AIEEE Solved Paper-2012
A) 150 sec and 200 sec
B) 0 and 50 sec
C) 50 sec and 100 sec
D) 100 sec and 150 sec
Correct Answer: D
Solution :
\[Q=c{{\varepsilon }_{0}}{{e}^{-t/cR}}\] \[4\varepsilon =4{{\varepsilon }_{0}}{{\varepsilon }^{-t/\tau }}\] \[\varepsilon ={{\varepsilon }_{0}}\,{{\varepsilon }^{-t/\tau }}\] When \[t=0\Rightarrow {{\varepsilon }_{0}}=25\] \[\varepsilon ={{\varepsilon }_{0}}=25\] when \[t=200\Rightarrow \] \[\varepsilon =5\] In \[5=25\,{{e}^{-\frac{200}{\tau }}}\] \[5=\frac{200}{\ell n5}=\frac{200}{\ell n10-\ell n2}\] \[=\frac{200}{\ell n10-0.693}\] Alternative : Time constant is the time in which 63% discharging is completed. So remaining charge \[=0.37\times 25=9.25\,V\] Which time in \[100<t<150\] sec.
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