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Question

Physics Question on Capacitors and Capacitance

The equivalent inductance of two inductors is 2.4H2.4 \,H when connected in parallel and 10H10 \,H when connected in series. What is the value of inductances of the individual inductors?

A

8H,2H8\,H, \,2\,H

B

6H,4H6\,H, \,4\,H

C

5H,5H5\,H, 5\,H

D

7H,3H7\,H, 3\,H

Answer

6H,4H6\,H, \,4\,H

Explanation

Solution

In series connection L1+L2=10H(i)L_{1}+L_{2}=10\,H\dots(i) and in parallel connection L1L2(L1+L2)=2.4H(ii)\frac{L_{1}L_{2}}{\left(L_{1}+L_{2}\right)}=2.4 \,H \ldots\left(ii\right) Substituting the value of (L1+L2)(L_{1}+ L_{2}) from (i) into (ii), we get L1L2=(2.4)(L1+L2)=2.4×10=24L_{1}L_{2}=\left(2.4\right)\left(L_{1}+L_{2}\right)=2.4\times10=24 (L1L2)2=(L1+L2)24L1L2\left(L_{1}-L_{2}\right)^{2}=\left(L_{1}+L_{2}\right)^{2}-4L_{1}L_{2} L1L2=[(10)24×24]12=2H(iii)L_{1}-L_{2}=\left[\left(10\right)^{2}-4\times24\right]^{1 2}=2H\ldots\left(iii\right) Solving (i)\left(i\right) and (iii)\left(iii\right), we get L1=6H,L2=4HL_{1}=6\,H, L_{2}=4\,H