Find the magnitude of charge stored in the entire network

[rstein66]

Problem: In the circuit shown in the figure below, the capacitors C1 and C2 have
a capacitance of 45µF. C3 and C4 have a capacitance of 90µF. The two
terminals A and B on the left have a potential difference of 25V

(Sorry - had to quickly draw it, could not save/find image)

(a) Find the magnitude of charge stored in the entire network.
(b) Find the magnitude of charges stored in each capacitor.
(c) Find the energy stored in each capacitor. Identify the capacitor with
the largest energy content.
(d) The C3 capacitor is now removed from the circuit, leaving a break
in the wire at its position. What is the voltage drop across the three
remaining capacitors?
(e) Describe how the maximal amount of energy can be stored using
these four capacitors. Draw the resulting circuit and state how much
energy is stored in your circuit using the same potential difference of
25V.

Equations/etc:
C=Q/V
Ceq series=(1/c1+1/c2...)^-1
Ceq parallel=c1+c2...
Series capacitors have equal charges and different voltages while parallel's have different charges and equal voltages.
U=(CV^2)/2 (Energy of capacitors)

Attempt:

a) Q1234=C1234(V)

Knowing that C1 and C3 are in series , C13=3e-5F
C2 and C4 in parallel, thus C24 = 1.34e-4F
C13 and C24 are in series thus C1234=2.45e-5F

Thus Q1234=C1234*V=2.45e-5F*25V=6.1e-4C

b)Since C1 and C3 are series, Q1=Q3=Q13=Q1234=C1234(V)=2.45e-5F*25V=6.1e-4C

I am not sure how to then solve Q2 and Q4 assuming I did Q1, Q3 correctly.

c) U=(CV^2)/2 , I realize that C for each is already given but am not sure what are the individual voltages.

d) I am unsure how to approach this

e) Since Ceq parallel is greater then the individual capacitance then it would make sense for it to be an all parallel circuit where the new Energy is equal to U=(CV^2)/2 where C=C1+C2+C3+C4.

Thanks.

 

[ehild]

Problem: In the circuit shown in the figure below, the capacitors C1 and C2 have
a capacitance of 45µF. C3 and C4 have a capacitance of 90µF. The two
terminals A and B on the left have a potential difference of 25V

(Sorry - had to quickly draw it, could not save/find image)

(a) Find the magnitude of charge stored in the entire network.
(b) Find the magnitude of charges stored in each capacitor.
(c) Find the energy stored in each capacitor. Identify the capacitor with
the largest energy content.
(d) The C3 capacitor is now removed from the circuit, leaving a break
in the wire at its position. What is the voltage drop across the three
remaining capacitors?
(e) Describe how the maximal amount of energy can be stored using
these four capacitors. Draw the resulting circuit and state how much
energy is stored in your circuit using the same potential difference of
25V.

Equations/etc:
C=Q/V
Ceq series=(1/c1+1/c2...)^-1
Ceq parallel=c1+c2...
Series capacitors have equal charges and different voltages while parallel's have different charges and equal voltages.
U=(CV^2)/2 (Energy of capacitors)

Attempt:

a) Q1234=C1234(V)

Knowing that C1 and C3 are in series , C13=3e-5F.

C1 and C3 are not in series.

 

[vela]

Problem: In the circuit shown in the figure below, the capacitors C1 and C2 have
a capacitance of 45µF. C3 and C4 have a capacitance of 90µF. The two
terminals A and B on the left have a potential difference of 25V

(Sorry - had to quickly draw it, could not save/find image)

(a) Find the magnitude of charge stored in the entire network.
(b) Find the magnitude of charges stored in each capacitor.
(c) Find the energy stored in each capacitor. Identify the capacitor with
the largest energy content.
(d) The C3 capacitor is now removed from the circuit, leaving a break
in the wire at its position. What is the voltage drop across the three
remaining capacitors?
(e) Describe how the maximal amount of energy can be stored using
these four capacitors. Draw the resulting circuit and state how much
energy is stored in your circuit using the same potential difference of
25V.

Equations/etc:
C=Q/V
Ceq series=(1/c1+1/c2...)^-1
Ceq parallel=c1+c2...
Series capacitors have equal charges and different voltages while parallel's have different charges and equal voltages.
U=(CV^2)/2 (Energy of capacitors)

Attempt:

a) Q1234=C1234(V)

Knowing that C1 and C3 are in series , C13=3e-5F

C1 and C3 are not in series. When two elements are in series, the current through one element has to go through the other. The current through C1, however, can go through C2 or C3. In contrast, C3 and C4 are in series. There's nowhere else for the current through C3 to go other than through C4.

C2 and C4 in parallel, thus C24 = 1.34e-4F
C13 and C24 are in series thus C1234=2.45e-5F

Thus Q1234=C1234*V=2.45e-5F*25V=6.1e-4C

b)Since C1 and C3 are series, Q1=Q3=Q13=Q1234=C1234(V)=2.45e-5F*25V=6.1e-4C

I am not sure how to then solve Q2 and Q4 assuming I did Q1, Q3 correctly.

c) U=(CV^2)/2 , I realize that C for each is already given but am not sure what are the individual voltages.

d) I am unsure how to approach this

e) Since Ceq parallel is greater then the individual capacitance then it would make sense for it to be an all parallel circuit where the new Energy is equal to U=(CV^2)/2 where C=C1+C2+C3+C4.

Thanks.

 

[rstein66]

Thanks Vela, that made a lot of sense. I'll give it another go.

 

[HalfLostAndSearching]

Your approach to part (a) is correct. For part (b), you can use the fact that in a parallel circuit, the voltage across each capacitor is the same, so you can use the same voltage of 25V to calculate the charge stored in each capacitor. Therefore, Q2=C2*V=45e-6F*25V=1.125e-3C and Q4=C4*V=90e-6F*25V=2.25e-3C.

For part (c), you can use the same approach as in part (b) to find the individual voltages across each capacitor. Once you have the voltages, you can use the equation U=(CV^2)/2 to calculate the energy stored in each capacitor. The capacitor with the largest energy content will be the one with the highest capacitance, which in this case is C4.

For part (d), you can use the equation Ceq parallel=C1+C2+C3 to find the equivalent capacitance of the remaining capacitors (C1, C2, and C4). Then, you can use the equation C=Q/V to find the voltage across the equivalent capacitance, which will be the voltage drop across the remaining capacitors.

For part (e), you are correct that the maximum amount of energy can be stored by using the capacitors in a parallel configuration. Your approach to calculating the energy is also correct. You can draw the resulting circuit with all four capacitors in parallel and calculate the energy using the equation U=(CV^2)/2.