#39 Transient Analysis #39 — Capacitor/inductor combinations and initial conditions

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

1. 1. Series and parallel capacitors

   

   Ideal capacitor equivalents

   Series: the same charge q

   $1/C_{\mathrm{e}}q=1/C_{1}+1/C_{2}$

   Two elements: C_eq=C₁C₂/(C₁+C₂)

   Parallel: the same voltage v

   $C_{\mathrm{e}}q=C_{1}+C_{2}$

   Capacitor rules are opposite to resistor rules

   Narration transcript

   When capacitors are connected in series, the equivalent capacitance is the reciprocal sum. One over C-eq equals one over C-one plus one over C-two — just like resistors in parallel. For two capacitors: C-eq equals C-one times C-two, divided by C-one plus C-two. When capacitors are in parallel, the equivalent capacitance is simply the sum: C-one plus C-two. This is opposite to resistors: series capacitors combine like parallel resistors, and parallel capacitors combine like series resistors.

2. 2. Capacitor voltage and current division

   

   Equal charge in series; equal voltage in parallel

   Series: V₁=q/C₁ and V₂=q/C₂

   $V_{1}=V_{\mathrm{s}}\cdot C_{2}/(C_{1}+C_{2})$

   The smaller C carries the larger voltage

   Parallel: i_k=C_k·dv/dt

   $I_{1}=I_{\mathrm{T}}\cdot C_{1}/(C_{1}+C_{2})$

   The larger C carries the larger current

   Narration transcript

   Voltage division with series capacitors has a surprising twist. V-one equals V-source times C-two over C-one plus C-two — notice it's the OTHER capacitor in the numerator. This is the INVERSE of resistor voltage division. The smaller capacitor gets MORE voltage, because it stores the same charge but with less capacitance. Current division in parallel capacitors is direct: I-one equals I-total times C-one over C-one plus C-two. The larger capacitor carries more current — this makes sense since I equals C times d-V-d-t, and both capacitors share the same voltage.

3. 3. Series and parallel inductors

   

   Ideal, magnetically uncoupled inductors: M=0

   Series: L_eq=L₁+L₂

   The same current flows in series

   The same voltage appears in parallel

   $1/L_{\mathrm{e}}q=1/L_{1}+1/L_{2}$

   Two elements: L_eq=L₁L₂/(L₁+L₂)

   Magnetic coupling introduces mutual inductance terms

   Narration transcript

   Inductors combine exactly like resistors. Series inductors add directly: L-eq equals L-one plus L-two. Parallel inductors use the reciprocal formula: one over L-eq equals one over L-one plus one over L-two. This is the exact dual of capacitors — swap series and parallel, and you swap the C and L combination rules. Series inductors are simple addition, just like series resistors.

4. 4. Capacitor–inductor combination duality

   

   Compare C and L combination rules

   Series C: reciprocal sum

   Series L: direct sum for M=0

   Parallel C: direct sum

   Parallel L: reciprocal sum for M=0

   Series-capacitor voltage divides inversely

   Series ↔ parallel and C ↔ L duality

   Narration transcript

   Let's compare capacitor and inductor combination rules side by side. Series capacitors use reciprocal sum, but series inductors add directly. Parallel capacitors add directly, but parallel inductors use reciprocal sum. Voltage divides inversely with capacitors, but directly with inductors. This beautiful duality extends to everything we've seen about C and L — it's a fundamental symmetry in circuit theory.

5. 5. Continuity at switching

   

   Energy-storage state variables cannot jump

   $v_{\mathrm{C}}(0⁻)=v_{\mathrm{C}}(0⁺)$

   A jump in v_C requires infinite current through i=C·dv/dt

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

   A jump in i_L requires infinite voltage through v=L·di/dt

   i_C and v_L may jump

   dv_C/dt and di_L/dt may be discontinuous

   Narration transcript

   The most fundamental rules for circuit transient analysis. Capacitor voltage cannot change instantaneously: V-C at zero-minus equals V-C at zero-plus. Why? Because an instantaneous voltage change would require infinite current: I equals C times d-V-d-t. Similarly, inductor current cannot change instantaneously: I-L at zero-minus equals I-L at zero-plus. An instantaneous current change would require infinite voltage: V equals L times d-I-d-t. However, the DERIVATIVES d-V-d-t and d-I-d-t CAN be discontinuous at t equals zero.

6. 6. Three-step switching method

   

   Find initial conditions systematically

   1) Solve the t<0 circuit in DC steady state

   At DC, C is open and L is short

   Find v_C(0⁻) and i_L(0⁻)

   2) Carry values to 0+ by continuity

   3) Solve the new t=0+ circuit

   dv_C/dt=i_C/C and di_L/dt=v_L/L

   Narration transcript

   Here is the three-step method for analyzing switch circuits. Step one: Analyze the circuit for t less than zero. Find the DC steady state. In DC steady state, capacitors are open circuits and inductors are short circuits. This gives you V-C at zero-minus and I-L at zero-minus. Step two: Apply the continuity rules. V-C at zero-plus equals V-C at zero-minus. I-L at zero-plus equals I-L at zero-minus. Step three: Analyze the NEW circuit at t equals zero-plus, using V-C and I-L as known initial values. Calculate d-V-d-t at zero-plus equals I at zero-plus divided by C, and d-I-d-t at zero-plus equals V at zero-plus divided by L. This three-step method is the foundation for all transient analysis.

7. 7. RC switching example

   

   $V_{\mathrm{s}}=12 V, R=4 k\Omega, C=2 µF$

   The switch closes at t=0; capacitor initially uncharged

   $v_{\mathrm{C}}(0⁻)=v_{\mathrm{C}}(0⁺)=0 V$

   At t=0+, all 12 V is across the resistor

   $i(0⁺)=12/4000$

   i(0⁺)=3 mA

   dv_C/dt(0⁺)=3 mA/2 µF=1500 V/s

   Narration transcript

   Let's apply our method to a simple RC circuit. A 12 volt source in series with a 4 kilohm resistor and a 2 microfarad capacitor. The switch closes at t equals zero. Before that, no current flows, so V-C at zero-minus equals zero. Step two: V-C at zero-plus equals V-C at zero-minus equals zero volts. Step three: At t equals zero-plus, the capacitor voltage is zero, so all 12 volts appear across the resistor. I at zero-plus equals 12 over 4000 equals 3 milliamps. The initial rate of voltage change: d-V-d-t at zero-plus equals I over C equals 3 milliamps over 2 microfarads equals 1500 volts per second. These initial conditions are everything we need to write the complete transient response.

8. 8. RL switching example

   

   $V_{\mathrm{s}}=24 V, R=8 \Omega, L=2 H$

   Switch long closed: L is a short

   $i_{\mathrm{L}}(0⁻)=i_{\mathrm{L}}(0⁺)=24/8=3 A$

   After source removal, current continues through R

   v_L(0⁺)=−iR=−24 V

   di_L/dt(0⁺)=v_L/L

   di_L/dt(0⁺)=−12 A/s

   Narration transcript

   Now an RL circuit with opposite behavior. A 24 volt source, 8 ohm resistor, and 2 henry inductor. The switch has been closed for a long time, so the inductor is a short circuit. I-L at zero-minus equals 24 over 8 equals 3 amps. At t equals zero, the switch opens, disconnecting the voltage source. Step two: I-L at zero-plus equals I-L at zero-minus equals 3 amps — the current must continue flowing. Step three: At t equals zero-plus, the 3 amp current flows through the resistor, creating a voltage drop. V-L at zero-plus equals minus I times R equals minus 24 volts. The initial rate of current change: d-I-d-t at zero-plus equals V-L over L equals minus 24 over 2 equals minus 12 amps per second. The negative sign tells us the current is decreasing — the inductor is de-energizing.

9. 9. Combinations and initial conditions summary

   

   Preparing for transient analysis

   Series C and parallel L: reciprocal sum

   Parallel C and series L: direct sum

   v_C and i_L are continuous at switching

   First solve the 0⁻ DC steady state

   Then apply continuity and solve the 0+ circuit

   Correct initial conditions anchor the full transient response

   Narration transcript

   Four key results from this lecture. First: series capacitors use reciprocal sum, parallel capacitors add directly — opposite to resistors. Second: capacitor voltage division is INVERSE — the smaller capacitor gets more voltage. Third: the continuity rules — capacitor voltage and inductor current cannot jump at t equals zero. Fourth: the three-step switch analysis method — find DC steady state, apply continuity, then analyze the new circuit. These initial condition skills are essential for the next lecture, where we solve the RC natural response — our first complete transient solution.