Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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- Help me please IN THE CIRCUIT SHOWN, THE VALUE OF THE CURRENT SOURCE IS IN RMS VALUE. DETERMINE: A) THE VOLTAGE (ONLY THE MAGNITUDE) AT THE INDUCTOR TERMINALS, VL B) THE VOLTAGE (ONLY THE MAGNITUDE) AT THE RESISTOR TERMINALS, VR C) THE AVERAGE POWER ABSORBED BY THE 40 Ω RESISTORarrow_forwardDetermine the equivalent impedance of the network shown in Figure if the frequency is f= 60 Hz. Then compute the current i(t) if the voltage source is v(t) = 50 cos (wt+30°) V. Finally, calculate the equivalent impedance if the frequency is f= .400 Hz i(t) R = 25 v(t) L=20 mH C=50 μFarrow_forwardAnswer subparts D,E,Farrow_forward
- Unless given, use the: POLAR FORM for voltage and current, RECTANGULAR OF 5 FORM for impedance. Consider the following circuit, supplied by a 10VRMS current source with a 0° phase shift (i.e. Vsource = 1020° V). R1 C (2 V sine 50 Hz 100 Q 16 μF L 320 mH a. What is the total impedance of the circuit? b. Is the circuit inductive, capacitive, or neither (resistive)? c. What is the total power (apparent) delivered by the current source? d. Draw the phasor diagram of the voltage and current at the source. e. What is the current passing through the R2 and L branch? f. Draw the phasor diagram of the voltage and current at the R2 and L branch. ww R2 100arrow_forwardA circuit consists of a series connected 84 Ω resistor and a 163 mH inductor. The circuit supply is a 121 V rms source with a frequency of 42 Hz. Ultimately, you need to calculate the average power dissipated in the circuit. You should have found the polar form of the complex current. The average power dissipated in the resistive and reactive circuit is: ?=??cosφ where φ is the phase of the current, and V and I are the rms values. Finally, calculate the circuit Power Factor (cos?) where φ is the phase of the current. Finally, will the circuit Power Factor cos? be lagging or leading?arrow_forwardSubject: Network Theory Please solve this qarrow_forward
- A three phase 390 volt (line voltage) distribution system feeds power to a household load. Each phase of the system (distribution line) has an impedance consisting of a resistance (0.602) and a reactance (0.40) (inductive). The system is connected to three-phase load with an impedance of j +62 in a delta connection. What is the active, reactive and apparent power of the load (notice that distribution line impedance is not part of the load)? Give the answers to two decimal places. a) Absolute value of current feeded by the voltage source. I = 3.62 A Your last answer was interpreted as follows: 3.62 b) Active power P = c) Reactive power Q = d) Apparent power S (absolute value) kVA = kW kV Ar 4arrow_forwardA 110V(rms), 60HZ AC source is connected to a motor, which can be represented by a resistor and an inductor connected in series. There is also a 56µF capacitor connected across the motor (parallel to the source). Source is providing 1KVA at 73% lagging power factor. Draw a well-labeled circuit diagram and determine: (1) Magnitude of the source Current (ii) Active and reactive power of the motor (iii) Power factor of the motor (iv) Magnitude of the motor current (v) Motor resistor and inductor valuearrow_forward
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