ch2sol

ch2sol - Chapter 2 Problems Solutions 2.1 The network input...

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Unformatted text preview: Chapter 2 Problems Solutions 2.1 The network input impedance is given by 240 30 12 90 12 20 60 V j R jX I ∠ ° = = = ∠ ° = + = + ∠ - ° Z Thus, R = ( 29 12 31.83 mH 2 60 X L ϖ π = = = 2.2 The network input admittance is 20 60 0.08333 90 0.08333 240 30 I j G jB V ∠ - ° = = = ∠ - ° =- =- ∠ ° Y Thus, 1/ 1/0 R G = = → ∞ (no resistor present) ( 29 ( 29 1 1 31.83 mH 2 60 0.08333 L B ϖ π = = = 2.3 If the source voltage and current are in phase, the circuit must appear pure resistive (res- onance condition). Hence, 1 L LC ϖ- = or ( 29 ( 29 1/ 2 6 6 1 1/ 11,952.3 rad /s 140 10 50 10 LC ϖ-- = = = × × 11,952.3 1.9 kHz 2 2 f ϖ π π = = = 2.4 A copy of the screen display when using csum.m is shown below. The three compon- ents of V were entered as maximum value phasors so that SUM in polar form gives dir- ectly m V φ ∠ . 1 2.5 ( a ) It is convenient to assume V on the reference. The two branch impedances are found as ( 29 1 6 1 1 1 90 90 265.258 90 120 10 10 j C C ϖ ϖ π- ∠ - ° = - = ∠ - ° = = ∠ - ° Ω × Z ( 29 ( 29 2 10 120 0.025 10 9.425 13.742 43.3 R j L j j ϖ π = + = + = + = ∠ ° Ω Z The branch current can now be determined. 1 1 240 0 0.905 90 A 265.258 90 V I ∠ ° = = = ∠ ° ∠ - ° Z 2 2 240 0 17.465 43.3 A 13.742 43.3 V I ∠ ° = = = ∠ - ° ∠ ° Z Applying KCL, 1 2 0.905 90 17.465 43.3 I I I = + = ∠ ° + ∠ - ° 12.710 11.073 16.857 41.06 A j =- = ∠ - ° Hence, 16.857 A I = ( b ) The power dissipated by R is ( 29 ( 29 2 2 2 17.465 10 3050.26 W R P I R = = = 2 2.6 Problem 2.5 was solved assuming V on the reference. If V shifts by 30- ° , so does I . Then, for this problem solution, 240 30 V V = ∠ - ° and 16.857 71.06 A I = ∠ - ° . ( a ) By [2.55], ( 29 ( 29 * 240 30 16.857 71.06 4045.68 41.06 VA V I = = ∠ - ° ∠ ° = ∠ ° S ( b ) The values of average power and reactive power supplied to the circuit are ( 29 ( 29 ( 29 cos 240 16.857 cos 30 71.06 3050.53 W P VI θ = =- ° + ° = ( 29 ( 29 ( 29 sin 240 16.857 sin 30 71.06 2657.40 VARs Q VI θ = =- ° + ° = The minute difference (0.18%) between the value of P above and the value of R P in part ( b ) of Prob. 2.5 is due to numerical accuracy. With S = S , known from part ( a ), the power triangle below results. 2.7 ( a ) The two branch impedances are ( 29 1 6 265.258 90 120 10 10 j j j C ϖ π--- = = = - ∠ - ° Ω × Z ( 29 ( 29 2 10 120 0.025 10 9.425 13.742 43.3 R j L j j ϖ π = + = + = + = ∠ ° Ω Z The input impedance is ( 29 ( 29 1 2 1 2 265.258 90 13.742 43.3 14.237 41.06 265.258 10 9.425 j j ∠ - ° ∠ ° = = = ∠ ° Ω +- + + Z Z Z Z Z ( 29 ( 29 cos cos 41.06 0.754 lagging PF = ∠ = ° = Z 3 ( b ) A power method offers the best approach. The required PF is satisfied if the power triangle seen by the source has an angle ( 29 1 cos 0.9 25.84 θ- = = ° . The power triangle for the R-L branch can be drawn....
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ch2sol - Chapter 2 Problems Solutions 2.1 The network input...

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