CBSE-XI-Physics
38: Electromagnetic Induction
- #10The switches in figure
(a) and
(b) are closed at t = 0 and reopened after a long time at t = t0.
Figure
(a) The charge on C just after t = 0 is εC.
(b) The charge on C long after t = 0 is εC.
(c) The current in L just before t = t0 is ε/R.
(d) The current in L long after t = t0 is ε/R.digAnsr: b,dAns : (b) The charge on C long after t = 0 is εC.
(d) The current in L long after t = t0 is ε/R.
The charge on the capacitor at time ''t'' after connecting it with a battery is given by,
`` Q=C\epsilon \left[1-{e}^{-t/RC}\right]``
Just after t = 0, the charge on the capacitor will be
`` Q=C\epsilon \left[1-{e}^{0}\right]=0``
For a long after time, `` t\to \infty ``
Thus, the charge on the capacitor will be
`` Q=C\epsilon \left[1-{e}^{-\infty }\right]``
`` \Rightarrow Q=C\epsilon \left[1-0\right]=C\epsilon ``
The current in the inductor at time ''t'' after closing the switch is given by
`` I=\frac{{V}_{\,\mathrm{\,b\,}}}{R}\left(1-{e}^{-tR/L}\right)``
Just before the time t0, current through the inductor is given by
`` I=\frac{{V}_{\,\mathrm{\,b\,}}}{R}\left(1-{e}^{-{t}_{0}R/L}\right)``
It is given that the time t0 is very long.
∴ `` {t}_{0}\to \infty ``
`` I=\frac{\epsilon }{R}\left(1-{e}^{-\infty }\right)=\frac{\epsilon }{R}``
When the switch is opened, the current through the inductor after a long time will become zero.
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