p33_037 - 37(a Now XC = 0 while R = 160 and XL = 86.7...

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37. (a) Now X C = 0, while R = 160 Ω and X L =86 . 7 Ω remain unchanged. Therefore, the impedance is Z = q R 2 + X 2 L = p (160 Ω) 2 +(86 . 7Ω) 2 = 182 Ω . The current amplitude is now found to be I = E m Z = 36 . 0V 182 Ω =0 . 198 A . The phase angle is, from Eq. 33-65, φ =tan 1 µ X L X C R =tan 1 µ 86 . 7Ω 0 160 Ω =28 . 5 . (b) We Frst Fnd the voltage amplitudes across the circuit elements: V R = IR =(0 . 198 A)(160 Ω) 32 V V L = IX L =(0 . 216 A)(86 . 7Ω) 17 V This is an inductive circuit, so E m leads I . The phasor diagram is drawn to scale below. .........................................................................................................
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