The switch $S$ shown in figure is kept closed for a long time and then opened at $t = 0$, then the current in the middle $20\, \Omega$ resistor at $t = 0.25\, ms$ is :-
Diffcult
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Switch is kept closed for a long time.
Current through $20\, \Omega$ resistor $i=\frac{12}{40}$
Charge on the capacitor at steady state.
$q_{0}=25 \times \frac{12}{40} \times 20=150\, \mu C$
at $t=0,$ $S$ witch is opened, $i = {i_0}{e^{ - t/\tau }}$
$\tau=\mathrm{RC}=20 \times 25=500\, \mu \mathrm{S}$
Current $\quad {\rm{i}} = \frac{{{{\rm{q}}_0}}}{\tau }{{\rm{e}}^{\frac{{0.25 + {{10}^{ - 3}}}}{{500 \times {{10}^{ - 6}}}}}}$
$\mathrm{i}=\frac{150}{500} \mathrm{e}^{-1 / 2}=0.189 \mathrm{\,A}$
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