Electric Circuits, Global Edition
10th Edition
ISBN: 9781292060545
Author: James W. Nilsson, Susan Riedel
Publisher: Pearson Education Limited
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Chapter 7, Problem 104P
To determine
Develop the expression for heat rate in beats per minute.
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-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 + R2
10.2 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◦ V
Apply 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) W
Chapter 7 Solutions
Electric Circuits, Global Edition
Ch. 7.1 - The switch in the circuit shown has been closed...Ch. 7.1 - Prob. 2APCh. 7.2 - Prob. 3APCh. 7.2 - Prob. 4APCh. 7.3 - Prob. 5APCh. 7.3 - Prob. 6APCh. 7.5 - Prob. 7APCh. 7.5 - Prob. 8APCh. 7.7 - There is no energy stored in the capacitor at the...Ch. 7.7 - Prob. 10AP
Ch. 7 - Prob. 1PCh. 7 - Prob. 2PCh. 7 - In the circuit shown in Fig. P 7.2, the switch...Ch. 7 - Prob. 4PCh. 7 - Prob. 5PCh. 7 - The two switches in the circuit seen in Fig. P...Ch. 7 - Prob. 7PCh. 7 - Prob. 8PCh. 7 - The switch shown in Fig. P 7.4 has been open for a...Ch. 7 - Prob. 10PCh. 7 - In the circuit in Fig. P 7.11, let Ig represent...Ch. 7 - Prob. 12PCh. 7 - Prob. 13PCh. 7 - Prob. 14PCh. 7 - Prob. 15PCh. 7 - Prob. 16PCh. 7 - Prob. 17PCh. 7 - For the circuit seen in Fig. P 7.19, find
the...Ch. 7 - Prob. 19PCh. 7 - Prob. 20PCh. 7 - Prob. 21PCh. 7 - Prob. 22PCh. 7 - Prob. 23PCh. 7 - Prob. 24PCh. 7 - Prob. 25PCh. 7 - In the circuit shown in Fig. P 7.26, both switches...Ch. 7 - The switch in the circuit in Fig. P 7.25 is closed...Ch. 7 - Prob. 28PCh. 7 - Prob. 29PCh. 7 - Prob. 30PCh. 7 - The switch in the circuit seen in Fig. P 7.30 has...Ch. 7 - In Problem 7.30 how many microjoules of energy are...Ch. 7 - Prob. 33PCh. 7 - Prob. 35PCh. 7 - The switch in the circuit shown in Fig. P 7.38 has...Ch. 7 - Prob. 37PCh. 7 - Prob. 38PCh. 7 - Prob. 39PCh. 7 - Prob. 40PCh. 7 - Prob. 41PCh. 7 - Prob. 42PCh. 7 - Prob. 43PCh. 7 - Prob. 44PCh. 7 - Prob. 45PCh. 7 - Prob. 46PCh. 7 - Prob. 47PCh. 7 - For the circuit in Fig. P 7.4, find (in...Ch. 7 - Prob. 49PCh. 7 - Prob. 50PCh. 7 - Prob. 51PCh. 7 - Prob. 52PCh. 7 - The switch in the circuit of Fig. P 7.55 has been...Ch. 7 - Prob. 54PCh. 7 - The switch in the circuit seen in Fig. P 7.56 has...Ch. 7 - Prob. 56PCh. 7 - Prob. 57PCh. 7 - Prob. 58PCh. 7 - Prob. 59PCh. 7 - Prob. 60PCh. 7 - Prob. 61PCh. 7 - Prob. 62PCh. 7 - Prob. 64PCh. 7 - Prob. 65PCh. 7 - Prob. 66PCh. 7 - Prob. 67PCh. 7 - Prob. 68PCh. 7 - Prob. 69PCh. 7 - Prob. 70PCh. 7 - Prob. 71PCh. 7 - Prob. 72PCh. 7 - For the circuit in Fig. P 7.73, how many...Ch. 7 - Prob. 74PCh. 7 - Prob. 75PCh. 7 - Prob. 76PCh. 7 - Prob. 77PCh. 7 - Prob. 78PCh. 7 - Prob. 79PCh. 7 - Prob. 80PCh. 7 - Prob. 81PCh. 7 - Prob. 82PCh. 7 - Prob. 84PCh. 7 - Prob. 85PCh. 7 - Prob. 86PCh. 7 - Prob. 87PCh. 7 - Prob. 88PCh. 7 - Prob. 90PCh. 7 - Prob. 91PCh. 7 - Prob. 92PCh. 7 - Prob. 93PCh. 7 - Prob. 94PCh. 7 - Prob. 95PCh. 7 - Prob. 102PCh. 7 - Prob. 103PCh. 7 - Prob. 104PCh. 7 - Prob. 105PCh. 7 - Prob. 106PCh. 7 - Prob. 107P
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- 10.8 In the network of Fig. P10.8, Za = Zb = Zc = (25+ j5) W.Determine the line currents.arrow_forwardUsing D flip-flops, design a synchronous counter. The counter counts in the sequence 1,3,5,7, 1,7,5,3,1,3,5,7,.... when its enable input x is equal to 1; otherwise, the counter count 0. Present state Next state x=0 Next state x=1 Output SO 52 S1 1 S1 54 53 3 52 53 S2 56 51 0 $5 5 54 S4 53 0 55 58 57 7 56 56 55 0 57 S10 59 1 58 58 S7 0 59 S12 S11 7 $10 $10 59 0 $11 $14 $13 5 $12 S12 $11 0 513 $15 SO 3 S14 $14 S13 0 $15 515 SO 0 Explain how to get the table step by step with drawing the state diagram and finding the Karnaugh map.arrow_forwardFor the oscillator resonance circuit shown in Fig. (5), derive the oscillation frequency Feedback and open-loop gains. L₁ 5 mH (a) ell +10 V R₁ ww R3 S C2 HH 1 με 1000 pF 100 pF R₂ 1 με RA H (b) +9 V R4 CA 470 pF C₁ R3 HH 1 με R₁ ww L₁ 000 1.5 mH R₂ ww Hi 1 μF L2 m 10 mHarrow_forward
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- A) Calculate the efficiency of the test transformer at the resistive loads (X-25%, 50%, 75%, 100%, 125% full load). B) From part (A) draw the plot (efficiency Vs power output) of the transformer. C) Discuss the plot of part (B).arrow_forwarda- Determine fH; and Ho b- Find fg and fr. c- Sketch the frequency response for the high-frequency region using a Bode plot and determine the cutoff frequency. Ans: 277.89 KHz; 2.73 MHz; 895.56 KHz; 107.47 MHz. 14V Cw=5pF Cwo-8pF Coc-12 pF 5.6kQ Ch. 40. pF C-8pF 68kQ 0.47µF Vo 0.82 kQ V₁ B=120 0.47µF www 3.3kQ 10kQ 1.2kQ =20µF Narrow_forwardUsing D flip-flops, design a synchronous counter. The counter counts in the sequence 1,3,5,7, 1,7,5,3,1,3,5,7,.... when its enable input x is equal to 1; otherwise, the counter. This counter is for individual settings only need the state diagram and need the state table to use 16 states from So to S15.arrow_forward
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