Answer to Question #167951 in Electrical Engineering for jim

Question #167951

An impedance coil having a resistance of 15 ohms and an inductance of 0.08 henry is connected in series with a variable capacitor. If a 117 volt 60 cps source is connected to the circuit, what will be the capacitance of the capacitor and the current

(a) when the circuit has a power factor of 0.866 leading and (b) when the voltage drop across the coil is equal to the voltage drop across the capacitor?

Expert's answer

Explanations & Calculations

  • Consider the sketch attached

  • Knowns are

"\\qquad\\qquad\n\\begin{aligned}\n\\small R_L&= \\small 15\\Omega\\qquad\\qquad\\qquad X_L=2\\pi fL=30.16\\Omega\\\\\n\\small \\cos\\theta&=\\small 0.866\\to\\theta =30^0 (for\\,part\\,a\\,only)\n\\end{aligned}"

  • All the resistance, inductance & capacitance are connected in series hence the same current flows through each.


  • From the phasor diagram for part (a),

"\\qquad\\qquad\n\\begin{aligned}\n\\small V_R&= \\small 117\\cos\\theta =117\\times 0.866=101.32V\n\\end{aligned}"

  • Then by "\\small V=iR",

"\\qquad\\qquad\n\\begin{aligned}\n\\small i&= \\small \\frac{101.32V}{15\\Omega}=\\bold{6.75A}\n\\end{aligned}"

  • Then the voltage drop across the inductive part is,

"\\qquad\\qquad\n\\begin{aligned}\n\\small V_L&= \\small iX_L=203.58\\,V\n\\end{aligned}"

  • Finally,

"\\qquad\\qquad\n\\begin{aligned}\n\\small 117\\sin\\theta&= \\small V_c-V_L\\\\\n\\small V_c&= \\small 117\\sin30+203.58=262.08V\\\\\n\\small C&= \\small \\frac{6.75A}{2\\pi\\times 60Hz\\times 262.08V}\\\\\n&= \\small \\bold{68.32\\,\\mu F}\n\\end{aligned}"


  • By refering to the phasor diagram for part (b),

"\\qquad\\qquad\n\\begin{aligned}\n\\small V_{coil}&= \\small V_c\\\\\n\\small iZ_{coil}&= \\small iX_c\\\\\n\\small \\sqrt{R^2+X_L^2}&=\\small X_c=\\frac{1}{2\\pi f C}\\\\\n\\small C&=\\small \\frac{1}{2\\pi \\times 60\\times \\sqrt{15^2+30.16^2}}\\\\\n&= \\small \\bold{78.75\\,\\mu F}\n\\end{aligned}"

  • Then capacitive reactance is

"\\qquad\\qquad\n\\begin{aligned}\n\\small X_c&= \\small\\frac{1}{2\\pi f C}\\\\\n&= \\small \\frac{1}{2\\pi\\times 60\\times 78.75\\times10^{-6} }\\\\\n&= \\small 33.68 \\,\\Omega \n\\end{aligned}"

  • FInally,

"\\qquad\\qquad\n\\begin{aligned}\n\\small 117^2 &= \\small V_R^2+(V_c-V_L)^2\\\\\n&=\\small i^2\\big[R^2+(X_c-X_L)^2\\big]\\\\\n\\small i&= \\small \\frac{117}{\\sqrt{15^2+(33.68-30.16)^2}}\\\\\n&= \\small \\bold{7.59\\,A}\n\\end{aligned}"

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