Three inductances are connected as shown in t | vedclass

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(A) The circuit consists of two inductors of $0.7 \ H$ each connected in parallel,which are then connected in series with an inductor of $0.85 \ H$.Fi

Vedclass · Accepted Answer

$1.20$

Vedclass · Answer

(A) The circuit consists of two inductors of $0.7 \ H$ each connected in parallel,which are then connected in series with an inductor of $0.85 \ H$.First,calculate the equivalent inductance $(L_p)$ of the two parallel inductors:$\frac{1}{L_p} = \frac{1}{0.7} + \frac{1}{0.7} = \frac{2}{0.7}$$L_p = \frac{0.7}{2} = 0.35 \ H$Now,this equivalent inductance is in series with the $0.85 \ H$ inductor.The total equivalent inductance $(L_{eq})$ is:$L_{eq} = L_p + 0.85 \ H$$L_{eq} = 0.35 \ H + 0.85 \ H = 1.20 \ H$Therefore,the equivalent inductance between $A$ and $B$ is $1.20 \ H$.

Three inductances are connected as shown in the figure. The equivalent inductance between $A$ and $B$ is (in $H$)

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