The capacitive network shown in the figure is assembled with initially uncharged capacitors.Assume that all the quantities in the figure are accurate to two significant figures.The switch S in the network is kept open throughout.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

The capacitive network shown in the figure is assembled with initially uncharged capacitors.Assume that all the quantities in the figure are accurate to two significant figures.The switch S in the network is kept open throughout.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

Quizplus · Accepted Answer

The total energy stored in a capacitor is given by the formula:

E = (1/2) * C * V^2

where E is the energy stored, C is the capacitance, and V is the voltage across the capacitor.

Since the capacitors are initially uncharged, the potential difference across each capacitor will be equal to the applied voltage V = 8 V.

The equivalent capacitance of the network can be found using the formula:

1/Ceq = 1/C1 + 1/C2 + ... + 1/C7

where C1, C2, ..., C7 are the individual capacitances.

Plugging in the values, we get:

1/Ceq = 1/2 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/16
1/Ceq = 1.875
Ceq = 0.533 μF

The total energy stored in the network can now be calculated as:

E = (1/2) * Ceq * V^2 * 7

Plugging in the values, we get:

E = (1/2) * 0.533 * 8^2 * 7
E = 72 mJ

Therefore, the total energy stored in the seven capacitors is 72 mJ. The correct answer is B.

The Capacitive Network Shown in the Figure Is Assembled with Initially