ch31-p045

# ch31-p045 - 45(a For a given amplitude Em of the generator...

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where R is the resistance, L is the inductance, C is the capacitance, and ω d is the angular frequency. To find the maximum, set the derivative with respect to ω d equal to zero and solve for ω d . The derivative is dI d = −E m ± R 2 +( ω d L 1 d C ) 2 ² 3 / 2 ³ ω d L 1 ω d C ´³ L + 1 ω 2 d C ´ . The only factor that can equal zero is ω d L (1 d C ) and it does for ω d =1 / LC .F o rt h e given circuit, ω d = 1 LC = 1 µ (1 . 00 H)(20 . 0 × 10 6 F) = 224 rad / s . (b) For this value of the angular frequency, the impedance is Z = R and the current amplitude is I = E m R = 30 . 0V 5 . 00 =6 . 00 A . (c) and (d) You want to find the values of ω d for which I = E m / 2 R . This means E m µ R 2 +( ω d L 1 d C ) 2 = E m 2 R . Cancel the factors E m that appear on both sides, square both sides, and set the reciprocals of the two sides equal to each other to obtain R 2 + ω d L 1 ω d C · 2 =4 R 2 . Thus ω d L 1 ω d C · 2 =3 R 2 . Now take the square root of both sides and multiply by ω d C to obtain ω 2 d ( LC ) ± ω d ¸ 3 CR ¹ 1=0 , where the symbol ± indicates the two possible signs for the square root. The last equation is a quadratic equation for ω d . Its solutions are ω d = ± 3 CR ± 3 C

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ch31-p045 - 45(a For a given amplitude Em of the generator...

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