Electric Circuits. (11th Edition)
11th Edition
ISBN: 9780134746968
Author: James W. Nilsson, Susan Riedel
Publisher: PEARSON
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Chapter 5, Problem 6P
To determine
Calculate the value of current
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A linear electrical load draws 11 A at a 0.72 lagging power factor./1 153. When a capacitor is connected, the line current dropped to 122 A and the power factor
improved to 0.98 lagging. Supply frequency is 50 Hz.
a. Let the current drawn from the source before and after introduction of the capacitor be 11 and 12
respectively. Take the source voltage as the reference and express 11 and 12 as vector
quantities in polar form.
b. Obtain the capacitor current, IC = 12 - 11, graphically as well as using complex number
manipulation. Compare the results.
c. Express the waveforms of the source current before (11(t)) and after (12(t)) introduction of the
capacitor in the form Im sin(2лft + 0). Hand sketch them on the same graph. Clearly label your
plots.
d. Analytically solve i2(t) – i1(t) using the theories of trigonometry to obtain the capacitor current
in the form, ¡C(t) = ICm sin(2πft + OC). Compare the result with the result in Part b.
Transmission line data:Data:• Active power of the load (P): 50 kW• Power factor of the load (PF): 0.8 (lagging)• Line-to-line voltage at the load (V_LC): 13.8 kV• Line resistance (R): 2 Ω• Line inductance (L): 0.8 H• Line capacitance (C): 0.0003 F• Required series compensation: 60% of the line impedance.• Line length: 250 kmDetermine:1. Characteristic impedance and propagation constant.2. The generalized long line constants A, B, C, D.3. Total voltage, current and power at the generating end.4. Voltage regulation.5. Parameters A, B, C, D of the compensation circuit.6. New generalized constants of the power system afterseries compensation.7. Conclusion of the results obtained.
3.18
In a single-phase half-wave ac-dc converter, the average value of the load
current is 1.78 A. If the converter is operated from a 240 V, 50 Hz supply
and if the average value of the output voltage is 27% of the maximum
possible value, calculate the following, assume the load to be resistive.
(a) Load resistance
(b) Firing angle
(c) Average output voltage
(d) The rms load voltage
(e) The rms load current
(f) DC power
(g) AC power
(h) Rectifier efficiency
(i) Form factor
(j) Ripple factor
Chapter 5 Solutions
Electric Circuits. (11th Edition)
Ch. 5.2 - Assume that the op amp in the circuit shown is...Ch. 5.3 - The source voltage vs in the circuit in Assessment...Ch. 5.4 - Find vo in the circuit shown if va = 0.1 V and vb...Ch. 5.5 - Assume that the op amp in the circuit shown is...Ch. 5.6 - In the difference amplifier shown, vb = 4.0 V....Ch. 5.6 - Suppose the 12kΩ resistor Rd in the difference...Ch. 5.7 - The inverting amplifier in the circuit shown has...Ch. 5 - The op amp in the circuit in Fig. P5.1 is ideal....Ch. 5 - Replace the 2.5 V source in the circuit in Fig....Ch. 5 - Find io in the circuit in Fig. P5.3 if the op amp...
Ch. 5 - The op amp in the circuit in Fig. P5.4 is...Ch. 5 - The op amp in the circuit in Fig. P5.5 is ideal....Ch. 5 - Find iL (in milliamperes) in the circuit in Fig....Ch. 5 - Prob. 7PCh. 5 - Design an inverting amplifier with a gain of 2.5,...Ch. 5 - Design an inverting amplifier with a gain of 4....Ch. 5 - The op amp in the circuit in Fig. P5.10 is...Ch. 5 - The op amp in the circuit shown in Fig. P5.11 is...Ch. 5 - The op amp in Fig. P5.12 is ideal.
What circuit...Ch. 5 - Design an inverting-summing amplifier using a 120...Ch. 5 - Prob. 14PCh. 5 - Design an inverting-summing amplifier so...Ch. 5 - The op amp in Fig. P5.16 is ideal. Find vo if va –...Ch. 5 - Prob. 17PCh. 5 - The op amp in the circuit of Fig. P5.18 is...Ch. 5 - Prob. 19PCh. 5 - The op amp in the circuit shown in Fig. P5.20 is...Ch. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - The op amp in the circuit of Fig. P5.23 is...Ch. 5 - The circuit in Fig. P5.24 is a noninverting...Ch. 5 - The op amp in the circuit of Fig. P5.25 is...Ch. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Select the values of Rb and Rf in the circuit in...Ch. 5 - The op amp in the adder-subtracter circuit shown...Ch. 5 - In the difference amplifier shown in Fig. P5.32,...Ch. 5 - Prob. 33PCh. 5 - The op amp in the circuit of Fig. P5.34 is...Ch. 5 - Assume that the ideal op amp in the circuit seen...Ch. 5 - Prob. 37PCh. 5 - Show that when the ideal op amp in Fig. P5.38 is...Ch. 5 - The op amps in the circuit in Fig. P5.39 are...Ch. 5 - The two op amps in the circuit in Fig. P5.40 are...Ch. 5 - The circuit inside the shaded area in Fig. P5.41...Ch. 5 - Assume that the ideal op amp in the circuit in...Ch. 5 - Derive Eq. 5.31.
(5.31)
Ch. 5 - Prob. 44PCh. 5 - Prob. 45PCh. 5 - Repeat Problem 5.45 assuming an ideal op...Ch. 5 - Assume the input resistance of the op amp in Fig....Ch. 5 - Prob. 48PCh. 5 - Suppose the strain gages in the bridge in Fig....Ch. 5 - For the circuit shown in Fig. P5.50, show that if...Ch. 5 - Prob. 51PCh. 5 - Prob. 52PCh. 5 - Prob. 53P
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- Disassemmble In another way f(t) = [sin(200πt) - cos(300πt)]2arrow_forward☑ 口 ☐ : Homework help starts here! Home | bartleby → https://www.bartleby.com Answered: Decompose using relationships 3 S(+) = 50 sin 3 (500πiz) | bartleby ASK AN EXPERT √ MATH SOLVER Decompose using relationships 3 S(+) = 50 sin 3 (500... Steps "Disassemble in another way." f(+)= 50 Sin³ (500+) Step 1 To decompose f(t)=50 Sin³ (500x+), use trigonometric indentity for Sin³ (e) Sin³ (0)- 3 Sin (e)-Sin (30) 4 Step2: Apply indentity to f(t). f(+)= 50 Sin³ (500+) = 50. 3 Sin (500+) - Sin (3*500π+) = 50. = = 12.5 4 3 Sin (500+)-Sin (1500+) 4 [3 Sin (580x+)-Sin (1500x+)] 37.5 Sin (500+)-12.5 Sin (1500π+) final decomposed form is, PM 05:2 19/12/2024 દ ))^ C W E × م اكتب هنا للبحثarrow_forward6) The principal of DIT hire you to design PBAX with 120 phones. Assuming the number of call is 3/hour/line, the average call duration is 4 minutes, and 55% of all call are made external via a T-1 trunk (24 channels) to the PSTN. Determine carried traffic and channel usage.arrow_forward
- Need a solufor number 2arrow_forwardDecompose using relationships 3 S(+) = 50 sin ³ (500πiz)arrow_forwardA linear electrical load draws I₁ A at a 0.72 lagging power factor. 11 = 153 When a capacitor is connected, the line current dropped to 122 A and the power factor improved to 0.98 lagging. Supply frequency is 50 Hz. a. Let the current drawn from the source before and after introduction of the capacitor be 1₁ and 12 respectively. Take the source voltage as the reference and express 11 and 12 as vector quantities in polar form. b. Obtain the capacitor current, Ic = I2 − I₁, graphically as well as using complex number manipulation. Compare the results. c. Express the waveforms of the source current before (i₁(t)) and after (i2(t)) introduction of the capacitor in the form Im sin(2лft + 0). Hand sketch them on the same graph. Clearly label your plots. d. Analytically solve i₂ (t) - i₁ (t) using the theories of trigonometry to obtain the capacitor current in the form, ic (t) = 1cm sin(2´ft + 0c). Compare the result with the result in Part b.arrow_forward
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