"Solution:\\\\A)\\;P=\\frac{U^2}{R}=\\frac{120^2}{160}=90W;\\\\B)\\;\\frac{1}{4}P=I^2R; P=4I^2R;I=\\frac{U}{Z}\\\\P=\\frac{4U^2}{Z^2}R; Z=2U\\sqrt{\\frac{R}{P}}=240\\times\\sqrt{\\frac{160}{90}}=320\n\\Omega;\\\\Z=\\sqrt{R^2+X_{L_{max}}^2};\\\\X_{L_{max}}=\\sqrt{Z^2-R^2}=277.13\\Omega;\\\\X_{L_{max}}=2\\pi fL_{max};\\\\L_{max}=\\frac{X_{L_{max}}}{2\\pi f}=\\frac{277.13}{6.28\\times60}=0.735H;\\\\Answer:L_{max}=0.735H"
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