Ee1 chapter10 capacitors_inseriesparallel

Apr 12, 2023

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Contact Number

IT2001PAEngineering Essentials (1/2)

Chapter 10 – Capacitors In Series and Parallel Circuits


IT2001PA Engineering Essentials (1/2) 2

Lesson Objectives

Upon completion of this topic, you should be able to: Calculate the equivalent capacitance of capacitors

connected in series and in parallel.


IT2001PA Engineering Essentials (1/2)

Specific Objectives

Students should be able to : Sketch the circuit diagram of 3 capacitors connected in series. Sketch the circuit diagram of 3 capacitors connected in parallel. State the equations for calculating the equivalent capacitance

for a maximum of three capacitors as

1 / Ce = 1 / C1 + 1 / C2 + 1 / C3 + ........ + 1 / Cn (series connection)

Ce = C1 + C2 + C3 + ........ + Cn (parallel connection)

Solve simple problems involving a maximum of three capacitors connected in series & parallel.



Capacitors in Series

C1 C2 C3

Charge Q is the same in all parts of series cct. Charge Q is the same in all parts of series cct.

QQTT = Q = Q11 = Q = Q2 2 = Q= Q33

as Vas VTT = V = V11 + V + V2 2 + V+ V33

And C=Q / V so V = Q / CAnd C=Q / V so V = Q / C

QQTT/C/CTT = Q = Q11/C/C11+ Q+ Q22/C/C2 2 + Q+ Q33/C/C33

therefore : therefore : 1/C1/CTT = 1/C = 1/C11 + 1/C + 1/C22 + 1/C + 1/C33



Characteristics of a Series Circuit

Total voltage = Sum of individual voltages The formula for calculating CT is given by:

321T C

1

C

1

C

1

C

1

The total capacitance CT is smaller than the lowest

individual capacitance.



Series Capacitance

The calculation of total series capacitance is similar to the calculation of total resistance of parallel resistors.

parallelresistors

seriescapacitors

1

RT

+ + + … … + 1 1 1 1

R1 R2 R3 Rn

=

1

CT

+ + + … … + 1 1 1 1

C1 C2 C3 Cn

=



Series Capacitance

If there are only 2 capacitors in series

If there are n equal-value capacitors in series

The total capacitance of a series circuit is always less than the value of the smallest capacitor.

CT = C1 C2

C1+ C2

C n

CT =



Example 10-1 Determine the total capacitance.

VS

C1 C2 C3

10µF 5µF 8µF

1CT =

+ + 1 1 1 C1 C2 C3

= 1 1 1 10µF 5µF 8µF

1

+ + = 2.35µF



Example 10-2

Determine the total capacitance.

C1 C2

VS

150pF 80pF

CT = C1 x C2

C1 + C2

= 150pF x 80pF150pF + 80pF = 52.2pF



Example 10-3

The combined capacitance of five equal capacitors is 10 F. Calculate the value of each capacitor.

C n

CT =

10 5 =

= 2 F



Capacitors in Parallel

C1

Charge Q is the sum of all in series Charge Q is the sum of all in series

circuit. circuit.

QQTT = Q = Q11 + Q + Q2 +2 + Q Q33

as Vas VTT = V = V11 = V = V2 2 = V= V33

And Q = C V And Q = C V

CCTTVVTT = C = C11VV11+ C+ C22VV2 2 + C+ C33VV33

Hence : Hence : CCTT = C = C11 + C + C22 + C + C33

C2

C3



Parallel Capacitance

The calculation of total parallel capacitance is similar to the calculation of total resistance of series resistors.

RT = R1 + R2 + R3 + ... ... + Rn

CT = C1 + C2 + C3 + ... ... + Cn

seriesresistors

parallelcapacitors



Parallel Capacitance

If there are n equal-value capacitors in parallel,

Total capacitance of a parallel circuit is equal to the sum of each individual capacitance.



Example 10-4

Determine the total capacitance.

CT = C1 + C2 + C3 + C4 + C5

VS C10.01µF

C20.02µF

C30.01µF

C40.05µF

C50.03µF

= 0.01µF + 0.02µF + 0.01µF + 0.05µF + 0.03µF= 0.12µF



Quiz

1. A 60 F capacitor is connected across a 200 V supply. What is the charge stores in the capacitor? What charge does it take if the voltage is increased to 400 V?

(Ans: 0.012C ; 0.024 C)

Q = CV = 60 x 200 = 12 C

Q = CV = 60 x 400 = 0.024 C



Quiz

2. If the charge of a 40 mF capacitor is 380 mC, find the voltage applied.

(Ans: 9.5 V)

Q = 380 mC, C = 40mF

V = Q/C = 380m/40m = 9.5 V



Quiz

3. Calculate the capacitance required to give a 680 C charge if the supply voltage is 35 V.

(Ans: 19.43 F)

Q = 680C, V = 35V

C = Q/V = 680/35 = 19.43 F



Quiz4. The charge of the circuit below is 600 C, find the

circuit capacitance, the supply voltage and the voltage across each capacitor.(Ans: 3.75 F ; 160 V ; 100 V ; 60 V)

Q = 600C

1/CT = 1/CC1 + 1/CC2

1/CT = 1/6 + 1/10 CT = 3.75 F

V = Q/C = 600/3.75 = 160V

VC1 = Q/CC1

= 600/6 = 100V

VC2 = Q/CC2

= 600/10 = 60V



Quiz5. The total capacitance of two identical capacitors

connected in series is 60 F. What is the value of each capacitor? (Ans: 120 F)

CT= C/n 60 = C/2 C = 120 F



Quiz6. If the voltage across the 5 mF capacitor is 10 V. Find

the charge on each capacitor, the voltage across the other two capacitor and the supply voltage.

(Ans: 50 mC ; 50 V ; 25 V ; 85 V)

Q = CV

Q = 5 mF x 10 V

= 50 mC

V = Q/C

V = 50 mC / 1 F

= 50 V

V = Q/C

V = 50 mC / 2 F

= 25 V

V = V1 + V2 + V3

V = 10 V + 50 V + 25 V



Summary

How to sketch the circuit diagram of capacitors connected in series & parallel.

Formulas to calculate the equivalent capacitance for: Series circuit : 1/Ce = 1/C1 + 1/C2 + 1/C3 + ...

Parallel circuit : Ce = C1 + C2 + C3 + …

Solve simple problems using above formulas.



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Ee1 chapter10 capacitors_inseriesparallel

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