Question

At t = 0 the switch S is closed with the capacitor uncharged. If C = 50 F,  = 20 V, and R = 4.0 k, what is the charge on the capacitor when I = 2.0 mA?

Answer 1

Hello!

Recall the equation for the current in an RC circuit as a capacitor is charging:
$i(t) = \frac{\epsilon}{R}e^{-\frac{t}{RC}}$

i(t) = Current as a function of time (2.0 mA)

ε = Emf of voltage source (20V)
R = Resistance of resistor (4kΩ)

C = Capacitance of capacitor (50 μF)

We can solve for the time at which the current in the circuit decreases to 2.0 mA by plugging in the given values and solving for 't'.

$0.002 = \frac{20}{4000} e^{-\frac{t}{(4000)(0.00005)}}\\\\0.002 = 0.005 e^{-\frac{t}{(4000)(0.00005)}$

Solve:
$.4 = e^{-\frac{t}{0.2}}\\\\ln(.4) = -\frac{t}{0.2}\\\\t = -0.2(ln(.4)) = 0.1833 s$

Now, we can solve for the charge on the capacitor using the equation for charge of a charging capacitor:
$q(t) = C\epsilon(1 - e^{\frac{-t}{RC}})$

Plug in the given values and the time we solved for.

$q(t) = (0.00005)(20)(1 - e^{\frac{-0.1833}{(4000)(0.00005)}}) \\\\q(t) = 0.0006 = \boxed{600 \mu C}$