1.

Three identical metal plates of area S are at equal distances b as shown. Initially metal plate A is uncharged, while metal plates B and C have respective charges +Q_(0) and -Q_(0) initially as shown. Metal plates A and C are connected by switch K through a resistor of resistance R. the key K is closed at time t = 0 Then the magnitude of current in amperes through the resistor at any later time is :

Answer»

`(Q_(0)b)/(RS epsilon_(0))e^((-bt)/(RS epsilon_(0)))`
`(Q_(0)b)/(RS epsilon_(0))e^((-2bt)/(RS epsilon_(0)))`
`(Q_(0)b)/(2RS epsilon_(0))e^((-2bt)/(RS epsilon_(0)))`
`(Q_(0)b)/(2RS epsilon_(0))e^((-bt)/(RS epsilon_(0)))`

Solution :At any TIME t, the change on right CAPACITOR be Q. APPLYING Kirchoff's law `(Q_0-q)/C=iR+q/C therefore(Q_0-2q)/(2CR)=(dq)/(dt)`
integrating and evaluating the constant we get
Hence `q=Q_0/2(1-e^(-(2t)/(RC)))`
or `i=(dq)/(dt)=Q_0/(RC)e^(-(2t)/(RC))`


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