More Problems

Problem 1:

A 6.00 μF capacitor that is initially uncharged is connected in series with a 5.00 Ω resistor and an emf source with ε = 50.0 V and negligible internal resistance.  At the instant when the resistor is dissipating electrical energy at a rate of 300 W, how much energy is been stored in the capacitor?

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:
  ε - IR - Q/C = 0.  P = I²R = 300 W, I = (300/5) ^½ A = 7.75 A.
  Q = Cε - IRC = 6.76*10^-5 C.
  U_C = ½Q²/C = 3.81*10^-4 J.

Problem 2:

A 45 V rms, 1000 rad/s power supply is connected in series with a 50.6 Ω resistor and a practical inductor which has 40 mH inductance and 30 Ω resistance combined.  Find
(a) the complex impedances of the inductor and the total circuit,
(b) the complex current in the circuit,
(c) the potential differences across the resistor and the inductor,
(d) the total power dissipated, and
(e) the power dissipated by the resistor and inductor.

Solution:

• Concepts:
  AC circuits
• Reasoning:
  The circuit given is a LR circuit with an AC voltage source.
• Details of the calculation:
  (a)  Z_L = (30 + i*1000*0.04) Ω = (30² + 40²)^1/2 exp(i*tan^-1(40/30)) Ω = 50*exp(i0.93) Ω.
  Circuit impedance Z_total = (50.6 + 50*cos(0.93) +i*50*sin(0.93)) Ω = (80.6 + i*40) Ω
  = (80.6² + 40²)^1/2 exp(i*tan^-1(40/80.6)) Ω = 90*exp(i0.46) Ω.
  (b)   I = V/Z = V₀exp(iωt)/Z_total = I₀exp(iωt), with V₀ = (45 V)*(√2) = 63.6 V and ω = 1000/s and the phase constant chosen to be zero.
  I₀ = (1/√2)exp(-i0.46) A.  I lags V by 26.4^o.
  (c)  V_R = RI = exp(iωt)(50.5/√2)exp(-i0.46) V (in phase with the current).
  V_L = Z_LI = exp(iωt)*50*exp(i0.93)* (1/(√2)exp(-i0.46)) V = exp(iωt)(50/√2)exp(i0.47) V.  V_L leads V by 26.9^o.
  (d)  Power dissipated P_total(avg) = I²_rmsR = 0.25*86.6 W = 20.16 W.
  (e)  Resistor power = I²_rmsR_R = 0.25*50.6 W = 12.66 W.
  Inductor power = I²_rmsR_L = 0.25*30 W = 7.5 W.

Problem 3:

For the ladder circuit shown below, assume that V_in = V₀exp(iωt).  The impedance for circuit may  be written as Z = Z_real + iZ_img.  Find Z_real.

Solution:

• Concepts:
  Ac circuits, ladder networks
• Reasoning:
  We 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. 
• Details of the calculation:
  (a)  An equivalent network with impedance Z is shown in the figure.
  
  Z = Z₁ + ZZ₂/(Z + Z₂),  Z² - Z₁Z - Z₁Z₂ = 0,  Z = Z₁/2 ± (Z₁²/4 + Z₁ Z₂)^½.
  Z₁ = 1/(iωC), Z₂ = R.
  Z = -i/(2ωC) ± (-1/(4ω²C²) - iR/(ωC))^½
  Z = -i/(2ωC) + [1/(ω²C²)(1/(16ω²C²) + R²)]^¼ exp(iφ/2), with tanφ = 4ωCR.
  Z_real = ±[1/(ω²C²)(1/(16ω²C²) + R²)]^¼cos(φ/2).
  We have to choose the plus or minus sign, so that Z_real is positive.