Mathcad 02-06 - V 120.208volt = V 170 2 volt ⋅ = a In phasor form the magnitude is given by Problem 2-6 Solution S 1.112 10 3 × volt amp ⋅ =

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V C I X c = Note the magnitude of the voltage across the inductance is larger than the source voltage magnitude. Also, the angle is 90 o ahead of the voltage across the resistance. arg V L ( 29 22.62deg = V L 231.17volt = V L 213.387 88.911j + volt = V L I X l = arg V R ( 29 67.38 - deg = V R 46.234volt = V R 17.782 42.677j - volt = V R I R = c. To find the voltage across each element, multiply current times impedance: arg I ( ) 67.38 - deg = I 9.247 amp = Note: "arg" is the angle of a complex number I 3.556 8.535j - amp = I V Z = Z 5 12j + ohm = Z R X l + X c + = X c 13j - ohm = X l 25j ohm = R 5 ohm = b. The current is the voltage divided by the impedance. Since the three elements are in series we can add them directly: We can let the voltage be our reference, so the angle is zero
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Unformatted text preview: V 120.208volt = V 170 2 volt ⋅ = a. In phasor form, the magnitude is given by Problem 2-6: Solution S 1.112 10 3 × volt amp ⋅ = reactive Q 1.026 10 3 × volt amp ⋅ = Q Im S ( ) = P 427.515 watt = P Re S ( ) = where S 427.515 1.026j 10 3 × + watt = S V I ⋅ = d. Find the complex power: Sum 120.208volt = Sum V R V L + V C + = As a check, add the voltages The capacitor voltage is 90 o behind the resistor voltage and the magnitude of the capacitor voltage is coincidentally the same as the source voltage arg V C ( 29 157.38-deg = V C 120.208volt = V C 110.961-46.234j-volt =...
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This note was uploaded on 03/26/2012 for the course ECET 231 taught by Professor Piller during the Spring '12 term at Purdue University-West Lafayette.

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Mathcad 02-06 - V 120.208volt = V 170 2 volt ⋅ = a In phasor form the magnitude is given by Problem 2-6 Solution S 1.112 10 3 × volt amp ⋅ =

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