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Find the current as a function of time in case of (a) charging and (b) discharging of a capacitor in a simple RC circuit |
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Answer» Solution :During charging,`q = q_(0 )(1 - E^ (-t//CR)` `i=(dq)/(dt)=-q_(0)((-1)/(CR))e^(-I//CR)=q_(0)/(CR)e^((t)/(CR)` where `q-(o) = CV_(O). V_(0)` being the EMF of the cell (b) During discharging. `q = q_(o) e^(t// CR)` `i=(dq)/(dt)=-(q_(0))/(CR)e^(-t//CR)` . where `q_(o)`=the initial charge on the CAPACITOR: |
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