Find the charge on the capacitor 1 text{ s} a | vedclass

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(B) When the switch is closed for a long time $(t = \infty)$,the capacitor acts as an open circuit. The steady-state current in the circuit is $i = \f

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$25 e^{-10} \mu 	ext{C}$

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(B) When the switch is closed for a long time $(t = \infty)$,the capacitor acts as an open circuit. The steady-state current in the circuit is $i = \frac{9 	ext{ V}}{12 	ext{ k}\Omega + 15 	ext{ k}\Omega} = \frac{9}{27 	imes 10^3} 	ext{ A} = \frac{1}{3} 	imes 10^{-3} 	ext{ A}$.The voltage across the capacitor $(V_c)$ is equal to the voltage across the $15 	ext{ k}\Omega$ resistor: $V_c = i 	imes 15 	imes 10^3 = (\frac{1}{3} 	imes 10^{-3}) 	imes 15 	imes 10^3 = 5 	ext{ V}$.The initial charge on the capacitor is $q_0 = C V_c = (5 	imes 10^{-6} 	ext{ F}) 	imes 5 	ext{ V} = 25 	imes 10^{-6} 	ext{ C} = 25 \mu 	ext{C}$.When the switch is opened,the capacitor discharges through the $15 	ext{ k}\Omega$ and $5 	ext{ k}\Omega$ resistors in series. The equivalent resistance is $R_{eq} = 15 	ext{ k}\Omega + 5 	ext{ k}\Omega = 20 	ext{ k}\Omega = 20 	imes 10^3 \Omega$.The time constant is $	au = R_{eq} C = (20 	imes 10^3 \Omega) 	imes (5 	imes 10^{-6} 	ext{ F}) = 0.1 	ext{ s}$.The charge at time $t$ is $q(t) = q_0 e^{-t/	au}$.For $t = 1 	ext{ s}$,$q = 25 e^{-1/0.1} \mu 	ext{C} = 25 e^{-10} \mu 	ext{C}$.

Find the charge on the capacitor $1 \text{ s}$ after opening the switch,given that the switch was closed for a long time $(t = \infty)$.

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