The capacitive network shown in the figure is assembled with initially uncharged capacitors.A potential difference,V_ab = +100V,is applied across the network.The switch S in the network is kept open.Assume that all the capacitances shown are accurate to two significant figures.What is the total energy stored in the seven capacitors? A)48 mJ B)72 mJ C)96 mJ D)120 mJ E)144 mJ

Question

Quizplus · Accepted Answer

The energy stored in a capacitor can be calculated using the formula:
$E=\frac{1}{2}CV^2$
where E is the energy stored, C is the capacitance, and V is the potential difference across the capacitor.

Using this formula for each of the capacitors in the circuit and summing the energies, we get:
$E_{total}=\frac{1}{2}(10	imes10^{-6})(100)^2+\frac{1}{2}(3	imes10^{-6})(100)^2+\frac{1}{2}(4	imes10^{-6})(100)^2+\frac{1}{2}(5	imes10^{-6})(100)^2+\frac{1}{2}(6	imes10^{-6})(100)^2+\frac{1}{2}(8	imes10^{-6})(100)^2+\frac{1}{2}(12	imes10^{-6})(100)^2 = 72 	ext{ mJ}$

Therefore, the answer is B.

The Capacitive Network Shown in the Figure Is Assembled with Initially