Problem 1:

In the below circuit, the switch has been closed for a long time before t = 0.  The switch then opens at t = 0, and then closes at t = 100 μs.  Find and plot v_out(t).

Solution:

• Concepts:
  Transient circuits, Thevenin equivalent circuits
  
• Reasoning:
  We are asked to analyze the transient behavior of an RC circuit.
  
• Details of the calculation:
  There are different approaches to analyzing this circuit, Kirchhoff's rules or Thevenin equivalent circuits.
  
  Use SI units, measure resistance in Ω, voltages in V and inductances in H.  Let positive currents flow clockwise.  Let I₁ be the current flowing through the battery, I₂ be the current flowing through the inductor and I₃ be the current flowing through the resistor on the right.
  For t < 0 the inductor acts like a short circuit, the voltage across the inductor v_out = 0.  The current flowing through the inductor is 0.1 A.
  
  For 0 < t < 100 μs:
  After the switch is opened the inductor and resistor on the right form a simple RL circuit. 
  A current  I₂(0) = 0.1 A flows counterclockwise in this circuit at t = 100 μs and decays exponentially with a time constant τ = 1mH/100 Ω = 10 μs.
  I₂(0) = 0.1 A exp(t/(10 μs)),  I₃(0) = -0.1 A exp(t/(10 μs)), using the sign convention established above.
  v_out = I₃ 100 Ω = -10 V exp(t/(10 μs)).
  At t = 100 μs we have  v_out = 4.54*10^-4 V ≈ 0 ,  I₂ = 4.54*10^-6 A ≈ 0 .
  
  (c)  Using Kirchhoff's rules:
  For t > 100 μs or t' = t - 100 μs > 0:
  We can assume that no currents are flowing, v_out = 0.
  10 - 100 I₁ - 100 I₃ = 0.
  LdI₂/dt' - 100 I₃ = 0.
  I₁ = I₂ + I₃.
  Eliminate  I₁ from the first equation:
  10 - 100(I₂ + I₃) + 100 I₃ = 0.  I₂ = 0.1 - 2 I₃.
  Use  the second equation to eliminate I₂:
  2LdI₃/dt' + 100 I₃ = 0.  dI₃/dt = -5*10⁴ I₃.
  I₃ = I₃(0)exp(-5*10⁴ t').
  Initial condition:  I₃(0) = 10/200 A = 0.05 A.
  v_out = I₃100 V = 5 exp(-5*10⁴ t') V.
  The time constant for the circuit is 20 μs.
  
  (ii)  Using Thevenin equivalent circuits:
  Consider the inductor to be the load, v_out(t) = V_L(t).
  ε_eff = (10 V) 100/200 = 5 V.
  Z_eff = 50 Ω.  (We have 2 resistors in parallel)
  
  The equivalent circuit is just an simple Rl circuit.
  v_out(t') = V_L('t) = 5 exp(-Z_efft'/L) V = 5 exp(-5*10⁴ t') V.

Problem 2:

In the below circuit, the switch has been open for a long time before t = 0.  The switch is then closed at t = 0.  Find the voltage across the capacitor at t = 60 ms.

Solution:

• Concepts:
  Transient RC circuit
  
• Reasoning:
  We are asked to analyze the transient behavior of an RC circuit.
  
• Details of the calculation:
  There are different approaches to analyzing this circuit.
  (i)  Using Kirchhoff's rules:
  Use SI units, measure resistance in Ω, voltages in V and C in Farad.  Let positive currents flow clockwise.  Let I₁ be the current flowing through R₁, I₂ be the current flowing through R₂ and I₃ be the current flowing into the capacitor.
  For t > 0 we have
  20 - 5*10³I₁ - 5*10²I₂ = 0.
  Q/C - 5*10²I₂ = 0.
  I₂ + I₃ = I₁.  I₃ = dQ/dt.
  
  Eliminate  I₁ from the first equation:
  20 - 5000(I₂ + I₃) - 500I₂ = 0.  20 - 5000I₃ - 5500I₂ = 0.
  Use  the second equation to eliminate I₂:
  I₂ = Q/(500C) = Q/(5*10^-2).
  20 - 5000I₃ - 1.1*10⁵Q = 0.  I₃ = 4*10^-3 - 22Q = dQ/dt
  
  Assume Q = A - Bexp(-t/x).
  dQ/dt = B/x = -Q/x + A/x
  x = 1/22, A = 4*10^-3/22
  Q(t) = 4*10^-3/22 - Bexp(-22t).
  Q(0) = 0, B = 4*10^-3/2.2.
  V_C(t) = 1.82 V -  1.82 V exp(-22t).
  At t = 60 ms we have V_C(60) = (1.82 V -  1.82 exp(-22*0.06)) V = 1.33 V.
  
  (ii)  Using Thevenin equivalent circuits:
  ε_eff = VR₂/(R₁ + R₂ + R₃) = 20*0.5/5.5 = 1.82.
  Z_eff = (R₁ + R₃)R₂/(R₁ + R₂ + R₃) = 25*10²/5.5.
  
  The equivalent circuit is just an simple RC circuit.
  V_C(t) = 1.82V - 1.82 V exp(-t/0.0455).

Problem 3:

(a) The circuit shown below is at steady state.  The input currents are i₁(t) =  10 cos(25 t) mA  and i₃(t) =  10 cos(25t + 135^o) mA.  Determine the voltage v₂(t).

(b) The circuit shown below is at steady state.  The inputs to this circuit are the current source current i₁(t) =  0.12 cos(100 t + 45^o) A and the voltage source v₂(t) =  24 cos(100 t -  60^o) V.  Determine the current i₂(t).

Solution:

• Concepts:
  AC circuits. Kirchhoff's laws
  
• Reasoning:
  All currents and voltages vary sinusoidaly
  
• Details of the calculation:
  (a)  i₁(t) = I₁e^iωt, i₂(t) = I₂e^iωt, i₃(t) = I₃e^iωt,v₂(t) = V₂e^iωt.
  I₁ = 10, I₃ = 10e^i3π/4 = 10^-2 cos(3π/4) + i10^-2 sin(3π/4), ω = 25.  (All quantities are in SI units.)
  Junction rule:  I₂ = I₁ - I₃ = 10^-2(1 - cos(3π/4)) - i10^-2 sin(3π/4) = 17.07*10^-3 - i0.707*10^-3 = 18.478*10^-3 e^-i0.393.
  V₂ = 250*I₂ = 4.619e^-i0.393.  v₂(t) = 4.619 cos(25t - 22.5^o).
  
  (b)  i₁(t) = I₁e^iωt, i₂(t) = I₂e^iωt, i₃(t) = I₃e^iωt,v₂(t) = V₂e^iωt.
  I₁ = 0.12e^iπ/4, V₂ = 24e^-iπ/3 = 24 cos(π/3) - i24 sin(π/3), ω = 100. ( All quantities are in SI units.)
  Junction rule:  I₂ =  I₁ - V₂/96 = 0.12 cos(π/4) - 0.25 cos(π/3) + i[0.12 sin(π/4) - 0.25 sin(π/3)].
  I₂ = -0.04 - i0.301 = 0.304*e^i1.703 =  0.304 cos(25t + (180 + 97.59)^o)

Problem 4:

Let V_in = Re(V₀exp(iωt)).  Derive an expression for V_out/V_in as a function of frequency for the resonator circuit below and determine the resonant frequency and the phase shift of V_out at resonance from this expression.

Solution:

• Concepts:
  AC circuits
• Reasoning:
  We are asked to analyze an AC circuit.
• Details of the calculation:
  Let V = V₀e^iωt.  
  I = V_in/Z_total,  V_out = IZ_LC = V_inZ_LC/Z_total.
  1/Z_LC  = 1/Z_C + 1/Z_L =  1/iωL + iωC = -i(1 - ω²LC)/(ωL).
  Z_LC =  iωL/(1 - ω²LC).
  Z_total = R + Z_LC,  Z_total/Z_LC = 1 + R/Z_LC = 1 - iR(1 - ω²LC)/(ωL)
  = [1 + (R/(ωL))²(1 - ω²LC)²]^½exp(-iφ) with tanφ = R(1 - ω²LC)/(ωL).
  Z_LC/Z_total = ωL/[(ωL)²+ R²(1 - ω²LC)²]^½exp(iφ).
  V_out = V_inωL exp(iφ)/[(ωL)²+ R²(1 - ω²LC)²]^½.
  |V_out| is maximum at resonance, when 1 - ω²LC = 0, ω² = 1/(LC)^½, f = 1/[2π(LC)^½].
  Then tanφ = 0, the phase shift is zero.
  V_out = V_in at resonance.  No current flows through the resistor.