#40 Transient Analysis #40 — RC natural response and capacitor discharge

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Physical setup of the RC natural response

   

   Zero-input RC natural response

   The capacitor is initially charged to V₀

   At t=0 it begins discharging through R

   There is no external source for t≥0

   The only energy source is w_C(0)=½CV₀²

   We seek v_C(t)

   The natural response is decay of initial energy

   Narration transcript

   Consider a capacitor charged to V-zero volts, connected to a resistor through a switch. At t equals zero, the switch closes. No external source is connected — so this is a zero-input response, also called the natural response. The only energy comes from the capacitor's initial charge. The question is: how does the capacitor voltage change over time?

2. 2. Derivation of the differential equation

   

   Apply KVL to the loop or KCL to the node

   $i_{\mathrm{R}}+i_{\mathrm{C}}=0$

   v/R+C·dv/dt=0

   dv/dt+v/(RC)=0

   First-order homogeneous differential equation

   Initial condition v(0)=V₀

   v(t)=V₀e⁽−t/τ), τ=RC

   Narration transcript

   Apply KVL around the loop. The voltage across the resistor is i times R, and the voltage across the capacitor is v. Since i equals C times d-v-d-t, we substitute: R-C times d-v-d-t plus v equals zero. Dividing both sides by R-C gives d-v-d-t plus v over R-C equals zero. This is a first-order linear ODE. The solution is an exponential: v of t equals V-zero times e to the minus t over R-C. We define the time constant tau equals R times C.

3. 3. Time constant and graphical meaning

   

   τ=RC is the circuit's speed scale

   $\Omega\cdot F=s$

   $v(\tau)=V₀e⁻¹=0.368V₀$

   Larger R or C ⇒ slower discharge

   Initial slope dv/dt(0)=−V₀/τ

   The t=0 tangent hits the axis at t=τ

   Only 0.674 percent remains at 5τ

   Narration transcript

   The time constant tau equals R-C has units of seconds — ohms times farads gives seconds. At t equals tau, the voltage drops to e to the minus one times V-zero — that's about 36.8 percent of the initial value. The time constant tells us HOW FAST the circuit responds. A larger tau means slower decay — more resistance or more capacitance slows the discharge. A smaller tau means faster decay. The tangent line at t equals zero would reach zero at exactly t equals tau.

4. 4. Current, power, and energy expressions

   

   Derive all quantities from v(t)

   i_C=C·dv/dt

   $i_{\mathrm{C}}(t)=-V₀/R\cdot e^{(}-t/\tau)$

   The minus sign indicates discharge in the passive reference

   $p_{\mathrm{R}}(t)=V₀²/R\cdot e^{(}-2t/\tau)$

   w_C(t)=½CV₀²e⁽−2t/τ)

   Power and energy decay at twice the exponent rate

   Narration transcript

   From the voltage, we derive all other quantities. Current: i of t equals C times d-v-d-t, which gives minus V-zero over R times e to the minus t over tau. The negative sign means current flows opposite to the charging direction — the capacitor is discharging. Power dissipated in the resistor: p of t equals i-squared times R, which equals V-zero-squared over R times e to the minus two-t over tau. Notice power decays twice as fast as voltage, because it depends on i-squared.

5. 5. Numerical RC natural-response example

   

   $R=2 k\Omega, C=100 µF, V₀=50 V$

   $\tau=(2000)(100\times 10⁻⁶)$

   $\tau=0.2 s$

   $v(t)=50e^{(}-t/0.2)=50e^{(}-5t) V$

   i_C(t)=−25e⁽−5t) mA

   At t=τ, e⁻¹≈0.368

   $v(0.2)=18.4 V$

   Narration transcript

   Example: A 100 microfarad capacitor charged to 50 volts discharges through a 2 kilohm resistor. Step one: calculate tau equals R-C equals 2000 times 100 times ten to the minus six, which equals 0.2 seconds — 200 milliseconds. Step two: v of t equals 50 times e to the minus t over 0.2 volts. Step three: i of t equals minus 50 over 2000 times e to the minus t over 0.2, which equals minus 25 milliamps times e to the minus 5-t. At t equals 0.2 seconds, v equals 50 times 0.368 equals 18.4 volts.

6. 6. Five-time-constant rule

   

   Remaining voltage V/V₀=e⁽−t/τ)

   1τ: 36.8 percent

   2τ: 13.5 percent

   3τ: 5.0 percent

   4τ: 1.8 percent

   5τ: 0.674 percent; practically zero

   In the example, 5τ=1.0 s

   Narration transcript

   The 5-tau rule is a practical guideline for engineers. At t equals one-tau, voltage is 36.8 percent of V-zero. At two-tau, 13.5 percent. At three-tau, 5.0 percent. At four-tau, 1.8 percent. And at five-tau, only 0.7 percent remains. Engineers consider the transient effectively DEAD after 5 time constants. In our example with tau equals 200 milliseconds, the capacitor is fully discharged after one second — 5 times 200 milliseconds.

7. 7. Energy conservation and graphical method

   

   All initial energy is dissipated in R

   w_C(0)=½CV₀²

   $p_{\mathrm{R}}(t)=V₀²/R\cdot e^{(}-2t/\tau)$

   ∫₀∞p_R(t)dt=½CV₀²

   Electric-field energy becomes heat

   v(t) approaches zero asymptotically

   The initial tangent estimates τ from measurements

   Narration transcript

   The energy initially stored in the capacitor is one-half C times V-zero squared. As the capacitor discharges, ALL of this energy is dissipated as heat in the resistor. We can verify: integrate p of t from zero to infinity. The integral of V-zero-squared over R times e to the minus two-t over tau gives one-half C times V-zero-squared — exactly the initial stored energy. Energy is conserved: nothing is lost, nothing is created. The capacitor's electric field energy converts entirely to thermal energy in the resistor. The voltage curve starts at V-zero and decays exponentially toward zero, never quite reaching it. The initial slope is minus V-zero over tau — draw the tangent line at t equals zero, and it crosses the time axis at exactly t equals tau. This gives us a simple graphical method to estimate the time constant from measurements. The current curve is a mirror image — it starts at its most negative value and decays toward zero with the same time constant.

8. 8. RC natural-response summary

   

   Essential RC natural-response results

   v(t)=V₀e⁽−t/RC)

   τ=RC determines the decay speed

   $i_{\mathrm{C}}(t)=-V₀/R\cdot e^{(}-t/\tau)$

   The transient is practically over after 5τ

   Initial energy becomes heat in the resistor

   Natural response: no source, but initial energy

   Narration transcript

   Three essential results from this lecture. First: the RC natural response is v of t equals V-zero times e to the minus t over tau, where tau equals R-C. Second: the time constant tau determines the speed of decay — the tangent at t equals zero hits the axis at t equals tau, and after 5-tau the transient is essentially zero. Third: all energy stored in the capacitor, one-half C V-zero-squared, is dissipated as heat in the resistor. In the next lecture, we'll see the RL natural response — the dual of everything we learned today.