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RC Circuits and Transient Behavior

RC circuits consist of resistors and capacitors connected together. These circuits exhibit transient behavior, meaning their voltage and current change over time when the circuit is switched on or off. Understanding RC circuits is crucial for analyzing timing circuits, filters, and energy storage systems.

Time Constant

🎯 Time Constant Definition

The time constant τ (tau) determines how quickly an RC circuit responds to changes.

It represents the time required for the voltage or current to change by 63.2% of its total change.

$$\tau = RC$$

The time constant is the product of resistance and capacitance. It has units of seconds and characterizes the speed of the circuit's response to changes.

Physical Interpretation

Example: Calculating Time Constant

Problem: Find the time constant of an RC circuit with R = 10kΩ and C = 100μF.

Step 1: Use Time Constant Formula

$$\tau = RC$$ $$\tau = (10kΩ)(100μF)$$

Step 2: Convert Units

$$\tau = (10 \times 10^3Ω)(100 \times 10^{-6}F)$$ $$\tau = 10^4 \times 10^{-4}s = 1s$$

Answer

The time constant is 1 second.

Mathematical Derivation

Differential Equation for RC Circuit

For an RC circuit with battery voltage V₀, we can write Kirchhoff's voltage law:

$$V_0 = V_R + V_C = IR + \frac{Q}{C}$$

Since I = dQ/dt, we get the differential equation:

$$V_0 = R\frac{dQ}{dt} + \frac{Q}{C}$$

Rearranging:

$$\frac{dQ}{dt} + \frac{Q}{RC} = \frac{V_0}{R}$$

This is a first-order linear differential equation with solution:

$$Q(t) = CV_0(1 - e^{-t/RC})$$

Deriving V(t) from Q(t)

Since V = Q/C:

$$V_C(t) = \frac{Q(t)}{C} = \frac{CV_0(1 - e^{-t/RC})}{C} = V_0(1 - e^{-t/RC})$$

With τ = RC:

$$V_C(t) = V_0(1 - e^{-t/\tau})$$

Deriving I(t) from Q(t)

Since I = dQ/dt:

$$I(t) = \frac{d}{dt}[CV_0(1 - e^{-t/RC})] = CV_0 \cdot \frac{d}{dt}(1 - e^{-t/RC})$$

Using the chain rule:

$$I(t) = CV_0 \cdot \frac{1}{RC}e^{-t/RC} = \frac{V_0}{R}e^{-t/RC}$$

With τ = RC:

$$I(t) = \frac{V_0}{R}e^{-t/\tau}$$

Charging Process

⚡ Charging Behavior

When a capacitor charges, voltage increases exponentially toward the battery voltage.

Current decreases exponentially as the capacitor fills up.

Voltage During Charging

$$V_C(t) = V_0(1 - e^{-t/\tau})$$

Current During Charging

$$I(t) = \frac{V_0}{R}e^{-t/\tau}$$

During charging, the capacitor voltage starts at zero and approaches the battery voltage exponentially. The current starts at its maximum value and decreases exponentially to zero.

Example: Charging Analysis

Problem: In an RC circuit with τ = 2s and V₀ = 12V, find the capacitor voltage after 4 seconds.

Step 1: Use Charging Formula

$$V_C(t) = V_0(1 - e^{-t/\tau})$$ $$V_C(4s) = 12V(1 - e^{-4s/2s})$$

Step 2: Calculate Exponential

$$V_C(4s) = 12V(1 - e^{-2})$$ $$V_C(4s) = 12V(1 - 0.135)$$ $$V_C(4s) = 12V(0.865) = 10.38V$$

Answer

The capacitor voltage is 10.38V after 4 seconds.

Discharging Process

Differential Equation for Discharging

When the switch opens, the battery is disconnected. Kirchhoff's voltage law becomes:

$$0 = V_R + V_C = IR + \frac{Q}{C}$$

Since I = dQ/dt:

$$R\frac{dQ}{dt} + \frac{Q}{C} = 0$$

Rearranging:

$$\frac{dQ}{dt} + \frac{Q}{RC} = 0$$

This is a first-order linear differential equation with solution:

$$Q(t) = Q_0e^{-t/RC}$$

Where Q₀ is the initial charge on the capacitor.

Deriving V(t) for Discharging

Since V = Q/C and Q₀ = CV₀:

$$V_C(t) = \frac{Q(t)}{C} = \frac{Q_0e^{-t/RC}}{C} = \frac{CV_0e^{-t/RC}}{C} = V_0e^{-t/RC}$$

With τ = RC:

$$V_C(t) = V_0e^{-t/\tau}$$

Deriving I(t) for Discharging

Since I = dQ/dt:

$$I(t) = \frac{d}{dt}[Q_0e^{-t/RC}] = Q_0 \cdot \frac{d}{dt}(e^{-t/RC})$$

Using the chain rule:

$$I(t) = Q_0 \cdot (-\frac{1}{RC})e^{-t/RC} = -\frac{Q_0}{RC}e^{-t/RC}$$

Since Q₀ = CV₀:

$$I(t) = -\frac{CV_0}{RC}e^{-t/RC} = -\frac{V_0}{R}e^{-t/RC}$$

With τ = RC:

$$I(t) = -\frac{V_0}{R}e^{-t/\tau}$$

The negative sign indicates current flows in the opposite direction during discharging.

🎯 Discharging Behavior

When a capacitor discharges, voltage decreases exponentially toward zero.

Current also decreases exponentially but flows in the opposite direction.

Voltage During Discharging

$$V_C(t) = V_0e^{-t/\tau}$$

Current During Discharging

$$I(t) = -\frac{V_0}{R}e^{-t/\tau}$$

During discharging, the capacitor voltage starts at its initial value and decreases exponentially to zero. The current flows in the opposite direction and also decreases exponentially.

Example: Discharging Analysis

Problem: A capacitor initially charged to 9V discharges through a resistor. If τ = 3s, find the voltage after 6 seconds.

Step 1: Use Discharging Formula

$$V_C(t) = V_0e^{-t/\tau}$$ $$V_C(6s) = 9Ve^{-6s/3s}$$

Step 2: Calculate Exponential

$$V_C(6s) = 9Ve^{-2}$$ $$V_C(6s) = 9V(0.135) = 1.22V$$

Answer

The capacitor voltage is 1.22V after 6 seconds.

Key Time Points

Important Time Intervals

Example: Time Point Analysis

Problem: A 10V capacitor discharges with τ = 1s. Find the voltage at t = 2τ.

Step 1: Calculate Time

$$t = 2\tau = 2(1s) = 2s$$

Step 2: Use Discharging Formula

$$V_C(2s) = 10Ve^{-2s/1s} = 10Ve^{-2}$$ $$V_C(2s) = 10V(0.135) = 1.35V$$

Verification

At t = 2τ, the voltage should be 13.5% of the initial value, which matches our calculation.

Answer

The voltage is 1.35V at t = 2τ.

This video really helps sum everything up:

Interactive RC Circuit Simulation

RC Circuit Parameters

12V
10kΩ
100μF

Circuit Values

Time Constant (τ): 1.0s

Capacitor Voltage: 0.0V

Current: 0.0mA

Time Elapsed: 0.0s

Status: Ready

Real-time Graphs

Voltage vs Time
Current vs Time
Charge vs Time

Key Takeaways