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RC Time Constant worksheet

Grades 11–12 · CCSS HSF.LE.A.4, HSF.LE.B.5

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

1.
Problem 1.
- Capacitor is: Discharging
- Find: Voltage at a time
- Resistance (R): 14 kΩ
- Capacitance (C): 110 µF
- Supply voltage (charging) or starting voltage (discharging): 4 V
- Time since the switch closed: 2 s
Find: Capacitor voltage after 2 s
Answer
2.
Problem 2.
- Capacitor is: Charging
- Find: Time to reach a voltage
- Resistance (R): 7 kΩ
- Capacitance (C): 105 µF
- Supply voltage (charging) or starting voltage (discharging): 6 V
- Target capacitor voltage: 5 V
Find: Time to reach 5 V
Answer
3.
Problem 3.
- Capacitor is: Charging
- Find: Time to reach a voltage
- Resistance (R): 14 kΩ
- Capacitance (C): 116 µF
- Supply voltage (charging) or starting voltage (discharging): 6 V
- Target capacitor voltage: 4 V
Find: Time to reach 4 V
Answer
4.
Problem 4.
- Capacitor is: Discharging
- Find: Voltage at a time
- Resistance (R): 10 kΩ
- Capacitance (C): 100 µF
- Supply voltage (charging) or starting voltage (discharging): 5 V
- Time since the switch closed: 1 s
Find: Capacitor voltage after 1 s
Answer
5.
Problem 5.
- Capacitor is: Charging
- Find: Voltage at a time
- Resistance (R): 10 kΩ
- Capacitance (C): 65 µF
- Supply voltage (charging) or starting voltage (discharging): 5 V
- Time since the switch closed: 2 s
Find: Capacitor voltage after 2 s
Answer
6.
Problem 6.
- Capacitor is: Discharging
- Find: Time to reach a voltage
- Resistance (R): 6 kΩ
- Capacitance (C): 116 µF
- Supply voltage (charging) or starting voltage (discharging): 6 V
- Target capacitor voltage: 3 V
Find: Time to reach 3 V
Answer
7.
Problem 7.
- Capacitor is: Discharging
- Find: Voltage at a time
- Resistance (R): 12 kΩ
- Capacitance (C): 100 µF
- Supply voltage (charging) or starting voltage (discharging): 4 V
- Time since the switch closed: 2 s
Find: Capacitor voltage after 2 s
Answer
8.
Problem 8.
- Capacitor is: Charging
- Find: Time to reach a voltage
- Resistance (R): 8 kΩ
- Capacitance (C): 110 µF
- Supply voltage (charging) or starting voltage (discharging): 5 V
- Target capacitor voltage: 4 V
Find: Time to reach 4 V
Answer
9.
Problem 9.
- Capacitor is: Discharging
- Find: Time to reach a voltage
- Resistance (R): 8 kΩ
- Capacitance (C): 100 µF
- Supply voltage (charging) or starting voltage (discharging): 5 V
- Target capacitor voltage: 3 V
Find: Time to reach 3 V
Answer
10.
Problem 10.
- Capacitor is: Charging
- Find: Time to reach a voltage
- Resistance (R): 9 kΩ
- Capacitance (C): 85 µF
- Supply voltage (charging) or starting voltage (discharging): 5 V
- Target capacitor voltage: 3 V
Find: Time to reach 3 V
Answer

Answer key: RC Time Constant worksheet

1. Capacitor voltage after 2 s: 1.0915 V
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $14,000 Ω \times 0.00011 F$ $= 1.54 s$
2. Divide the time by τ to count how many time constants have passed
  $\frac{2 s}{1.54 s}$ $= 1.2987013$
3. Find the discharging voltage: the starting voltage times the share still left
  $4 V \times e^{− 1.2987013}$ $= 1.09154384 V$
2. Time to reach 5 V: 1.3169 s
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $7,000 Ω \times 0.000105 F$ $= 0.735 s$
2. Work the charging curve backward to find the time: use the gap still left to the supply voltage
  $− 0.735 \times ln (1 - \frac{5}{6})$ $= 1.3169432099 s$
3. Count the same time in time constants, to see how far along the curve it is
  $\frac{1.3169432099}{0.735}$ $= 1.791759 τ$
3. Time to reach 4 V: 1.7841 s
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $14,000 Ω \times 0.000116 F$ $= 1.624 s$
2. Work the charging curve backward to find the time: use the gap still left to the supply voltage
  $− 1.624 \times ln (1 - \frac{4}{6})$ $= 1.7841463568 s$
3. Count the same time in time constants, to see how far along the curve it is
  $\frac{1.7841463568}{1.624}$ $= 1.098612 τ$
4. Capacitor voltage after 1 s: 1.8394 V
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $10,000 Ω \times 0.0001 F$ $= 1 s$
2. Divide the time by τ to count how many time constants have passed
  $\frac{1 s}{1 s}$ $= 1$
3. Find the discharging voltage: the starting voltage times the share still left
  $5 V \times e^{− 1}$ $= 1.83939721 V$
5. Capacitor voltage after 2 s: 4.7695 V
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $10,000 Ω \times 0.000065 F$ $= 0.65 s$
2. Divide the time by τ to count how many time constants have passed
  $\frac{2 s}{0.65 s}$ $= 3.07692308$
3. Find the charging voltage: the supply voltage times the share filled so far
  $5 V \times (1 - e^{− 3.07692308})$ $= 4.76949556 V$
6. Time to reach 3 V: 482.4304 ms
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $6,000 Ω \times 0.000116 F$ $= 0.696 s$
2. Work the discharging curve backward to find the time: use the share of the starting voltage left
  $− 0.696 \times ln (\frac{3}{6})$ $= 0.4824304377 s$
3. Show the time in ms, which is easier to read
  $0.4824304377 s$ $= 482.430438 ms$
4. Count the same time in time constants, to see how far along the curve it is
  $\frac{0.4824304377}{0.696}$ $= 0.693147 τ$
7. Capacitor voltage after 2 s: 0.7555 V
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $12,000 Ω \times 0.0001 F$ $= 1.2 s$
2. Divide the time by τ to count how many time constants have passed
  $\frac{2 s}{1.2 s}$ $= 1.66666667$
3. Find the discharging voltage: the starting voltage times the share still left
  $4 V \times e^{− 1.66666667}$ $= 0.75550241 V$
8. Time to reach 4 V: 1.4163 s
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $8,000 Ω \times 0.00011 F$ $= 0.88 s$
2. Work the charging curve backward to find the time: use the gap still left to the supply voltage
  $− 0.88 \times ln (1 - \frac{4}{5})$ $= 1.4163053629 s$
3. Count the same time in time constants, to see how far along the curve it is
  $\frac{1.4163053629}{0.88}$ $= 1.609438 τ$
9. Time to reach 3 V: 408.6605 ms
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $8,000 Ω \times 0.0001 F$ $= 0.8 s$
2. Work the discharging curve backward to find the time: use the share of the starting voltage left
  $− 0.8 \times ln (\frac{3}{5})$ $= 0.408660499 s$
3. Show the time in ms, which is easier to read
  $0.408660499 s$ $= 408.660499 ms$
4. Count the same time in time constants, to see how far along the curve it is
  $\frac{0.408660499}{0.8}$ $= 0.510826 τ$
10. Time to reach 3 V: 700.9624 ms
1. Multiply the resistance (in ohms) by the capacitance (in farads) to get the time constant τ, in seconds
  $9,000 Ω \times 0.000085 F$ $= 0.765 s$
2. Work the charging curve backward to find the time: use the gap still left to the supply voltage
  $− 0.765 \times ln (1 - \frac{3}{5})$ $= 0.7009624099 s$
3. Show the time in ms, which is easier to read
  $0.7009624099 s$ $= 700.96241 ms$
4. Count the same time in time constants, to see how far along the curve it is
  $\frac{0.7009624099}{0.765}$ $= 0.916291 τ$