Find the equivalent impedance Zeq seen by the source when Vs = 6 cos (5t) v, C = 0.1 F, R = 12 and L = 0.2 H. (Give angles in degrees and round final answers to two decimal places.) Calculate the voltage across each component - capacitor, resistor and inductor. (Give angles in degrees and round fin answers to two decimal places.) Vs R The equivalent impedance seen by the source is 1±2% +j -1±2% 2 = 1.41 ± 2% at an angle of -45 ± 2% . The voltage across the capacitor is 8.49 ± 2% at an angle of -45 ± 2% v. The voltage across the resistor is 4.24 ± 2% at an angle of 45 ± 2% v. The voltage across the inductor is 4.24 ± 2% at an angle of 135 ± 2% v.
Find the equivalent impedance Zeq seen by the source when Vs = 6 cos (5t) v, C = 0.1 F, R = 12 and L = 0.2 H. (Give angles in degrees and round final answers to two decimal places.) Calculate the voltage across each component - capacitor, resistor and inductor. (Give angles in degrees and round fin answers to two decimal places.) Vs R The equivalent impedance seen by the source is 1±2% +j -1±2% 2 = 1.41 ± 2% at an angle of -45 ± 2% . The voltage across the capacitor is 8.49 ± 2% at an angle of -45 ± 2% v. The voltage across the resistor is 4.24 ± 2% at an angle of 45 ± 2% v. The voltage across the inductor is 4.24 ± 2% at an angle of 135 ± 2% v.
Introductory Circuit Analysis (13th Edition)
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Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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Transcribed Image Text:Find the equivalent impedance Zeq seen by the source when Vs = 6 cos (5t) v, C = 0.1 F, R = 1 §2 and L = 0.2 H.
(Give angles in degrees and round final answers to two decimal places.)
Calculate the voltage across each component - capacitor, resistor and inductor. (Give angles in degrees and round final
answers to two decimal places.)
Vs
R
The equivalent impedance seen by the source is 1±2% +j -1 ± 2% 2 = 1.41 ± 2% at an angle of -45 ± 2% 2.
The voltage across the capacitor is 8.49 ± 2% at an angle of -45±2% v.
The voltage across the resistor is 4.24 ± 2%
The voltage across the inductor is 4.24 ± 2%
at an angle of 45 ± 2% v.
at an angle of 135±2% v.
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