Question

The equivalent inductance of two inductor is 2.4 H
when connected in parallel and 10 H when
connecied in series. The difference between two
inductance is (Neglecting mutual induction between
coils)

Answer 1

Dear Student,

◆ Answer -

Difference between individual inductances = 2 H

● Explanation -

Let L1 and L2 be two individual inductors.

In series combination,

Ls = L1 + L2

L1 + L2 = 10

In parallel combination,

1/Lp = 1/L1 + 1/L2

1/2.4 = (L1 + L2) / (L1.L2)

1/2.4 = 10 / L1.L2

L1.L2 = 24

As we know,

(L1-L2)^2 = (L1+L2)^2 - 4L1.L2

(L1-L2)^2 = 10^2 - 4×24

(L1-L2)^2 =100 - 96

(L1-L2)^2 = 4

L1-L2 = 2

Therefore, difference between two inductors must be 2 H.

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Answer 2

Answer:

Let the two inductors be L1 and L2.

When the inductors are connected in series.

Equivalent resistance,

Ls=L1+L2⇒10=L1+L2⇒L1=10−L2When the inductors are connecetd in parallel1Lp=1L1+1L2⇒Lp=L1L2L1+L2⇒2.4=(10−L2)L210−L2+L2⇒10L2−L22=24⇒L22−10L2+24=0⇒L22−6L2−4L2+24=0⇒L2(L2−6)−4(L2−6)=0⇒(L2−6)(L2−4)=0therefore, L2=6 or 4

The two inductors are 6 H and 4 H, therefore the difference between them = 6 - 4 = 2 H

Explanation:

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