Question

If c1=c2=4.00μf and c4=8.00μf, what must the capacitance c3 be if the network is to store 2.70×10−3 j of electrical energy?

Answer 1

Missing detail in the text: total voltage of the circuit $\Delta V = 46.0 V$
Missing figure: https://www.physicsforums.com/attachments/prob-24-68-jpg.190851/

Solution:

1) The energy stored in a circuit of capacitors is given by
$U= \frac{1}{2} C_{eq} (\Delta V)^2$
where $C_{eq}$ is the equivalent capacitance of the circuit. We can find the value for $C_{eq}$ by using $\Delta V=46.0 V$ and the energy of the system, $U=2.7\cdot 10^{-3} J$
$C_{eq}= \frac{2U}{(\Delta V)^2}=2.55\cdot 10^{-6} F=2.55\mu F$

2) Then, let's calculate the equivalente capacitance of C1 and C2. The two capacitors are in series, so their equivalente capacitance is given by
$\frac{1}{C_{12}}= \frac{1}{C_1}+ \frac{1}{C_2}= \frac{1}{4 \mu F} + \frac{1}{4 \mu F}$
from which we find $C_{12}=2 \mu F$

3) Then let's find $C_{123}$, the equivalent capacitance of $C_{12}$ and C3. $C_{123}$ is in series with C4, therefore we can write
$\frac{1}{C_{eq}}= \frac{1}{C_{123}}+ \frac{1}{C_4}$
Since we already know $C_4=8 \mu F$ and $C_{eq}=2.55 \mu F$, we find
$C_{123}=3.70 \mu F$

4) Finally, we can find $C_{3}$, because it is in parallel with $C_{12}$, and the equivalent capacitance of the two must be equal to $C_{123}$:
$C_{123}=C_{12}+C_3$
So, using $C_{123}=3.70 \mu F$ and $C_{12}=2 \mu F$, we find
$C_3=1.70 \mu F$