Capacitors - 4

The stored energy of a capacitor is given by the formula

$\text{Stored Energy} = \frac{1}{2} C \times V^{2}$

At steady state, the potential difference across the capacitor is $6 \text{ V}$ . The capacitance of the capacitor is $5 \text{ µF} = 5 \times 10^{-6} \text{ F}$ . Thus the energy stored in the capacitor is

$\text{Stored Energy} = \frac{ 1}{2} \times (5 \times 10^{-6}) \times (6)^{2} = 9 \times 10^{-5} \text{ J} ~~~~~_\square$

The voltage across the capacitor in an RC-circuit is given by the function:

$V_{C}(t) = V_{s} (1 - e^{-\frac{t}{RC}})$

where $V_{s} = 6V$ and the time constant equals $RC = (4 \Omega)(5 \times 10^{-6}F) = 2 \times 10^{-5} s$ . At steady-state (i.e. $t \rightarrow \infty$ ), $V_{C}(t) \rightarrow 6V$ . Thus the energy stored computes to $E = \frac{CV^{2}}{2} = \frac{(5 \times10^{-6} F)(6V)^{2}}{2} = \boxed{9 \times 10^{-5} J}$ .