Exhaustive Guide to Capacitor Charging and Discharging Through a Resistor

Fundamentals of Capacitor Circuits

  • Circuit Components and Configuration:     * Power Supply: The circuit operates with a voltage source of +9 V+9\,V.     * Resistor (RR): The resistor in the circuit has a value of 100 kΩ100\,k\Omega.     * Capacitor (CC): The capacitor has a capacitance of 300 μF300\,\mu F.     * Switch 1 (S1S_1): This switch is used to initiate the charging process when closed.     * Switch 2 (S2S_2): This switch is used to discharge the capacitor; pressing it briefly connects the capacitor to 0 V0\,V, discharging it instantly.

The Time Constant (RCRC)

  • Definition and Calculation:     * The time constant, designated by the Greek letter tau (τ\tau), is the product of the resistance and the capacitance in a circuit.     * Formula: Time Constant=R×C\text{Time Constant} = R \times C     * Calculation for the provided circuit: 100 kΩ×300 μF=100,000 Ω×0.0003 F=30,000 ms=30 s100\,k\Omega \times 300\,\mu F = 100,000\,\Omega \times 0.0003\,F = 30,000\,ms = 30\,s.     * Therefore, 1 RC=30 s1\,RC = 30\,s.

  • Key Thresholds during Charging:     * 1 RC1\,RC (30 s30\,s): The capacitor charges to approximately 63%63\% of the supply voltage.     * 0.69 RC0.69\,RC: This represents the time required for the capacitor to reach 50%50\% of the supply voltage.     * 5 RC5\,RC: After five time constants, the capacitor is considered effectively fully charged (approximately 100%100\%).     * Intermediate Stages: Each subsequent time constant allows the capacitor to charge an additional 63%63\% of the remaining voltage difference between its current state and the supply voltage.

Capacitor Charging through a Resistor

  • Voltage Growth Calculation:     * The formula used to calculate the voltage across the capacitor (VCV_C) at any given time (tt) while charging is:     * VC=V×(1−e−tRC)V_C = V \times (1 - e^{-\frac{t}{RC}})     * In this context, the power of ee corresponds to the number of time constants that have elapsed.

  • Example 1: Charging at t=60 st = 60\,s:     * Given: t=60 st = 60\,s, V=+9 VV = +9\,V, RC=30 sRC = 30\,s.     * VC=+9 V×(1−e−60 s100 kΩ×300 μF)V_C = +9\,V \times (1 - e^{-\frac{60\,s}{100\,k\Omega \times 300\,\mu F}})     * VC=+9 V×(1−e−60 s30 s)V_C = +9\,V \times (1 - e^{-\frac{60\,s}{30\,s}})     * VC=+9 V×(1−e−2)V_C = +9\,V \times (1 - e^{-2})     * VC=+9 V×(1−0.135)V_C = +9\,V \times (1 - 0.135)     * VC=+9 V×0.865=7.78 VV_C = +9\,V \times 0.865 = 7.78\,V

  • Example 2: Charging at t=100 st = 100\,s:     * Given: t=100 st = 100\,s, V=+9 VV = +9\,V, RC=30 sRC = 30\,s.     * VC=+9 V×(1−e−100 s100 kΩ×300 μF)V_C = +9\,V \times (1 - e^{-\frac{100\,s}{100\,k\Omega \times 300\,\mu F}})     * VC=+9 V×(1−e−3.33)V_C = +9\,V \times (1 - e^{-3.33})     * VC=+9 V×(1−0.0357)V_C = +9\,V \times (1 - 0.0357)     * VC=+9 V×0.9643=8.68 VV_C = +9\,V \times 0.9643 = 8.68\,V

  • Determining Time from Voltage (Charging):     * To find the time required for a capacitor to reach a specific voltage, the formula is rearranged:     * t=−R×C×ln⁡(1−VCV)t = -R \times C \times \ln(1 - \frac{V_C}{V})

  • Application: Reaching 7 V7\,V while charging:     * t=−100 kΩ×300 μF×ln⁡(1−7 V9 V)t = -100\,k\Omega \times 300\,\mu F \times \ln(1 - \frac{7\,V}{9\,V})     * t=−30 s×ln⁡(1−0.777)t = -30\,s \times \ln(1 - 0.777)     * t=−30 s×ln⁡(0.222)t = -30\,s \times \ln(0.222)     * t=−30 s×(−1.5)=45 st = -30\,s \times (-1.5) = 45\,s

  • Application: Reaching 8 V8\,V while charging:     * t=−30 s×ln⁡(1−8 V9 V)t = -30\,s \times \ln(1 - \frac{8\,V}{9\,V})     * t=−30 s×ln⁡(0.111)t = -30\,s \times \ln(0.111)     * t=−30 s×(−2.2)=66 st = -30\,s \times (-2.2) = 66\,s

Capacitor Discharging through a Resistor

  • Process Initiation: Set the circuit by pressing switch S1S_1 to charge the capacitor instantly, then release S1S_1 to allow the capacitor to discharge through the resistor RR.

  • Key Thresholds during Discharging:     * 1 RC1\,RC (30 s30\,s): The voltage across the capacitor drops by 63%63\% (meaning it falls to 37%37\% of its original value).     * 0.69 RC0.69\,RC: The voltage drops by 50%50\%.     * 5 RC5\,RC: The capacitor is considered effectively fully discharged (≈100%\approx 100\% drop).

  • Voltage Decay Calculation:     * The formula for the voltage across the capacitor while discharging is:     * VC=V×e−tRCV_C = V \times e^{-\frac{t}{RC}}

  • Example 1: Discharging at t=60 st = 60\,s:     * Given: t=60 st = 60\,s, V=+9 VV = +9\,V, RC=30 sRC = 30\,s.     * VC=+9 V×e−60 s30 sV_C = +9\,V \times e^{-\frac{60\,s}{30\,s}}     * VC=+9 V×e−2V_C = +9\,V \times e^{-2}     * VC=+9 V×0.135=1.22 VV_C = +9\,V \times 0.135 = 1.22\,V

  • Example 2: Discharging at t=100 st = 100\,s:     * Given: t=100 st = 100\,s, RC=30 sRC = 30\,s.     * VC=+9 V×e−3.33V_C = +9\,V \times e^{-3.33}     * VC=+9 V×0.0357=0.32 VV_C = +9\,V \times 0.0357 = 0.32\,V

  • Determining Time from Voltage (Discharging):     * To find the time required for a capacitor to decay to a specific voltage:     * t=−R×C×ln⁡(VCV)t = -R \times C \times \ln(\frac{V_C}{V})

  • Application: Reaching 2 V2\,V while discharging:     * t=−100 kΩ×300 μF×ln⁡(2 V9 V)t = -100\,k\Omega \times 300\,\mu F \times \ln(\frac{2\,V}{9\,V})     * t=−30 s×ln⁡(0.222)t = -30\,s \times \ln(0.222)     * t=−30 s×(−1.5)=45 st = -30\,s \times (-1.5) = 45\,s

  • Application: Reaching 1 V1\,V while discharging:     * t=−30 s×ln⁡(1 V9 V)t = -30\,s \times \ln(\frac{1\,V}{9\,V})     * t=−30 s×ln⁡(0.111)t = -30\,s \times \ln(0.111)     * t=−30 s×(−2.2)=66 st = -30\,s \times (-2.2) = 66\,s

Comparative Graphical Analysis

  • Symmetry in Charging and Discharging:     * Comparison at 66 s66\,s: While charging, it takes 66 s66\,s for the capacitor to reach 8 V8\,V (within 1 V1\,V of the supply voltage). Conversely, while discharging starting from 9 V9\,V, it takes 66 s66\,s for the capacitor to reach 1 V1\,V (within 1 V1\,V of zero).     * Comparison at 45 s45\,s: While charging, the capacitor reaches 7 V7\,V in 45 s45\,s. While discharging, it takes 45 s45\,s to reach 2 V2\,V.     * Specific Voltage Levels at 2 RC2\,RC (60 s60\,s):         * Charging: VC=7.78 VV_C = 7.78\,V         * Discharging: VC=1.22 VV_C = 1.22\,V

Exercises

  • Upon reviewing these notes, complete the following from Chapter 2 of the 1A curriculum:     * Exercise 2.1     * Exercise 2.2     * Exercise 2.3
  • Submissions must be made via three separate, clear scans and uploaded to the Teams platform.