Two inductors A and B when connected in paral | vedclass

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(D) Let the inductances of the two inductors be $L_1$ and $L_2$.When connected in parallel,the equivalent inductance $L_p$ is given by $\frac{1}{L_p}

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(D) Let the inductances of the two inductors be $L_1$ and $L_2$.When connected in parallel,the equivalent inductance $L_p$ is given by $\frac{1}{L_p} = \frac{1}{L_1} + \frac{1}{L_2}$.Given $L_p = 1.5 \ H$,so $\frac{1}{1.5} = \frac{1}{L_1} + \frac{1}{L_2} \Rightarrow \frac{2}{3} = \frac{L_1 + L_2}{L_1 L_2} \quad (1)$.When connected in series,the equivalent inductance $L_s$ is $L_s = L_1 + L_2$.Given $L_s = 8 \ H$,so $L_1 + L_2 = 8 \quad (2)$.Substituting $(2)$ into $(1)$: $\frac{2}{3} = \frac{8}{L_1 L_2} \Rightarrow L_1 L_2 = 12 \quad (3)$.We have $L_1 + L_2 = 8$ and $L_1 L_2 = 12$. These are roots of the quadratic equation $x^2 - (L_1 + L_2)x + L_1 L_2 = 0$,which is $x^2 - 8x + 12 = 0$.Solving the quadratic: $(x - 6)(x - 2) = 0$,so $L_1 = 6 \ H$ and $L_2 = 2 \ H$.The difference in inductances is $|L_1 - L_2| = |6 - 2| = 4 \ H$.

Two inductors $A$ and $B$ when connected in parallel are equivalent to a single inductor of inductance $1.5 \ H$,and when connected in series are equivalent to a single inductor of inductance $8 \ H$. Find the difference in the inductances of $A$ and $B$. (in $H$)

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