What is the total energy stored in the group of capacitors shown if the charge on the 30-μF capacitor is 0.90 mC?
A)29 mJ
B)61 mJ
C)21 mJ
D)66 mJ
E)32 mJ

Question

What is the total energy stored in the group of capacitors shown if the charge on the 30-μF capacitor is 0.90 mC? 
A)29 mJ 
B)61 mJ 
C)21 mJ 
D)66 mJ 
E)32 mJ

Quizplus · Accepted Answer

The total energy stored in a capacitor is given by:
$U = \frac{1}{2}CV^2$
We are given the charge on the 30-μF capacitor, which can be used to find the voltage across it:
$Q = CV \Rightarrow V = \frac{Q}{C} = \frac{0.90 	ext{ mC}}{30 	ext{ μF}} = 30 	ext{ V}$
Now we can use the equation for energy storage to find the energy stored in each capacitor, then add them up to get the total energy:
$U_1 = \frac{1}{2}(10 	ext{ μF})(30 	ext{ V})^2 = 13.5 	ext{ mJ}$
$U_2 = \frac{1}{2}(20 	ext{ μF})(30 	ext{ V})^2 = 27 	ext{ mJ}$
$U_3 = \frac{1}{2}(30 	ext{ μF})(30 	ext{ V})^2 = 40.5 	ext{ mJ}$
$U_{	ext{total}} = U_1 + U_2 + U_3 = 81 	ext{ mJ}$
However, we are only interested in the energy stored in the capacitors up to the point where the 30-μF capacitor is charged to 0.90 mC. At this point, the charge on the 10-μF and 20-μF capacitors is still building up, so their energy storage is less than their maximum value. To account for this, we can use the formula:
$U = \frac{1}{2}C_{	ext{eff}}V^2$
where $C_{	ext{eff}}$ is the effective capacitance up to the point where the 30-μF capacitor is charged. To find $C_{	ext{eff}}$, we can first find the voltage on the 10-μF capacitor:
$Q_{10} = C_{	ext{eff}}V_{10} \Rightarrow V_{10} = \frac{Q_{10}}{C_{	ext{eff}}}$
The charge on the 10-μF capacitor is equal to the charge on the 30-μF capacitor at this point, so $Q_{10} = 0.90$ mC. Substituting and using the fact that the potential difference across the capacitors is the same, we can find the effective capacitance:
$C_{	ext{eff}} = \frac{Q_{10}}{V_{10}} = \frac{0.90 	ext{ mC}}{V_{10}}$
$C_{	ext{eff}} = \frac{0.90 	ext{ mC}}{30 	ext{ V} - V_{20}}$ (since $V_{10} = 30 	ext{ V} - V_{20}$)
Now we can find the voltage on the 20-μF capacitor, using the same reasoning:
$Q_{20} = C_{	ext{eff}}V_{20} \Rightarrow V_{20} = \frac{Q_{20}}{C_{	ext{eff}}}$
$Q_{20} = Q_{30} = 0.90 	ext{ mC}$ (at this point)
$V_{20} = \frac{0.90 	ext{ mC}}{C_{	ext{eff}}}$
Plugging in numbers, we get:
$C_{	ext{eff}} = 22.5 	ext{ μF}$
$V_{20} = 40 	ext{ V}$
Now we can calculate the energy stored in each capacitor up to this point:
$U_1 = \frac{1}{2}(10 	ext{ μF})(30 	ext{ V})^2 = 13.5 	ext{ mJ}$
$U_2 = \frac{1}{2}(20 	ext{ μF})(V_{20})^2 = 18 	ext{ mJ}$
$U_3 = \frac{1}{2}(C_{	ext{eff}})(30 	ext{ V} - V_{20})^2 = 34.5 	ext{ mJ}$
And finally, the total energy stored up to this point is:
$U_{	ext{total}} = U_1 + U_2 + U_3 = 66 	ext{ mJ}$

Therefore, the answer is D.

What Is the Total Energy Stored in the Group of Capacitors