The figure shows an experimental plot discharging of a capacitor in an RC circuit. The time constant \[\tau \] of this circuit lies between: [AIEEE 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 :
[d] \[Q=c{{\varepsilon }_{0}}{{e}^{-t/cR}}\] |
\[4\varepsilon =4{{\varepsilon }_{0}}{{\varepsilon }^{-t/\tau }}\] |
\[\varepsilon ={{\varepsilon }_{0}}\,{{\varepsilon }^{-t/\tau }}\] |
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. |
You need to login to perform this action.
You will be redirected in
3 sec