Electric Circuits (10th Edition)
10th Edition
ISBN: 9780133760033
Author: James W. Nilsson, Susan Riedel
Publisher: PEARSON
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Chapter 13, Problem 31P
(a)
To determine
Find the s-domain expressions of
(b)
To determine
Find the time domain expression of
(c)
To determine
Find the s-domain expression of
(d)
To determine
Find the time domain expression of
(e)
To determine
Find the time taken to turn on the current source when the one of the capacitors is breakdown.
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Section 15-3 reversing motors using magnetic starters.Section 15-3Use data sheet B on page 383 to draw the wiring diagram. Note: use only the number of contacts required.
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Use data sheet B on page 383 to draw the wiring diagram. Note: use only the number of contacts required.
First 1. Wire the motor to operate in forward and reverse at 115 VAC.
Chapter 13 Solutions
Electric Circuits (10th Edition)
Ch. 13.2 - Prob. 1APCh. 13.2 - The parallel circuit in Example 13.1 is placed in...Ch. 13.3 - Prob. 3APCh. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.3 - The dc current and dc voltage sources are applied...Ch. 13.3 - Prob. 6APCh. 13.3 - Using the results from Example 13.7 for the...Ch. 13.3 - The energy stored in the circuit shown is zero at...Ch. 13.4 -
Derive the numerical expression for the transfer...Ch. 13.5 - Find (a) the unit step and (b) the unit impulse...
Ch. 13.5 - The unit impulse response of a circuit is
υo(t) =...Ch. 13.7 - The current source in the circuit shown is...Ch. 13.7 - For the circuit shown, find the steady-state...Ch. 13 - Prob. 1PCh. 13 - Prob. 2PCh. 13 - Prob. 3PCh. 13 - Prob. 4PCh. 13 - An 8 kΩ resistor, a 25 mH inductor, and a 62.5 pF...Ch. 13 - Prob. 6PCh. 13 - Find the poles and zeros of the impedance seen...Ch. 13 - Find the poles and zeros of the impedance seen...Ch. 13 - Prob. 9PCh. 13 - Prob. 10PCh. 13 - Prob. 13PCh. 13 - Prob. 15PCh. 13 - There is no energy stored in the circuit in Fig....Ch. 13 - There is no energy stored in the circuit in Fig....Ch. 13 - Prob. 25PCh. 13 - Prob. 28PCh. 13 - The switch in the circuit seen in Fig. P13.32 has...Ch. 13 - Prob. 31PCh. 13 - Prob. 33PCh. 13 - Prob. 35PCh. 13 - Prob. 46PCh. 13 - Prob. 47PCh. 13 - Find the transfer function H(s) − Vo/Vi for the...Ch. 13 - Prob. 49PCh. 13 - Prob. 50PCh. 13 - Prob. 51PCh. 13 - Prob. 53PCh. 13 - Prob. 54PCh. 13 - The operational amplifier in the circuit in Fig....Ch. 13 - Find the transfer function Io/Ig as a function of...Ch. 13 - Prob. 58PCh. 13 - Prob. 59PCh. 13 - Prob. 60PCh. 13 - Prob. 61PCh. 13 - Assume the voltage impulse response of a circuit...Ch. 13 - Prob. 68PCh. 13 - The input voltage in the circuit seen in Fig....Ch. 13 - Find the impulse response of the circuit shown in...Ch. 13 - Prob. 73PCh. 13 - Prob. 74PCh. 13 - Prob. 75PCh. 13 - The op amp in the circuit seen in Fig. P13.81 is...Ch. 13 - Prob. 78PCh. 13 - The transfer function for a linear time-invariant...Ch. 13 - Prob. 80PCh. 13 - Prob. 81PCh. 13 - Prob. 82PCh. 13 - Prob. 84PCh. 13 - Prob. 85PCh. 13 - The parallel combination of R2 and C2 in the...Ch. 13 - Show that if R1C1 = R2C2 in the circuit shown in...Ch. 13 - The switch in the circuit in Fig P13.91 has been...Ch. 13 - Prob. 90PCh. 13 - Prob. 91P
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- I need help checking if its correct -E1 + VR1 + VR4 – E2 + VR3 = 0 -------> Loop 1 (a) R1(I1) + R4(I1 – I2) + R3(I1) = E1 + E2 ------> Loop 1 (b) R1(I1) + R4(I1) - R4(I2) + R3(I1) = E1 + E2 ------> Loop 1 (c) (R1 + R3 + R4) (I1) - R4(I2) = E1 + E2 ------> Loop 1 (d) Now that we have loop 1 equation will procced on finding the equation of I2 current loop. However, a reminder that because we are going in a clockwise direction, it goes against the direction of the current. As such we will get an equation for the matrix that will be: E2 – VR4 – VR2 + E3 = 0 ------> Loop 2 (a) -R4(I2 – I1) -R2(I2) = -E2 – E3 ------> Loop 2 (b) -R4(I2) + R4(I1) - R2(I2) = -E2 – E3 -----> Loop 2 (c) R4(I1) – (R4 + R2)(I2) = -E2 – E3 -----> Loop 2 (d) These two equations will be implemented to the matrix formula I = inv(A) * b R11 R12 (R1 + R3 + R4) -R4 -R4 R4 + R2arrow_forward10.2 For each of the following groups of sources, determineif the three sources constitute a balanced source, and if it is,determine if it has a positive or negative phase sequence.(a) va(t) = 169.7cos(377t +15◦) Vvb(t) = 169.7cos(377t −105◦) Vvc(t) = 169.7sin(377t −135◦) V(b) va(t) = 311cos(wt −12◦) Vvb(t) = 311cos(wt +108◦) Vvc(t) = 311cos(wt +228◦) V(c) V1 = 140 −140◦ VV2 = 114 −20◦ VV3 = 124 100◦ Varrow_forwardApply single-phase equivalency to determine the linecurrents in the Y-D network shown in Fig. P10.13. The loadimpedances are Zab = Zbc = Zca = (25+ j5) Warrow_forward
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