07.1 Capacitors in Parallel and Series Practice Problems
CAPACITORS IN PARALLEL AND SERIES | PRACTICE PROBLEMS Complete the following problems to reinforce your understanding of the concept covered in this module.
PROBLEM 1:
What is the equivalent capacitance for the circuit shown below?
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PROBLEM 2:
What is the total capacitance between terminals A and B in the circuit shown below?
PROBLEM 3:
Determine the total capacitance for the circuit shown below.
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CAPACITORS IN PARALLEL AND SERIES | SOLUTIONS
SOLUTION 1:
In this problem we are looking to calculate the equivalent or total capacitance of the circuit. We will need to identify pairs or groups of capacitors that we can combine and simplify, until we are able to simplify the circuit to have one capacitor or total equivalent capacitance. We will start with the group of 3 capacitors on the right had side of the circuit that are in series. We will use the formula for capacitors in series to calculate the equivalent capacitance.
CS =
1 1 1 1 + + C1 C 2 C 3
The formula for CAPACITORS IN SERIES can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Given capacitor values of 4 µF , 5 µF , and 10 µF , we calculate the first equivalent capacitance:
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1 C EQ,1
=
1 1 1 + + 4 µF 5 µF 10 µF
C EQ,1 = 1.8 µF
Hint: Use the equation solve function on your calculator to avoid using the various algebra or fractional calculations in the equation. This is helpful for saving time and avoiding simple algebra errors. Next, we will use the calculated capacitance equivalence and the 8 µF capacitor to further simplify the circuit. These capacitors are in parallel, so we will use the following formula:
CP = C1 + C2 The formula for CAPACITORS IN PARALLEL can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Plugging in the capacitor values of 8 µF and the first equivalent capacitance 1.8 µF , we can calculate the second equivalent capacitance:
C EQ,2 = 1.8 µF + 8 µF = 9.8 µF
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Next, we will simplify the second capacitance equivalence we just calculated and the 2
µF capacitor that is in series with it. As there are only two capacitors in a series, we can calculate the third and total equivalent capacitance value of the circuit by using the PRODUCT-OVER-THE-SUMRULE.
CT=
C1 x C2 C1 +C2
Note: This formula is not provided in the Reference Handbook. Plugging in the capacitor values of 8.44 µF and 1 µF , we can calculate the third and total equivalent capacitance:
CT=
2 µF x 9.8 µF 2 µF +9.8 µF
C T = 1.66 µF
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SOLUTION 2: In this problem we are looking to calculate the equivalent or total capacitance of the circuit. We will need to identify pairs or groups of capacitors that we can combine and simplify until we are able to simplify the circuit to have one capacitor or total equivalent capacitance. This circuit is fairly easy, as all the capacitors are in series, so there will not be multiple groups of capacitors that you will have to simplify. This circuit has 4 capacitors that are in series. We will use the formula for capacitors in series to combine the 4 capacitors and calculate the equivalent capacitance.
CS =
1 1 1 1 1 + + + C1 C2 C 3 C 4
The formula for CAPACITORS IN SERIES can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Given the capacitor values of 10 pF , 10 pF , 10 pF , and 10 pF , we calculate the total equivalent capacitance of the circuit:
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1 C EQT
=
1 1 1 1 + + + 10 pF 10 pF 10 pF 10 pF
C EQ,1 = 2.5 pF
Hint: Use the equation solve function on your calculator to avoid using the various algebra or fractional calculations in the equation. This is helpful for saving time and avoiding simple algebra errors.
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SOLUTION 3:
In this problem we are looking to calculate the equivalent or total capacitance of the circuit. We will need to identify pairs or groups of capacitors that we can combine and simplify until we are able to simplify the circuit to have one capacitor or total equivalent capacitance. This circuit is fairly easy, as all the capacitors are in parallel, so there will not be multiple groups of capacitors that you will have to simplify. This circuit has 3 capacitors that are in parallel with each other. We will use for the formula for capacitors in parallel to calculate the equivalent capacitance of the circuit. We will use for the formula for capacitors in parallel to calculate the equivalent capacitance.
C P = C1 + C2 + C 3 + C 4 The formula for CAPACITORS IN PARALLEL can be referenced under the topic of CAPACITORS AND INDUCTORS IN SERIES AND PARALLEL on page 201 of the NCEES Supplied Reference Handbook, 9.3 Version for Computer Based Testing. Given the capacitor values of 8 pF , 6 pF , and 9 pF , we calculate the total equivalent capacitance:
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CEQT = 3 pF + 6 pF + 9 pF CEQT = 18 pF
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