Problem 1:

Consider the LCR circuit shown. The circuit is driven by a sinusoidal voltage V(t) = V₀exp(iωt), where ω = ω₀ = 1/√(LC) is the resonant frequency of the circuit.
(a)  Solve for the current I(t) in the circuit.
(b)  Find the voltage V_AB(t) where V_AB = V_A - V_B.

Solution:

• Concepts:
  AC circuits
• Reasoning:
  We are asked to analyze an AC circuit.
• Details of the calculation:
  (a)  V = IZ.
  Z = R + iωL + 1/(iωC) = R + i(ωL - 1/(ωC)) = (R² + (ωL - 1/(ωC))²)^½exp(iφ).
  φ = tan^-1((ω²LC - 1)/(ωRC)).
  I = V/Z = [V₀/(R² + (ωL - 1/(ωC))²)^½]exp(i(ωt - φ)).
  If ω = ω₀ = 1/√(LC) then I = [V₀/R]exp(iω₀t).
  
  (b)  V_AB = I*Z_C = [V₀/R]exp(iω₀t)Z_C = -i[V₀/(ωRC)]exp(iω₀t) = V₀/(ωRC)] exp(iωt - π/2).
   

Problem 2:

(a)  Derive the expression and calculate the capacitance per unit length for a piece of coaxial cable in SI units.  The coaxial cable has a diameter of an inner wire of 1 mm and an inner diameter (ID) of an outer shield of 5 mm. The dielectric constant of the insulator between the two conductors is κ_e = 1.5.  The dielectric permeability of the free space is  ε₀ = 8.8*10^-12 F/m.
(b)  Derive the expression and calculate the inductance for this cable in SI units, assuming that the relative magnetic permeability is unity κ_m  = 1.  The magnetic permeability of the free space is
μ₀ = 4π*10^-7 H/m.  For problems (a) and (b) assume that the skin depth of the wire and shield are zero (a high-frequency limit).
(c)  Derive the expression and calculate the wave impedance for this cable.

Solution:

• Concepts:
  Gauss' law, Ampere's law, ladder circuits
• Reasoning:
  The problem has cylindrical symmetry.  Treat the circuit as an infinite ladder network with characteristic impedance Z.
• Details of the calculation:
  (a)  For a given charge on the cylinder we find D from Gauss' law for D.  We then use D to find E, V, and C.  Assume the inner cylinder carrier a free charge per unit length λ_f.
  D = (λ_f/2πr)(r/r) (a < r < b) from Gauss' law for D.  Here r is a cylindrical coordinate and we use SI units.
  E = (λ_f/2πεr)(r/r).
  ΔV = ∫_a^bEdr = ∫_a^b(D/ε)dr = (λ_f /(2πε))ln(b/a).
  C' = λ_f/ΔV = (2πε)/ln(b/a) is the capacitance per unit length.  Here ε = 1.5, a = 0.5 mm and b = 2.5 mm.
  C' = 53 pF/m.
  
  (b)  Now assume the inner cylinder carries a free current I.
  We find the magnetic field produce by the current from Ampere's law for H, use H to find B, and solve ½LI² = (1/(2μ))∫ B²dV for the self inductance L.
  Ampere's law:  H = I/(2πr) between the conductors, B = μI/(2πr)  B encircles the inner conductor according to the right hand rule. 
  U_{per unit lngth} = ½L'I² = (μI²/(4π)) ∫_a^b (1/r) dr = (μI²/(4π)) ln(b/a).  Here μ = μ₀.
  L' = (μ/(2π)) ln(b/a) is the self-inductance per unit length.  L' = 3.22*10^-7 H/m.
  
  (c)  Treat the circuit as an infinite ladder network with characteristic impedance Z.
  Since the ladder is infinite, the impedance Z will not change if an additional section is added to the front of ladder. 
  An equivalent network with impedance Z is  shown in the figure.  
  
  Here Z₁ = iωL'dl and Z₂ = -i/(ωC'dl).
  Z = Z₁ + ZZ₂/(Z + Z₂) = iωL'dl - iZ/(ωC'dl)/ (Z - i/(ωC'dl))
  = iωL'dl  + Z/(iωC'dlZ + 1).
  Since dl --> 0, we may expand (1 + iωC'dlZ)^-1 =  (1 - iωC'dlZ).
  Z = iωL'dl  + Z(1 - iωC'dlZ).  0 = L'  -  C'Z².  Z² = L'/C'.
  The wave impedance of the cable is (L'/C')^½ = [ln(b/a)/(2π)] (μ/ε)^½ = 78 Ω.

Problem 3:

In the circuit shown, the switch has been in position a for a very long time.  At t = 0, it abruptly switches to position b.  Find the expression for the capacitor voltage V_C as a function of time.

Let V = 50 V, R₁ = 500 Ω, R₂ = 125 Ω,  R₃ = 25 Ω, C = 2*10^-6 F, I₁ = 0.1 A, and I₂ = 0.2 A.

Solution:

• Concepts:
  RC circuits
• Reasoning:
  For t < 0, the portion of the circuit to the left of the switch  is not connected to the capacitor. The circuit to the right of the switch determines V_C.  Since the switch was in position a for a long time, the capacitor current is zero and V_C = I₂R₃ = 5 V.
  After the flip, the circuit to the left will determine V_C.
• Details of the calculation:
  For t > 0, the portion of the circuit to the right of the switch  is not connected to the capacitor.
  The Thevenin voltage V_TH is an ideal voltage source equal to the open circuit voltage at the terminals.  The Thevenin resistance R_TH is the resistance measured at the terminals with all voltage sources replaced by short circuits and all current sources replaced by open circuits.
  Assume the negative terminal of the voltage source is at ground.
  Assume that point b is at voltage V_b = V_Th.
  Use Kirchhoff's junction rule for the node above R₂.
  (V - V_b)R   + I₁ = V_b/R₂.
  The Thevenin equivalent voltage and resistance of the circuit on the left are
  V_Th = VR₂/(R₁ + R₂) + I₁R₁R₂/(R₁ + R₂) = 20 V.
  R_Th = R₁R₂/(R₁ + R₂) = 200 Ω.
  
  Now we just have a simple RC circuit.
  V_C(t) = V_f - (V_f - V_i)exp(-t/RC) = 20V - 15Vexp(-t/2*10^-4s).
  The capacitor starts at 5 V and charges to 20 V with a time constant of 200 μs.

Problem 4:

In the circuit shown below, two conducting spheres of radius R each are located far away from each other and are connected by thin wires through a solenoid with inductance L.  Initially, one of the spheres is charged and the other is neutral.  How long after the switch is closed do the charges on the spheres become equal?

Solution:

• Concepts:
  LC tank circuits
• Reasoning:
  We treat the spheres as capacitors, each with C_sp = 4πε₀R.  The capacitors are in series (the connection is at infinity), so we have an LC tank circuit with C = C_sp/2 = 2πε₀R. 
  
  The circuit is not neutral, it carries an excess charge Q approximately uniformly distributed over the two spheres.  Denote the oscillating charge by Q' .
• Details of the calculation:
  
  Kirchhoff's loop rule:  Q'/C + LdI/dt = 0.  d²Q'/dt² + Q'/(LC) = 0.
  Q'  = (Q/2)cosωt,  ω = (LC)^-½.
  Q/2 is the oscillating charge on the initially charged sphere.
  The oscillation period is T = 2π/ω = 2π(LC)^½ = (2π)^3/2(ε₀RL)^½.
  The charges on the spheres become equal for the first time after one quarter period, T/4 = (π/2)(2πε₀RL)^½