A circuit shown consists of two inductors of inductances $L_{1}$ and $L_{2},$ a capacitor of capacitance C, a battery of electromotive force E and internal resistance r and a switch. Initially the switch was in position 1 for a long time. Find the maximum current (in A) in the inductor $L_{2}$ after the switch is thrown to position 2?(Given : $E = 5 V, L_{1} = 4 m H, L_{2} = 1 m H, r = 2 Ω$ )

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Detailed Solution

t = 0

General t = t

So $L_{1} \frac{d i_{1}}{d t} = - L_{2} \frac{d i_{2}}{d t} = \frac{- q}{C}$

for maximum current in $L_{2} :$

$\Rightarrow L_{1} ({i^{'}}_{1} - \frac{E}{r}) = - L_{2} (i_{2}^{'} - 0) .... (i)$

$\frac{1}{2} L_{1} {(\frac{E}{r})}^{2} = \frac{1}{2} L_{1} i_{1}^{' 2} + \frac{1}{2} L_{2} i_{2}^{' 2} .... (i i) \Rightarrow \frac{L_{1} E^{2}}{r^{2}} = L_{2} i_{2}^{' 2} + L_{1} {(\frac{E}{r} - \frac{L_{2} i_{2}^{' 2}}{L_{1}})}^{2} \Rightarrow i_{2}^{'} = \frac{2 E L_{1}}{r (L_{1} + L_{2})}$