ and L4 are in series. So, its equivalent L’’ will be-
$
   \Rightarrow L'' = L' + {L_4} \\
    \\
   \Rightarrow L'' = \dfrac{{700}}{{19}} + 20 \\
    \\
   \Rightarrow L'' = \dfrac{{1080}}{{19}} \\
    \\
   \Rightarrow L'' = 56.84mH \\
$
As we can see that L’’ and L3 are in parallel. So, its equivalent L’’’ will be-
$
   \Rightarrow L''' = \dfrac{{L'' \times {L_3}}}{{L'' + {L_3}}} \\
    \\
   \Rightarrow L''' = \dfrac{{56.84 \times 30}}{{56.84 + 30}} \\
    \\
   \Rightarrow L''' = 19.6mH \\
$
Now, L’’’ and L2 are in series. So, its equivalent L’’’’ will be-
$
   \Rightarrow L'''' = L''' + L2 \\
    \\
   \Rightarrow L'''' = 19.6 + 40 \\
    \\
   \Rightarrow L'''' = 59.6mH \\
$
Since L’’’’ and L1 are in parallel, the total equivalent ${L_{EQ}}$ will be-
$
   \Rightarrow {L_{EQ}} = \dfrac{{L'''' \times {L_1}}}{{L'''' + {L_1}}} \\
    \\
   \Rightarrow {L_{EQ}} = \dfrac{{59.6 \times 20}}{{59.6 + 20}} \\
    \\
   \Rightarrow {L_{EQ}} = 15mH \\
$

Note: As told, the induced emf over a coil is directly proportional to the pace of change of current through it. The constant proportionality in this relation is called inductance. Remember the formula for circuits in parallel and circuits in series.