question_answer

        In the circuit shown,                     the switch \[{{S}_{1}}\] is closed at time t = 0 and the  switch\[{{S}_{1}}\]is opened and \[{{S}_{2}}\] is closed. The behavior of the current I as a function of time ?t? is given by [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

A)                

B)

C)                

D)

Correct Answer: B

Solution :

Initially, when the switch\[{{S}_{1}}\]is closed the current increases \[i={{i}_{0}}(1-{{e}^{-t/\tau }})\]and when the switch \[{{S}_{2}}\]is closed the current starts decreasing exponentially i.e., according to \[{{i}_{0}}{{e}^{-t/\tau }}\]. So, the behaviour of the current can be depicted from the following graph: *As per official answer key. Value \[{{t}_{0}}\]is not specified hence a, b and c may be correct option.