Question

An initially uncharged  6.6 µC capacitor is in series with a 10 kΩ resistor as shown in the figure above.

Immediately after the switch is closed the current in the circuit is initially 10 mA. How long does it take the current in the circuit to drop to 0.5 mA? (to 2 s.f and in s)

Answer 1

Gives

Capacitance (C)=6.6$\mu F$

Resistance (R)=10$K\Omega$

Initial current $I_i=10mA$

Final current $I_f=0.5mA$

We know that

$I_f=I_i e^{-\frac{t}{RT}}\rightarrow(1)$

Put value

$0.5\times10^{-3}=10\times 10^{-3} e^{-\frac{t}{10\times10^3\times6.6\times10^{-6}}}\rightarrow(2)$

$20=e^\frac{t}{66\times10^{-3}}$

Take log both side base of e

$ln20=\frac{t}{66\times10^{-3}}$

t=0.19sec