916_PartUniversity Physics Solution

# 916_PartUniversity Physics Solution - 31-14 Chapter 31...

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31-14 Chapter 31 (b) EXECUTE: power factor cos φ equals 1 so 0 = and . CL XX = Calculate the present value of to see how much more L X is needed: cos (60.0 )(0.720) 43.2 RZ = = Ω tan so tan LC XXR R φφ =− = cos 0.720 gives 43.95 == ° ( is negative since the voltage lags the current) Then tan (43.2 )tan( 43.95 ) 41.64 . −= = Ω ° = Ω Therefore need to add 41.64 of . L X Ω 41.64 2 and 0.133 H, 2 2 (50.0 Hz) L L X XLf L L f ωπ ππ Ω = = = amount of inductance to add. EVALUATE: From the information given we can&t calculate the original value of L in the circuit, just how much to add. When this L is added the current in the circuit will increase. 31.46. IDENTIFY: Use rms rms VI Z = to calculate Z and then find R. 2 av rms PIR = SET UP: 50.0 C X EXECUTE: () 2 22 2 rms rms 240 V 80.0 50.0 . 3.00 A C V ZR X R I =+ Ω Thus, 80.0 50.0 62.4 . R Ω = Ω The average power supplied to this circuit is equal to the power dissipated by the resistor, which is 2 2 rms 3.00 A 62.4 562 W. Ω = EVALUATE: 50.0 tan 62.4 R −− Ω Ω and 38.7 °. av rms rms cos (240 V)(3.00 A)cos( 38.7 ) 562 W, PV I = ° which checks. 31.47. IDENTIFY: The voltage and current amplitudes are the maximum values of these quantities, not necessarily the instantaneous values. SET UP: The voltage amplitudes are V R = RI , V L = X L I , and V C = X C I , where I = V/Z and 2 2 1 . L C ω ⎛⎞ =+− ⎜⎟ ⎝⎠ EXECUTE: (a) = 2 π f = 2 π (1250 Hz) = 7854 rad/s. Carrying extra figures in the calculator gives X L = L = (7854 rad/s)(3.50 mH) = 27.5 ; X C = 1/ C = 1/[(7854 rad/s)(10.0 & F)] = 12.7 ; ZRX X = (50.0 ) (27.5 12.7 ) Ω+ Ω Ω = 52.1 ; I = V/Z = (60.0 V)/(52.1 ) = 1.15 A; V R = RI = (50.0 )(1.15 A) = 57.5 V; V L = X L I = (27.5 )(1.15 A) = 31.6 V; V C = X C I = (12.7 )(1.15 A) = 14.7 V. The voltage amplitudes can add to more than 60.0 V because these voltages do not all occur at the same instant of time. At any instant, the instantaneous voltages all add to 60.0 V. (b) All of them will change because they all depend on . X L = L will double to 55.0 , and X C = 1/ C will decrease by half to 6.35 . Therefore (50.0 ) (55.0 6.35 ) Z = Ω Ω = 69.8 ; I = V/Z = (60.0 V)/(69.8 ) = 0.860 A; V R = IR = (0.860 A)(50.0 ) = 43.0 V; V L = IX L = (0.860 A)(55.0 ) = 47.3 V; V C = IX C = (0.860 A)(6.35 ) = 5.47 V. EVALUATE: The new amplitudes in part (b) are not simple multiples of the values in part (a) because the impedance and reactances are not all the same simple multiple of the angular frequency. 31.48. IDENTIFY and SET UP: 1 . C X C = . L XL = EXECUTE: (a) 1 1 1 L C = and 2 1 1 . LC = At angular frequency 2 , 2 21 2 1 (2 ) 4. 1/ L C LC XC ωω = = . L C > (b) At angular frequency 3 , 2 2 1 3 2 1 11 .

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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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916_PartUniversity Physics Solution - 31-14 Chapter 31...

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