Problem 2:

An AC source provides a voltage with amplitude 230 V with frequency 50 Hz.  A circuit contains one coil, two same capacitors and a resistor connected as shown in the figure.  All circuit elements have an unknown impedance.  With the switch open, the voltage lags behind current of the generator by 20^o.  If the switch is in the position 1, the voltage leads the current of by 20^o.  With the switch is in position 2, the current in the circuit has amplitude 2 A.

Find the resistance of the resistor, the inductance of the coil and the capacitance of the capacitor.

Solution:

• Concepts:
  AC circuits
• Reasoning:
  Depending on the switch position, we have a series RLC or LC circuit and a AC generator, generating a sinusoidal voltage.
• Details of the calculation:
  In general: V = IZ = I |Z| exp(iφ).
  (i)  Switch open:  Z = R + iωL + 1/(iωC) = R + i(ωL - 1/(ωC)) = (R² + (ωL - 1/(ωC))²)^½exp(iφ)
  tan φ = ((ωL - 1/(ωC))/R) = tan(-20^o) = -0.364.
  
  (ii)  Switch in position 1:  The two capacitors are in parallel.
  tan φ' = ((ωL - 1/(2ωC))/R) = tan(20^o) = +0.364.
  
  (iii)  Switch in position 2:  We have a LC circuit. 
  Z = iωL + 1/(iωC) = i(ωL - 1/(ωC)).
  The voltage and current are out of phase by ±π/2, depending on the sign of ωL - 1/(ωC).
  |I| = 230 V/|ωL - 1/(ωC)| = 2A.
  We have 3 equations for 3 unknowns.
  
  from (iii):  ωL - 1/(ωC) = ±115 Ω.
  from (i): (ωL - 1/(ωC))/R = -0.364  --> R = 316 Ω,  ωL < 1/(ωC).
  from (i) and (ii): ωL = 3/(4ωC).
  Using ωL = -115 Ω + 1/(ωC) we have
  C = 1/(4*ω*115) = 1/(3*50*2π*115) F = 6.91*10^-6 F, 
  L = 1.1 H.

Problem 3:

Consider the impedance bridge shown in the figure below.  Its purpose is to permit measurements of an unknown impedance Z_u in terms of the fixed resistance R_A and R_B, variable resistance R_S and variable capacitance C_S.  If Z_u is a pure resistance, then C_S may be removed from the circuit (shorted out) and the impedance bridge becomes a simple Wheatstone bridge.
(a)  For a purely resistive impedance Z_u find the value of R_S for which a balance is obtained (no current on the null detector).
(b)  For a complex impedance Z_u whose resistive component is R_u and whose reactive component is X_u, find the values of R_S and C_S for which balance is obtained.  Assume Z_u has zero inductance. 
(c)  Show that for a purely inductive component Z_u balance is not possible.

Solution:

• Concepts:
  AC circuits
• Reasoning:
  The circuit shown is a simple, two terminal, AC circuit.
• Details of the calculation:
  (a)  The circuit in part (a) is equivalent to the circuit shown below.
  
  We want V_b = V_a.
  V = I₁(R_B + R_S) = I₂(R_A + R_u),  We want I₁R_B = I₂R_A,  I₁R_S = I₂R_u.
  R_S = (R_B/R_A)R_u.
  (b)  The circuit in part (b) is equivalent to the circuit shown below.
  
  We want V_b = V_a.  We want to find Z_s in terms of Z_u.
  We use the same arguments as in part (a) to show
  Z_S = (R_B/R_A)Z_u = (R_B/R_A)(R_u + iX_u).
  Z_S = (R_B/R_A)R_u + iR_BX_u/R_A = R_S - i/(ωC_S).  R_S = (R_B/R_A)R_u,  C_s = -R_A/(R_BX_uω).
  If Z_u = (R_u + 1/(iωC_u)) then Xu = -1/(ωC_u), R_s = (R_B/R_A)R_u,  C_s = R_AC_u/R_B.
  (c)  If Z_u = iωL_u, then X_u = ωL_u and C_s = -R_A/(R_BL_uω²), which is impossible, since all constants are positive.

Problem 4:

A 10 μF capacitor that is initially uncharged is connected in series with a 5 Ω resistor and an emf source with ε = 50 V and negligible internal resistance.  At the instant when the energy stored in the capacitor is 5*10^-4 J, what is the rate at which the resistor is dissipating electrical energy.

Solution:

• Concepts:
  Simple transient circuits
• Reasoning:
  When any closed circuit loop is traversed, the algebraic sum of the changes in the potential must equal zero.
• Details of the calculation:
  U_C = ½Q²/C = 5*10^-4 J.  (Q/C)² = (10^-3J)/(10 μF) = 100 V².
  ε - IR - Q/C = 0.  IR = ε - Q/C = 50 V - 10 V = 40 V.
  P = I²R = (IR)2/R = 320 W.