A 300? resistor is in series with a 0.100H inductor and a 0.560 ? F capacitor. P

Question

A 300? resistor is in series with a 0.100H inductor and a 0.560?F capacitor.

Part A

Compute the impedance of the circuit at a frequency of f1 = 500Hz and at a frequency of f2 = 1000Hz .

Enter your answer as two numbers separated with a comma.

Part B

In each case, compute the phase angle of the source voltage with respect to the current.

Enter your answer as two numbers separated with a comma.

Part C

State whether the source voltage lags or leads the current at a frequency 500Hz .

Part D

State whether the source voltage lags or leads the current at a frequency 1000Hz.

Halp! Explanations plz

Problem 22.14 A 300 Ohm resistor is in series with a 0. 100H inductor and a 0.560MuF capacitor Part B In each case. compute the phase angle of the source voltage with respect to the current Enter your answer as two numbers separated with a comma. Part C State whether the source voltage lags or leads the current at a frequency 500Hz Part D State whether the source voltage lags or leads the current at a frequency 1000Hz Problem 22.14 A 300 Ohm resistor is in series with a 0. 100H inductor and a 0.560MuF capacitor Part A Compute the impedance of the circuit at a frequency of f1 = 500Hz and at a frequency of f2 = 1000Hz Enter your answer as two numbers separated with a comma. Part B In each case. compute the phase angle of the source voltage with respect to the current Enter your answer as two numbers separated with a comma.

Explanation / Answer

impedance due to C, XC = 1/2?fC

impedane due to L, XL = 2?fL

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at f1
impedance due to C, Xc = 1/2?*500*0.56*10^-6 = 568.41 ohms

impedane due to L, XL = 2?*500*0.100 = 314.159 ohms

total impedance, Z1 = [(XL - XC)2 + R2 ]1/2 = [(314.159 - 568.41)2 + (300)2]1/2 = 393.247 ohms

at f2
impedance due to C = 1/2?*1000*0.56*10^-6 = 284.2 ohms

impedane due to L = 2?*1000*0.100 = 628.318 ohms

total impedance, Z1 = [(XL - XC)2 + R2 ]1/2 = [(628.318 - 284.2)2 + (300)2]1/2 = 456.5 ohms

a) Z1 , Z2 = 393.247 ohms , 456.5 ohms

b) angle at f1 = cos-1(300/393.247 ) = 40.28o

    angle at f2 = cos-1(300/456.5) = 48.91o

c) first case capacitance is higher so voltage lags current.

d inductance impedance is higher so voltage leads current.

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