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Capacitor worksheet

Grades 11–12

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Connection: Series
- Capacitor values: 13, 28, 40
- Capacitance unit: Picofarads (pF)
- Voltage across the group (optional): 9 V
Find: Total capacitance (series)
Answer
2.
Problem 2.
- Connection: Series
- Capacitor values: 9, 20, 50
- Capacitance unit: Nanofarads (nF)
- Voltage across the group (optional): 14 V
Find: Total capacitance (series)
Answer
3.
Problem 3.
- Connection: Parallel
- Capacitor values: 9, 14, 60
- Capacitance unit: Nanofarads (nF)
- Voltage across the group (optional): 12 V
Find: Total capacitance (parallel)
Answer
4.
Problem 4.
- Connection: Series
- Capacitor values: 10, 22, 47
- Capacitance unit: Millifarads (mF)
- Voltage across the group (optional): 14 V
Find: Total capacitance (series)
Answer
5.
Problem 5.
- Connection: Series
- Capacitor values: 10, 22, 47
- Capacitance unit: Nanofarads (nF)
- Voltage across the group (optional): 15 V
Find: Total capacitance (series)
Answer
6.
Problem 6.
- Connection: Series
- Capacitor values: 10, 22, 45
- Capacitance unit: Farads (F)
- Voltage across the group (optional): 9 V
Find: Total capacitance (series)
Answer
7.
Problem 7.
- Connection: Parallel
- Capacitor values: 8, 18, 42
- Capacitance unit: Picofarads (pF)
- Voltage across the group (optional): 15 V
Find: Total capacitance (parallel)
Answer
8.
Problem 8.
- Connection: Series
- Capacitor values: 9, 20, 53
- Capacitance unit: Picofarads (pF)
- Voltage across the group (optional): 10 V
Find: Total capacitance (series)
Answer
9.
Problem 9.
- Connection: Parallel
- Capacitor values: 13, 14, 36
- Capacitance unit: Nanofarads (nF)
- Voltage across the group (optional): 13 V
Find: Total capacitance (parallel)
Answer
10.
Problem 10.
- Connection: Series
- Capacitor values: 12, 22, 50
- Capacitance unit: Nanofarads (nF)
- Voltage across the group (optional): 11 V
Find: Total capacitance (series)
Answer

Answer key: Capacitor worksheet

1. Total capacitance (series): 7.27 pF
1. In series, every capacitor holds the same charge. So you add 1 divided by each value (its reciprocal)
  $\frac{1}{C}$ $= \frac{1}{13} + \frac{1}{28} + \frac{1}{40}$ $= 0.1376373626 1/pF$
2. That sum is 1 divided by the total, so flip it over to get the total C
  $C$ $= \frac{1}{0.1376373626}$ $= 7.2654690619 pF$
3. Change to farads (1 pF = 0.000000000001 F), because the charge and energy formulas work in farads
  $7.2654690619 pF \times 0.000000000001$ $= 0.000000000007265 F$
4. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 0.000000000007265 \times 9$ $= 0.000000000065389 C$
5. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 0.000000000007265 \times 9^{2}$ $= 0.000000000294251 J$
2. Total capacitance (series): 5.52 nF
1. In series, every capacitor holds the same charge. So you add 1 divided by each value (its reciprocal)
  $\frac{1}{C}$ $= \frac{1}{9} + \frac{1}{20} + \frac{1}{50}$ $= 0.1811111111 1/nF$
2. That sum is 1 divided by the total, so flip it over to get the total C
  $C$ $= \frac{1}{0.1811111111}$ $= 5.5214723926 nF$
3. Change to farads (1 nF = 0.000000001 F), because the charge and energy formulas work in farads
  $5.5214723926 nF \times 0.000000001$ $= 0.000000005521472 F$
4. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 0.000000005521472 \times 14$ $= 0.000000077300613 C$
5. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 0.000000005521472 \times 14^{2}$ $= 0.000000541104294 J$
3. Total capacitance (parallel): 83 nF
1. In parallel, every capacitor gets the same voltage, so you add their values to get the total C
  $C$ $= 9 + 14 + 60$ $= 83 nF$
2. Change to farads (1 nF = 0.000000001 F), because the charge and energy formulas work in farads
  $83 nF \times 0.000000001$ $= 0.000000083 F$
3. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 0.000000083 \times 12$ $= 0.000000996 C$
4. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 0.000000083 \times 12^{2}$ $= 0.000005976 J$
4. Total capacitance (series): 6 mF
1. In series, every capacitor holds the same charge. So you add 1 divided by each value (its reciprocal)
  $\frac{1}{C}$ $= \frac{1}{10} + \frac{1}{22} + \frac{1}{47}$ $= 0.1667311412 1/mF$
2. That sum is 1 divided by the total, so flip it over to get the total C
  $C$ $= \frac{1}{0.1667311412}$ $= 5.9976798144 mF$
3. Change to farads (1 mF = 0.001 F), because the charge and energy formulas work in farads
  $5.9976798144 mF \times 0.001$ $= 0.005997679814385 F$
4. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 0.005997679814385 \times 14$ $= 0.083967517401392 C$
5. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 0.005997679814385 \times 14^{2}$ $= 0.587772621809745 J$
5. Total capacitance (series): 6 nF
1. In series, every capacitor holds the same charge. So you add 1 divided by each value (its reciprocal)
  $\frac{1}{C}$ $= \frac{1}{10} + \frac{1}{22} + \frac{1}{47}$ $= 0.1667311412 1/nF$
2. That sum is 1 divided by the total, so flip it over to get the total C
  $C$ $= \frac{1}{0.1667311412}$ $= 5.9976798144 nF$
3. Change to farads (1 nF = 0.000000001 F), because the charge and energy formulas work in farads
  $5.9976798144 nF \times 0.000000001$ $= 0.00000000599768 F$
4. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 0.00000000599768 \times 15$ $= 0.000000089965197 C$
5. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 0.00000000599768 \times 15^{2}$ $= 0.000000674738979 J$
6. Total capacitance (series): 5.96 F
1. In series, every capacitor holds the same charge. So you add 1 divided by each value (its reciprocal)
  $\frac{1}{C}$ $= \frac{1}{10} + \frac{1}{22} + \frac{1}{45}$ $= 0.1676767677 1/F$
2. That sum is 1 divided by the total, so flip it over to get the total C
  $C$ $= \frac{1}{0.1676767677}$ $= 5.9638554217 F$
3. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 5.963855421686747 \times 9$ $= 53.674698795180724 C$
4. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 5.963855421686747 \times 9^{2}$ $= 241.53614457831327 J$
7. Total capacitance (parallel): 68 pF
1. In parallel, every capacitor gets the same voltage, so you add their values to get the total C
  $C$ $= 8 + 18 + 42$ $= 68 pF$
2. Change to farads (1 pF = 0.000000000001 F), because the charge and energy formulas work in farads
  $68 pF \times 0.000000000001$ $= 0.000000000068 F$
3. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 0.000000000068 \times 15$ $= 0.00000000102 C$
4. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 0.000000000068 \times 15^{2}$ $= 0.00000000765 J$
8. Total capacitance (series): 5.56 pF
1. In series, every capacitor holds the same charge. So you add 1 divided by each value (its reciprocal)
  $\frac{1}{C}$ $= \frac{1}{9} + \frac{1}{20} + \frac{1}{53}$ $= 0.1799790356 1/pF$
2. That sum is 1 divided by the total, so flip it over to get the total C
  $C$ $= \frac{1}{0.1799790356}$ $= 5.5562026791 pF$
3. Change to farads (1 pF = 0.000000000001 F), because the charge and energy formulas work in farads
  $5.5562026791 pF \times 0.000000000001$ $= 0.000000000005556 F$
4. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 0.000000000005556 \times 10$ $= 0.000000000055562 C$
5. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 0.000000000005556 \times 10^{2}$ $= 0.00000000027781 J$
9. Total capacitance (parallel): 63 nF
1. In parallel, every capacitor gets the same voltage, so you add their values to get the total C
  $C$ $= 13 + 14 + 36$ $= 63 nF$
2. Change to farads (1 nF = 0.000000001 F), because the charge and energy formulas work in farads
  $63 nF \times 0.000000001$ $= 0.000000063 F$
3. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 0.000000063 \times 13$ $= 0.000000819 C$
4. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 0.000000063 \times 13^{2}$ $= 0.0000053235 J$
10. Total capacitance (series): 6.72 nF
1. In series, every capacitor holds the same charge. So you add 1 divided by each value (its reciprocal)
  $\frac{1}{C}$ $= \frac{1}{12} + \frac{1}{22} + \frac{1}{50}$ $= 0.1487878788 1/nF$
2. That sum is 1 divided by the total, so flip it over to get the total C
  $C$ $= \frac{1}{0.1487878788}$ $= 6.7209775967 nF$
3. Change to farads (1 nF = 0.000000001 F), because the charge and energy formulas work in farads
  $6.7209775967 nF \times 0.000000001$ $= 0.000000006720978 F$
4. Multiply the capacitance by the voltage to get the stored charge Q, in coulombs
  $Q$ $= 0.000000006720978 \times 11$ $= 0.000000073930754 C$
5. For the stored energy E, take half the capacitance times the voltage squared. The answer is in joules
  $E$ $= 0.5 \times 0.000000006720978 \times 11^{2}$ $= 0.000000406619145 J$