Question

For the circuit shown in the figure, the switch S is initially open and the capacitor is uncharged. The switch is then closed at time t = 0. How many seconds after closing the switch will the energy stored in the capacitor be equal to 50.2 mJ?

Answer 1

Hi there!

To start, we can use the equation for the charge of a capacitor versus time.
$q(t) = C\epsilon (1 - e^{-\frac{t}{\tau}})$

$q(t)$ = Charge on capacitor vs. time

$\epsilon$ = EMF of battery (40 V)

**Note: 'Cε' is equal to the MAXIMUM CHARGE on a capacitor.

Where the time constant is given as:
$\tau = RC$
$\tau$ = Time Constant (s)
$R$ = Resistance (0.5 MΩ)
$C$ = Capacitance (90 μF)

Let's first convert the given units to the necessary 'J',  'Ω' and 'F' to make things easier:

$J = 50.2 mJ$

$50.2 mJ * \frac{0.001J}{1mJ} = 0.0502 J$

$R = 0.5 M\Omega\\\\0.5M\Ohm * \frac{1000000\Omega}{1M\Ohm} = 500000\Omega$

$F = 90 \mu F$

$90 \mu F * \frac{0.000001 F}{1\mu F} = 0.00009 F$

Now, let's calculate the time constant.

$\tau = RC = (500000)(0.00009) = 45 s$

Time to use the equation for energy stored in a capacitor:
$E_C = \frac{1}{2}\frac{Q^2}{C}$

Our 'Q' varies with time, so:
$E(t)= \frac{1}{2}\frac{( C\epsilon (1 - e^{-\frac{t}{\tau}}))^2}{C}\\\\E(t) = \frac{1}{2}\frac{( C^2\epsilon^2 (1 - e^{-\frac{t}{\tau}})^2)}{C}\\\\E(t) = \frac{1}{2}( C\epsilon^2 (1 - e^{-\frac{t}{\tau}})^2)$

Let's now plug values into our equation.

$0.0502 = \frac{1}{2}(0.00009)(40^2) (1 - e^{-\frac{t}{45}})^2$

Simplifying:
$0.697 = (1 - e^{-\frac{t}{45}})^2$

Take the square root of both sides.

$0.835 = 1 - e^{-\frac{t}{45}}\\\\ e^{-\frac{t}{45}} = 0.165$

Take the natural log of both sides.

$-\frac{t}{45} = ln(0.165) \\\\\frac{t}{45} = 1.802\\\\t = \boxed{81.08 s}$