1.

In the arrangement as shown in the figure, S_(1) and S_(3) are already closed for a long time. Now at t=0 the S_(2) is closed and S_(1), S_(3) are opened [ epsilon=1 volt, L=1 henry, R=1Omega & C=1 Faraday ]

Answer»

After `S_(2)`s is CLOSED the maximum current through the INDICTOR is `2A`
After `S_(2)` is closed the maximum current through the indictor is `sqrt(2)A`
Maximum charge on the capacitor is `2+sqrt(2)` coulomb
Maximum charge on the capacitor is `sqrt(2)` coulomb

Solution :`Q_(i)^(2)+1/2Li_(i)^(2)=1/2 Li_("MAX")^(2)`
`I_("max")=sqrt(2)` amp
Also, `(Q_(i)^(2))/(2c)+1/2Li_(i)^(2)=(Q_("max")^(2))/(2c)`


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