#41 Transient Analysis #41 — RL natural-response theory

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

1. 1. Physical setup of the RL natural response

   

   Zero-input RL natural response

   Initial inductor current is I₀

   $i_{\mathrm{L}}(0⁻)=i_{\mathrm{L}}(0⁺)=I₀$

   There is no external source for t≥0

   The only energy source is w_L(0)=½LI₀²

   The inductor releases energy through R

   We seek i(t)

   Current cannot fall to zero instantaneously

   The natural response is decay of magnetic energy

   Narration transcript

   Now we examine the RL natural response. We have an inductor with initial current I-zero and a resistor. The switch opens at t equals zero. Before that, the inductor current is steady at I-zero. At the moment the switch opens, no external source drives the circuit. Only the energy stored in the inductor's magnetic field exists. The inductor will discharge its energy through the resistor, and we need to find i of t for t greater than zero.

2. 2. RL equation and exponential solution

   

   Apply KVL to the RL loop

   $v_{\mathrm{L}}+v_{\mathrm{R}}=0$

   L·di/dt+R·i=0

   di/dt+(R/L)i=0

   Separate: di/i=−(R/L)dt

   $\ln i=-t/\tau+K$

   $i(t)=A\cdot e^{(}-t/\tau)$

   $i(0)=I₀ \Rightarrow A=I₀$

   $i(t)=I₀e^{(}-t/\tau), \tau=L/R$

   Narration transcript

   Apply Kirchhoff's voltage law around the loop: V-L plus V-R equals zero. Substituting V equals L times d-i-d-t and V equals i-R, we get L times d-i-d-t plus i times R equals zero. Rearrange: d-i-d-t plus R-over-L times i equals zero. This is a first-order linear homogeneous ODE. Let's define the time constant tau equals L over R — notice it has units of seconds, just like R-C. Separate variables: d-i over i equals negative one-over-tau times d-t. Integrate both sides: natural log of i equals negative t over tau plus a constant. Solve for i: i of t equals A times e to the power negative t over tau. Apply the initial condition: at t equals zero, i equals I-zero. Therefore A equals I-zero. Our final solution: i of t equals I-zero times e to the negative t over tau.

3. 3. RL time constant

   

   τ=L/R is the current-decay speed scale

   $H/\Omega=(V\cdot s/A)/(V/A)=s$

   $i(\tau)=I₀e⁻¹=0.368I₀$

   Initial slope di/dt(0)=−I₀/τ

   The initial tangent hits the axis at t=τ

   Increasing L increases τ

   Larger L means slower current change

   Increasing R decreases τ

   Larger R dissipates energy faster

   Narration transcript

   The time constant tau equals L over R determines the rate of decay. At t equals tau, the current falls to thirty-six point eight percent of its initial value — that's one over e. The unit check: L has units of henries, which is volt-seconds per ampere. R has units of volts per ampere. So L over R gives seconds. The initial slope of the curve is d-i-d-t at t equals zero, which equals negative I-zero over tau. We can use this graphically: draw a tangent line to the exponential curve at t equals zero. That tangent line will cross the time axis exactly at t equals tau. This is a practical way to determine tau from experimental measurements. Larger L means the inductor stores more energy and resists rapid current change, so tau is larger and decay is slower. Larger R means faster dissipation, so tau is smaller and decay is quicker.

4. 4. RL voltage, power, and energy

   

   Derive all quantities from current

   $i(t)=I₀e^{(}-t/\tau)$

   $v_{\mathrm{R}}(t)=I₀R\cdot e^{(}-t/\tau)$

   $v_{\mathrm{L}}(t)=-I₀R\cdot e^{(}-t/\tau)$

   v_L+v_R=0 at every instant

   $p_{\mathrm{R}}(t)=I₀²R\cdot e^{(}-2t/\tau)$

   w_L(t)=½LI₀²e⁽−2t/τ)

   ∫₀∞p_Rdt=½LI₀²

   All magnetic-field energy becomes heat in R

   Narration transcript

   Here are all the response quantities for the RL natural response. The inductor current decays as i of t equals I-zero times e to the negative t over tau. The voltage across the resistor is V-R equals i times R equals I-zero times R times e to the negative t over tau. We can write this as V-zero times e to the negative t over tau, where V-zero equals I-zero times R is the initial voltage. The voltage across the inductor is V-L equals negative V-zero times e to the negative t over tau. Notice the negative sign — the inductor opposes the decrease in current by creating a reverse voltage. Kirchhoff's voltage law holds: V-L plus V-R equals zero at all times. The power dissipated in the resistor is p equals i-squared times R equals I-zero-squared times R times e to the negative two-t over tau. Power decays twice as fast as current because it depends on i-squared. Finally, the magnetic energy stored in the inductor is w equals one-half L times i-squared equals one-half L times I-zero-squared times e to the negative two-t over tau. Over time, all this magnetic energy is converted to thermal energy — heat dissipated in the resistor.