Calculations About Capacitors in Series and Parallel
Table of Contents
Capacitors in Series
Just like a resistor, capacitors can also be installed in a series and in parallel. When different capacitors are installed on a series, then aggregate capacitance declines. Its reason is that the internal distance between the capacitors’ plates increases, due to which overall capacitance of the circuit reduces. The formula which is being used to determine the total capacitance of multiple capacitors installed in a series is similar to the formula being used for determining the total resistance of resistors installed in a series.
Figure 6.30; Capacitor in series
Let’s assume that three capacitors C1, C2, C3 are installed in a series, which are being provided supply voltages V. Suppose, that V1, V2, V3 voltages are found parallel to each capacitor. This has been shown in Figure 6.30. In this situation, charge found in each of the capacitors will be the same.
Q = Q1 = Q2 = Q3 … (1)
If capacitance of the equivalent capacitor is is C, then;
Q = CV
∴ V = Q/ C
We know that;
V1 = Q1/ C1, V2= Q2/ V2, V3 = Q3/ C3
After applying Kirchhoff’s second law in close loop,
V = V1 + V2 + V3
Thus, Q/C = Q1/ C1 + Q2/ C2 + Q3/ C3
After dividing both sides by Q according to equation 1,
1/ C = 1/C1 + 1/C2 + 1/ C3 … (2)
If two capacitors are installed in a series, then the formula to determine their total capacitance will be as follows;
C = C1 C2 / C1 + C2 … (3)
If several capacitors of same value are fixed in a series, then their total capacitance can be ascertained with the help of the following formula;
1/ CT = 1/C + 1/ C + 1/ C+ … + 1/ C or CT = C/ n … (4)
Remember that the total capacitance of a circuit can be determined just in the same way, as total parallel resistance is determined.
Capacitors in Parallel
When several capacitors are installed in parallel, then an increase occurs in the charge area of the capacitor. that’s effective plate area increases. Thus, by the way of adding individual capacitance together, total capacitance tends to increase.
Figure 6.31; Capacitor in parallel
Let’s suppose that three capacitors C1, C2, and C3 are attached to the supply voltage V in a parallel, as has been shown via figure 6.31. If the charge found on all the three capacitors be Q1, Q2, Q3 respectively, then the total charge Q will be equal to the sum of individual charges, i.e.,
Q = Q1 + Q2 + Q3 … (5)
If the capacitance of the equivalent capacitor be C, then;
Q = C V
Thus, Q1 = C1 V, Q2 = C2 V, Q3 = C3 V … (6)
By substituting the values of equation 6 into equation 5, we get;
CV = C1 V + C2 V+ C3 V
As voltages in an equivalent circuit are always equal, that’s;
V = V1 = V2 = V3
C = C1 + C2+ C3 … (7)
If two capacitors are in parallel, then their total capacitance can be determined with the help of the following formula; It should be remembered that equal voltages V are received by both the capacitors in parallel.
V = Q1/ C1 = Q2 /C2
Q1/ C1 = Q2 / C2, or Q1/ Q2 = C1 / C2
If several capacitors of the same value are fixed in parallel, then the formula for determining their total capacitance is as follows;
CT = bc
Remember that total parallel capacitance of any circuit can be found in the same way as the total resistance of any circuit in a series is determined.
Comparison of Series and Parallel Capacitors Circuits
| Capacitors in Parallel | Capacitors in Series |
| 1. Charge found in parallel to every capacitor tends to be different, and its value is directly proportional to the capacitance of the capacitor, i.e., Q = CV |
Q = Q1 = Q2 = Q3 |
| 2. The potential difference found across a capacitor tends to be the same, the value of which is equal to the applied voltage, i.e., V = V1 = V2 =V3 | 2. The potential difference found across a capacitor tends to be different, the value of which is exactly inverted to the capacitance value of each capacitor, i.e., V = Q / C |
| 3. The sum of individual charges found on every capacitor equals to the total charges provided by a power source, i.e., Q = Q1 + Q2 + Q3 + … | 3. The sum of total voltages found across all capacitors equals to the applied voltage, i.e., V = V1 + V2 + V3 + … |
| 4. Total capacitance equals to the sum of individual capacitances, i.e., C = C1 + C2 + C3 + … | 4. The following reciprocal formula is used to determine total capacitance; 1/ C = 1/C1 + 1/ C2 + 1/ C3 + … |
| 5. Active plate area increases, due to which total capacitance also increases. | 5. Active plate area reduces, due to which total capacitance also decreases. |
Calculations About Capacitors in Series and Parallel
Example 1; Determine the total capacitance if figure 6.32
Fig. 6.22
Solution;
1/ C1 = 1 /C1 + 1/ C2 + 1/ C3 = 1/ 10 μϝ + 1/ 5 μϝ + 1/ 8 μϝ
Taking the reciprocal of both sides;
CT = 1 ÷ 1/ μϝ + 1/ μϝ + 1/8 μϝ = 1 / 0.425 μϝ = 2.35 μϝ
Example 2; Find CT in figure 6.33
Figure 6.34
Solution;
CT = C1 C2 / C1 + C2 = (100 pϝ) (300 Pϝ)/ 400 pϝ = 75 pϝ
Example 3; Determine CT for the series capacitors in figure 6.34
Solution;
C1 = C2 = C3 = C4 = C
C1 = C/ n = 0.02 μϝ/4 = 0.005 μϝ
Example 4: Find the voltage across each capacitor in figure 6.35
Figure 6.35
Solution;
1/CT = 1/ C1 + 1/ C2 + 1/ C3 = 1 / 0.1 μϝ + 1/ 0.5 μϝ + 1/ 0.02 μϝ
CT = 1/7 μϝ = 0.0588
VS = VT = 25 V
V1 = [CT/ C1] VT = [0.0588 μϝ/ 0.1 μϝ] 25 V = 14.71 V
V2 = [CT / C2] VT = [0.0588 μϝ / 0.5 μϝ] 25 μϝ = 2.94 V
V3 = [CT / C3] VT = [0.0588 μϝ / 0.02 μϝ] 25 V = 7.35 V
Example 5; For the circuit of figure 6.36;
(a). find the total capacitance
(b). Determine the charge on each plate
©. Find the voltage across each capacitor
Figure 6.36
Solution;
(a). 1/CT = 1/C1 + 1/ C2 + 1 / C3
= 1/ 200 x 10-6 F + 1 / 50 x 10-6 F + 1/ 10 x 10-6 F
= 0.005 x 106 + 0.02 x 106 + 0.1 x 106 = 0.125 x 106
= 0.005 x 106 + 0.02 x106 + 0.1 x 106 = 0.125 x 106
And CT = 1 / 0.25 x 106 = 8 μϝ
(b). QT = Q1 = Q2 =Q3
QT = CTE = (8 x 10-6F) (60 V) = 480μC
©. V1 = Q1 /C1 = 480 x 10-6 C/ 200 x 10-6 F = 2.4 V
V2 = Q2 / C2 = 480 x 10-6 C/ 50 x 10-6 F = 9.6 V
V3 = Q3 / C3 = 480 x 10-6 C/ 10 x 10-6 F = 48.0 V
And E = V1 + V2 + V3 = 2.4 V + 9.6 V + 48 V
= 60 V (checks)
Example 6; What is the total capacitance in figure 6.37? what is the voltage across each capacitor?
Figure 6.37
Solution;
CT = C1 + C2 = 330 pF + 220 pF = 550 pF
V1 = V2 + = 5V
Example 8; For the network of figure 6.39;
Figure 6.39
(a). Find the total capacitance
(b). Determine the charge on each plate
©. Find the total charge
Solution;
(a). CT = C1 + C2 + C3 = 800μF + 60μF + 1200μF = 2060μF
(b). Q1 = C1E = (800 x 10-6 F) (48 V) = 38.4 mC
Q2 = C2E = (60 x 10-6F) (48V) = 2.88 mC
Q3 = C3 E = (1200 x 10-6 F) (48V) = 57.6 mC
©. QT = Q1 + Q2 + Q3 = 38.4 mC + 2.88 mC + 57.6 mC
Example 9: Three capacitors 10μF, 25μF, and 50μF are connected (a) in series (b) in parallel. Find the equivalent capacitance and the energy stored for each of the cases (a) and (b) when the capacitors are connected across a 500 volts supply.
Solution;
In series
1/C = 1/C1 + 1/ C2 + 1/C3 = 1/ 10 + 1/ 25 + 1/ 50
= 10 + 4+ 2 /100 = 16 / 100
C = 100/16 = 6.25μF
Energy stored = ½ CV2 = ½ 6.25 / 106 x 5002 joules = 0.78 joules
(b). In parallel
C = C1 + C2 + C3 = 10 + 25 + 50 = 85μF
Energy stored = ½ CV2 = ½ 85/ 106 x 5002 joules = 10.625 joules
Example 10; Find the voltage across and charge on each capacitor for the network of fig. 6.40
Figure 6.40
Solution;
CT = C2 +C3 = 4 μF + 2 μF = 6 μF
CT = C1 CT/ C1 + CT = (3μF) (6μF) / 3 μF + 6 μF = 2 μF
QT = CT E = (2 x 10-6 F) (120 V) = 240 μC
An equivalent circuit (fig. 6.41) has QT = Q1 = QT
And therefore, Q1 = 240 μC
Figure 6.41
And V1 = Q1 / C1 = 240 x 10-6 C / 3 x 10-6 F = 80 V
QT = 240 μC
And therefore, VT = Q’T / C’T = 240 x 10-6 C / 6 x 10-6 F = 40 V
And Q2 = C2 V’T = (4 x10-6 F) (40 V) = 160 μC
Q3 = C3 V’T = (2 x 10-6 F) (40 V) = 80 μC
Example 11: A 50 μF capacitor is connected across a 220 V supply. What charge in coulombs does it take? What charge does it take if the voltage is increased to 400 volts?
Solution;
The charge = capacity x voltage
= 50 x 10-6 x 220 = 11000 x 10 -6 coulomb
= 11000 μC Ans.
Now, if the voltage is increased then the new charge will be;
= 50 x 10-6 x 400
= 20000 x 10-6 coulomb = 20000 μC Ans.
Example 12; Three capacitors of 5, 10, and 15 μF are connected in series across 100 V supply. Find the equivalent capacity. Find also the voltage across each.
Figure 6.42
Solution;
Here, all three capacitors are joined in series across 100 V supply. In series, the equivalent capacity is;
1/Ceq = 1/ C1 + 1/ C2 + 1/ C3 = 1/5 + 1/10 + 1/15 = 11/ 30
Ceq = 30 / 11 = 2.73 μF
In series connection, the voltage across each capacitor can be given as;
V1 = Q / C1; V2= Q / C2; V3 = Q / C3
V1, V2, V3 are the voltage across individual capacitors
So, Q1 = the charge = V x C
= Voltage x capacity
= 100 x 30 / 11 = 3000 / 11 μC
V1 = 3000/ 11 x 1/5 =54.5 Volt
V2 = Q / C2 =3000 x 1 / 11 x 10 = 27 .3 Volts
V3 = Q / C3 = 3000 x 1 / 11 x15 = 18.2 Volts
So, the voltages are 54.5 volts, 27.3 volts, and 18.2 volts, Ans.
Example 13; In the above question if the capacitors after being charged in series are disconnected and then connected in parallel with likely polarity together, find;
(a). Total charge of the combination
(b). Charge on each capacitor
©. Voltage across parallel combination
Solution;
As it was connected in series (as in example 12) the charge will be 3000/ 11 μC; If now these are connected in parallel so the total charge will be added up now charge of all three capacitors.
Q = q1 + q2 + q3
=3000 / 11 + 3000 / 11 + 3000 / 11 = 9000 / 11
Q = 818.2 μC. Ans
The total capacity will be added up, as they are connected in parallel’ so
Ctotal = C1 + C2 + C3 + ….
= 5 + 10 + 15 = 30 μF
Now, the new voltage so resulted will be across the parallel combination;
∴ V = Charge / Capacity = q / C = 818.2 x 10-6 / 30 x 10-6 = 27.273 volts
∴ Voltage across = 27.273 volts
The new charge will be;
q1= V x C1 = 27.273 x 5 = 136.4 μC
q2= 27.273 x 10 = 272.73 μC
q3= 27.273 x 15 = 409.1 μC
so, the charges are 136.4 μC, 273.73 μC, 409.1 μC Ans.
And voltage across the combination = 27.273 volts, Ans.
Example 14; Three capacitors are connected in series across a 120 volts supply. The voltage across them is 30, 40, and 50 and the charge on each is 4500 μC. What is the value of each capacitor and of the series combination?
Solution;
All three capacitors are connected in series so the charge on each capacitor will be same and voltage across each will be according to their capacity, charge is given = 4500 μC., and voltage across each respectively be;
V1 across C1 capacitor = Q/C which is given as 30 V.
So, capacity = Q/V1 = 4500 x 10-6 / 30 = 150 x 10-6
= C1 = 150 μF
In the same way, C2 = Q / V2 = 4500 x10-6 / 40 = 112.5 μF
C3 = Q/V3 = 4500 x 10-6 / 50 = 90 μF
So, different capacities are; 150 μF, 112 μF, and 90 μF, Ans.
Now, the value of series combination can easily be calculated being voltage equal to 120 volts and equivalent charge as 4500 μC.
C = Q / V = 4500 x 10-6 / 120 = 37.5 μF
Example 15: Four capacitors are connected in parallel across a 250 V supply, the charge taken by them are 750, 1000, 1500, and 2000 μC. What is the equivalent capacitance of the combination?
Solution;
In parallel combination, the charge on the capacitors will be added up and will be equal to the charge of the equivalent combination;
Q = q1 + q2 + q3 + …
= 750 + 1000 + 1500 + 2000 = 5250 μC
Now, the voltage across the combination, is 250 V, so the equivalent value of capacitance;
C = Q / V = 5250 / 250 = 21 μF, Ans.
Example 16; A 100 μF capacitor is charged from a 200 V supply. The charged capacitor is now connected across three uncharged capacitors connected in parallel. The charges on these are 4000, 5000, and 6000 μC. Find,
(a). Capacitance of each capacitor
(b). Voltage across combination
Solution;
The charge acquired by the capacitor when connected to 200 V supply;
q = C x V = 100 x 10 -6 x 200 = 20000 μC
whenever connected across the combination of three capacitors connected in parallel, it will deliver the charge according to their capacities, so the voltage across the combination remains the same. Now, the total charge’
= q1 + q2 + q3
= 4000+ 5000+ 6000= 15000 μC
The remaining charge = 20000 – 15000 = 5000 μC will remain with the capacitor of 100 μF. Now, the voltage across the whole combination can be calculated as;
V = Q/C = 5000 x 10-6 / 100 x 10-6 = 50 Volts
Now, the capacity;
C1 = q / V = 4000 x 10-6 / 50 = 80 μF
C2 = 5000 x 10-6 / 50 = 100 μF
C3 = 6000 x 10-6 / 50 = 120 μF
Now, capacities are 80 μF, 100 μF, and 120 μF.
And voltage across combination is 50 V, Ans.
Example 17; Three capacitors A, B, and C are charged as follows; A – 10 μF, B – 15 μF, C – 25 μF at 200 volts. They are then connected in parallel with terminals of like polarity together. Find the voltage across the combination?
Solution;
The capacitors are charged and they acquire the charges as follows;
q1 = C1 V1 = 10 x 10 -6 x 100 = 1000 μC
q2 =C2 V2 = 15 x 10 -6 x 150 = 2250 μC
q3 = C3 V3 = 25 x 10-6 x 200 = 5000 μC
in parallel combination, their charge will be added up and the total charge (q) is;
q = q1 + q2 + q3 + …
= (1000 + 2250 + 5000) μC = 8250 μC
The capacities are also added in case of parallel combination;
C = C1 + C2 + C3 = 10 + 15 + 25 = 50 μF
Here, of this combination the capacity is 50 μF, and charge amounting 8250 μC. Then, voltage across the combination is;
V = Q / C = 8250 x 10 -6 / 50 x 10 -6 = 165 Volts. Ans
Example 18; Two capacitors when connected in parallel across a 250 V supply have charges of 2000 and 2500 μC. Find the energy stored in each and total energy.
Solution;
The energy stored by a capacitor = ½ CV2 as we know that;
C = q /V
So, ½ CV2 = ½ q / V. V2 = ½ q.v
Energy stored = ½ charge x voltage
In first capacitor, i.e., E1 = ½ q1 v = ½ x 2000 x 10-6 x 250 = 0.25 J.
And E2 = ½ q2. V = ½ x 2500 x 10-6 x 250 = 0.3125 J
The total energy = (0.25 + 0.3125) = 0.5625 J
0.25 J, 0.3125 J, and 0.5625 J, Ans.
Example 19: Three capacitors A, B, C have capacitances 10, 50, and 25 μF respectively. Calculate (i) charges on each when connected in parallel to 250 V supply (ii) total capacitances and (iii) p.d across each when connected in series.
Figure 6.44
Solution;
(i). Parallel connection is shown in figure (6.44) (a). Each capacitor has a p.d of 250 V across it.
Q1 = C1 V = 10 x 250 = 2500 μC; Q2 = 50 x 250 = 12500 μC; Q3 = 25 x 250 = 6750 μC
(ii). C = C1 + C2 + C3 = 10 + 50 + 25 = 85 μF
(iii). Series connection is shown if figure 6.44 (b). Here, charge on each capacitor is the same and is equal to that on the equivalent single capacitor.
1 / C = 1/ C1 + 1 /C2 + 1/ C3 Or 1/C = 1/10 + 1/ 50 + 1/25 = 8 / 50 Or C = 25 /4 μF
Q = CV = 25 x 250 /4 = 1562.5 μF
Q = C1 V1 and V1 = 1562.5 / 10 = 156.25 V
V2 = 1562.5 / 25 = 62.5 V; V3 = 1562.5 / 50 =31.25 V
Example 20: Find the charges on capacitors in figure 6.45 and the p.d across them;
Figure 6.45
Solution;
Equivalent capacitance between points A and B is
C2 + C3 = 5 + 3 = 8 μϝ
Capacitance of the whole combination;
C = 8 x 2 / 8 + 2 = 1.6 μϝ
Charge on the combination is;
Q1 = CV = 100 X 1.6 = 160 μC
V1 = Q1 / C1 = 160 / 2 = 80 V; V2 = 100 – 80 = 20 V
Q2 = C2 V2 = 3 x 10 -6 x 20 = 60 μC
Q3 = C3 V2 = 5 x 10-6 x 20 = 100 μC
Example 21; Two capacitors A and B are connected in series across a 100 V supply and it is observed that the p.ds across them are 60 V and 40 V respectively. A capacitor of 2 μϝ capacitance is now connected in parallel with A and the p.d across B rises to 90 volts. Calculate the capacitance of A and B in microfarads.
Solution;
Since they are connected in series (fig. 6.46 a), the charge across each is the same;
60 C1 = 40 C2 or C1/C2 = 2/3 … (i)
In figure 6.46 (b) is shown a capacitor 2 μϝ connected across capacitor A. Their combined capacitance = (C1 + 2) μϝ
(C1 + 2) 10 = 90 C2 or (C1 + 2) / C2 = 9 … (ii)
Fig. 6.46
Putting the value of C2 = 3C1 / 2 from equation (i) in (ii), we get
C1 +2 / 3C1 / 2 = 9 ∴ C1 + 2 = 13.5 C1
Or C1 = 2 / 12.5 = 0.16 μϝ and
C2 = (3/2) x 0.16 = 9.24 μϝ
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