4.2.1  Parallel Combinations

Let's start by looking at the parallel combination. In this case, we have capacitors which have the same potential difference across their plates.

Figure 4.10: Two capacitors in parallel and a circuit with an equivalent capacitance.

For such combinations, we can simplify matters a bit by finding an equivalent capacitor whose characteristics make exactly the same demands from the EMF source in charging up the capacitors (presuming they had the same initial charge before the EMF source was connected). For this to be the case, the EMF source will have to push the same amount of charge onto the equivalent capacitor as it does for the two original capacitors it replaces. The work done by the emf source should be the same as well. This means that the potential difference created on the equivalent capacitor should be the same as for the two capacitors it replaces. We can now ask what the capacitance of the C_eq  must be for these conditions to hold. Since the potentials across C₁, C₂, and C_eq must all be the same and the charge on the equivalent capacitor must be the same as on the two individual capacitors in the original circuit, we have

Qeq = Q1 + Q2 CeqVb = C1V1 + C2V2 = C1Vb + C2VbÞ Ceq = C1 + C2 (4.2.1.12)

So the equivalent capacitor needs to have a capacitance which is the sum of the two capacitors it replaces. This result can easily be extended to any number of capacitors so that the result of n capacitors connected in parallel is

Ceq = C1 + C2 + C3 + ... + Cn

(4.2.1.13)

Figure 4.11: Extending the parallel capacitor formula to any number of capacitors.

4.2.2 Series combinations

Figure 4.12: A series arrangement of capacitors and a circuit with equivalent capacitance.

As with the parallel combination, we want to find the equivalent capacitance that can replace the pair of capacitors here without changing anything that the EMF source must do to produce full charge on the combination. In this case, it is the charges that must be equal magnitude on the combination since we note that the positive pole of the EMF source contacts one plate of one capacitor in the pair while the negative pole contacts one plate of the other capacitor. In each case, the plate connected to the battery pole induces the equal magnitude charge on the other plate for that capacitor. Since the inner plates are electrically isolated from the world, they maintain a net charge of zero (assuming all capacitors were initially uncharged before the EMF source was connected). Thus, q₁ = q₂ = q_eq  for the combination. The potential in going from the bottom plate of C₂ to the top plate of C₁ is V₂ + V₁, hence

Vb = V1 + V2 qeq Ceq = q1 C1 + q2 C2 Þ 1 Ceq = 1 C1 + 1 C2 (4.2.2.14)

This can also be extended to any number of capacitors in parallel. For example, for n capacitors in series,

1 Ceq = 1 C1 + 1 C2 + 1 C3 + ... + 1 Cn .

(4.2.2.15)

4.2.3 Examples of Capacitor Combinations

1. Consider the combinator of 4 parallel-plate capacitors as shown in figure 4.13. A battery maintains a potential difference of 20 volts between its poles. The values of the capacitances are:
   

   C1 = 20 nF,   C2 = 5 nF,    C3 = 10 nF,   C4 = 6 nF

   (4.2.3.16)

   1. Find the equivalent capacitance of the circuit.
   2. Find the charge on and potential across each of the capacitors.
   
   Figure 4.13: A combination of 4 capacitors.

   Solution:

   The equivalent capacitance is gotten by realizing that the definition of a series combination is that the negative plate of one capacitor is connected only to the positive plate of another capacitor. In this case, the net charge on the capacitor plates (assuming they were initially uncharged) must have the same magnitude and opposite sign for both plates. By definition this is a series combination. Looking at the diagram, we realize that capacitors C₁ and C₂  are in parallel. The equivalent capacitor for this pair (call it C₁₂) is in parallel with C₄. Therefore, call this equivalent capacitance C₁₂₄. Finally, this equivalent capacitor is in series with C₃. Hence we can replace all these by one equivalent capacitor with equivalent capacitance
   

   1 C12 = 1 C1 + 1 C2 Þ C12 = C1C2 C1 + C2 = 20·5 nF2 20 + 5 nF C12 = 4 nF (4.2.3.17) C124 = C12 + C4 = 4 nF + 6 nF C124 = 10 nF (4.2.3.18) 1 Ceq = 1 C124 + 1 C3 Þ Ceq = C3C124 C3 + C124 = 10·10 nF2 10 + 10 nF Ceq = 5 nF (4.2.3.19)

   
   Figure 4.14: The equivalent circuit for figure 4.13.

   To find the charge across each capacitor, first note that the charge across C₁₂₄  and C₃ must be the same since they are in series. This charge must be the same magnitude as the charge on the equivalent capacitance for the circuit, so
   

   Q3 = Q124 = Qeq = CeqVb = (5 nF)(20 V) Q3 = Q124 = Qeq = 100 nC = 0.1 mC (4.2.3.20)

   The voltage across C₃ and across C₁₂₄ can then be derived as follows.
   

   V3 = Q3 C3 = 100 nC 10 nF = 10 V V124 = Q124 C124 = 100 nC 10 nF = 10 V (4.2.3.21)

   Notice in this case that the voltages happen to be equal, but this is strictly a consequence of the capacitor values. For any capacitor values it would be necessary for
   

   V3 + V124 = Vb

   (4.2.3.22)

   as it does in this case. So we are done with C₃. To get the voltage across C₄ note that it must be the same as the voltage across V₁₂₄. So,
   

   V4 = 10 V.

   (4.2.3.23)

   The charge on C₄ is then
   

   Q4 = C4V4 = (6 nF)(10 V) = 60 nC

   (4.2.3.24)

   The charge on capacitors C₁ and C₂  must be the same since they are in series. Their equivalent capacitance, C₁₂, must also have this magnitude of charge and the same potential difference as C₄ since it is in parallel with it. Hence,
   

   Q1 = Q2 = Q12 = C12V124 = (4 nF)(10 V) = 40 nC

   (4.2.3.25)

   Finally, the voltages across C₁ and C₂ are
   

   V1 = Q1 C1 = 40 nC 20 nF = 2 V V2 = Q2 C2 = 40 nC 5 nF = 8 V (4.2.3.26)

   ---

2. In figure 4.15, the switch S is first thrown to the left. The capacitor C₁  acquires a potential difference V_b. Capacitors C₂ and C₃ are initially uncharged. The switch is then thrown to the right. Find the magnitudes of the final charges on each capacitor after the switch is thrown to the right.
   
   Figure 4.15: Capacitors in a switch-operated circuit.

   Solution:

   When the switch is closed to the left, capacitor C₁ charges up. The voltage of this capacitor becomes V_b  and the charge on this capacitor becomes
   

   Q1 = C1Vb

   (4.2.3.27)

   When the switch is then thrown to the right, C₁ charges the other two capacitors. The equivalent capacitance of C₂ and C₃ is
   

   1 C23 = 1 C2 + 1 C3 Þ C23 = C2C3 C2 + C3 (4.2.3.28)

   The final voltage across C₁ and C₂₃ must be the same since the top and bottom plates are connected by conducting lines. The equivalent capacitance of the entire circuit with the switch closed to the right is
   

   Ceq = C1 + C23 = C1C2 + C1C3 + C2C3 C2 + C3

   (4.2.3.29)

   Since the charge is conserved, the charge is shared among the capacitors such that
   

   Q1 = CeqV¢Þ V¢ = Q1 Ceq = C1 Ceq Vb V¢ = C1C2 + C1C3 C1C2 + C1C3 + C2C3 (4.2.3.30)

   where V¢ is the voltage across C₁ and C₂₃ after the switch is closed to the right. Once the voltage is known, we can get the new charge on each capacitor.
   

   Q1¢ = C1V¢ Q2 = Q3 = C23V¢ (4.2.3.31)

   Finally, the voltages across C₂ and C₃ are
   

   V2 = Q2 C2 = C3 C2 + C3 V¢ V3 = Q3 C3 = C2 C2 + C3 V¢ (4.2.3.32)

Send comments to larryg@upenn5.hep.upenn.edu.
This page was last modified on 02/04/2003 at 12:55:30 (EST).
Current date/time is Monday, 23-Nov-2009 15:01:39 EST
03338 hits since