用節點怎麼算

CHAPTER 4 BASIC NODAL AND MESH AU sepa 92 W our structure answer. Each component of the structure is accessed rately by name as shown. and at node 2 PRACTICE 4.2 For the circuit of Fig. 4.5, compute the voltage across each current source. 212 Unfortunately, we ha is a direct result of the it is not controlled by a equation that relates i In this case, we find 12 w 432 w 3 A 352 52. TA which upon substituti Reference node FIGURE 4.5 and Eq. [10] simplifi Ans: V3A = 5.235 V; v7A = 11.47 V. Solving, we findt The previous examples have demonstrated the basic approach to nodal analysis, but it is worth considering what Thus, the power sup (-60)(-75) = 4.5k

請問一下這題怎麼算😭好難

Vout - 5 volt D2 +5 vol o Vout D OVORI = Vout == 4.310 v ok suming the diodes are | (b) tions shown in Figure P9.23 D -9.3 大 li (c) WO Do V 12 +7 V o bin 88. I won T08 . 9.24 Sketch the output waveform and the voltage transfer characteristic for the circuit of Figure P9.24. Assume an ideal diode, us (t) = & sin (it), V, = XV, Ri = 812, and R2 = $ 12. 10 4 lo lo mm ogiloc www bools boRi t V. -Vo 15 k12 15 k92 1 vs borbo R2 V 4V V c) (d) bagnozib Figure P9.24 OU 29,22, find the range of Vin 25 Repeat Problem 9.24, using the offset diode model

二極體電路問題

Given the cirsuit. All resistors are as specified. Give the voltage of r14. vsrc 11 = 4V, vsrc 12 = 4V vsrc12 vsrc11 r11 40 r13 40 r15 20 Hw d1 w r12 r14 20 40 HII

各位 有沒有人可以幫助我🙏 看我錯的這幾題 用最上面的公式 感謝感謝🥰

1. 等加速度運動公式: 一切只疫反應。 ② V = Votat ② X= Volt hat (3) v= W 120K 2. 一位自行車選手在速度為20m/s時,以等減速度2m/s? 煞車,滑行100公尺後停下 來,則從煞車到停止共歷時多少秒? A{os 3. 物體由靜止以等加速度2m/s? 開始運動,經過若干秒後物體的速度為10 m/s? | 2 10=24 t=5 4. 某公車由靜止開始作等加速度運動,10秒後的速度為25公尺/秒,該車的加速度為 幾公尺/秒2 35 a 2.5 m/s² Im/s 5. 一物體由靜止作等加速度直線運動,最初2秒走了10m,若全部行程為250m,走完 全程尚需若干秒? 250 25=50s 8 6. 某個廠牌的歐規車,能在4秒內從靜止加速至100m/s,則此車子在4秒內的平均加 速度大小為多少m/s?? ( 100-0 4 4 = 25 m/s 7. 在筆直公路上以12公尺/秒速度行駛的車子,以-2公尺/秒的加速度煞車,第3 秒時的車速為何? A = 6 m/s 8. 一飛機自靜止加速至25 m/s,花費4秒鐘,若此飛機作等加速度運動,求該飛機經 過的距離為多少? 25+4=100 so 9. 一物體以2公尺/秒之速度滑進一粗糙的平面,經5秒鐘後即完全停止,若減速過 摩擦力一定,則言 體自滑進粗糙面到完全停止的滑行距離為多少公尺? Airom 5 10 假日時,段老師以初速20m/s 開車帶家人出去玩、如果他在行駛的過程中保持等加速 度 5m/s? 前進,則經過5秒後,車子總共前進了約多少公尺? 20 4.162.5m

請問第十三題為什麼可以用藍筆的圖解 同樣的座標為什麼要拆開寫

abo 2 G 1. 答:11.10[道明模考】 13Exi 4 (-4, - 8) • B(-1,1). C(1, 7) teria L : 8x=y+47 74, y=3x09 feb! 一多多 上,設動點P在直線L上,求3PA + 2PB + PC之最小值為1。 PA =}x+++(32+12)? PB = T[qx+1) Jolx+4 PC? Tiox+1 3JpX44 21012H14 voltz-4, Winx to no 14試問下列各選項之加值,何者會使得方程式+3\x + 4-2 = 0 無解?(1)5 (2) - -3 (4) -6 (5) (6) 3 (30)x= 10 10 3° 答:(3) 【北區聯合模考 y=-3x+4+2 之圖形交點, y = m=- go 1 作圖如右,則無解台-3<m <- 2 m=-3 Ay 【解】:考慮 = mx -4-4-4 t ( (3+3+50510 | [ 1 - 115io!

求解第四第五題!!!電機的 謝謝😭🥺🥺🥺 拜託看到盡快><

2. The circuit in Fig. 2 represents an energy storage bank with vi(0)=-3 V, v2(0) = 5 V, v3(0)=-5 V. (a) If the switch Si is closed at tı > 0, find the values of vi, v2, and v3 after ti. (b) Following (a), if the switch S2 is closed at 12 >ti. Find the final values of vi, vz, and va after 12. 3. Let iz(0) = 0. Find the current il(t) of the circuit shown in Fig. 3 for t> 0. 22 w 22 32 1 ] 322 + 8t u(t) A 22 2H 1:6 8u(1) V 台Fv() 3i Fig. 3 Fig. 4 4. Let vo(0-) = 0. Find the voltage ve() and current i(1) denoted in Fig. 4 for t> 0. 5. The switch shown in Fig. 5 is closed for a long time before t = 0. When the switch is open at t = 0, find i(t) for t < 0 and > 0. i=0 =0 212 622 W + 392 8 V 0.510) 2=vd) - Fig. 5 FE final Sophie de EES-B22.pdf 20210611_115335.jpg 20210611_115320.jpg 18.pdf

第四段第二行的pack是什麼意思？ 解答寫「電池」但是我查不到pack有這個意思

第31至35 題為題組 Since the early 1990s, the hihilin zion battery has been the most suitable battery for portation to y 31 electronic equipment. Today, they're commonly used for cellphones, computers, tablets, digital cameras, and other devices. Lithium-ion batteries have nearly twice the energy density of traditional nicket cadmium batteries. 31 This feature has important implications for cellphones and computers, because it makes these items more portable for consumers. It also makes power tools easier to use and allows workers to use them for longer periods of time, 32. 32 Lithium-ion batteries retain retain mo naman of their power capacity from previous charging cycles. Thus they require no scheduled cycling and can be fully re-fueled to their maximum capacity during each charging cycle. Other rechargeable battery types, in contrast, retain information from previous charging cycles, which wastes valuable storage space. Over time, this makes these rechargeable batteries hold less of a charge. 33 It is fragile and requires a protection circuit to maintain safe operation. A high load could overheat the pack and safety might be jeopardized. 34 F After 2-3 years of use, the pack often becomes unserviceable due to a large voltage drop caused by high internal resistance. It should be noted, however, that manufacturers are constantly making improvements on lithium-ion batteries. 35 With such rapid progress, the use of lithium-ion batteries will certainly expand further. (A (A) The lithium-ion battery is also a low maintenance battery. (B) Despite its overall advantages, the lithium-ion battery has its drawbacks. (C) New and enhanced chemical combinations are introduced every six months or so. (D) Attempts to develop rechargeable lithium-ion batteries failed due to memory problems. (E) That is, they carry more power in a smaller unit, helping to reduce overall weight and size. d (F). Another downside is the increase of the internal resistance that occurs with cycling and aging. -3-

想請教各位紅色圈起來的式子怎麼來的 感謝各位大大🙇

2 塔高=Hyg(一) 8 0 0 = V-zgh 如同」 v 2 類題 12.1 ( (1-2 co V2 初速鉛直上拋可上升最大高度h= 2g 而下行過程等同於由靜止自由落下 CIN ti It₂ 移 h 1 CIN gt;?, h ? 2 gt2 uu 2 2 ) h St= V2 2 g g (T 2h g g 利用對稱性,可知 鉛直上抛到最大高度的一半所需時間 時間 了。 =t-t=- (1 g 2 (另解]鉛直上拋至最大高度一半處的位移

看不懂他給的公式 位移怎麼算

mz 40 kg (3)施力期間的位移為 xz 1 1 = Volt + at = 0+,×6×0.5° = 0.75 m(向東) 2 2 (4)因兩人受力作用的時間應相同,則 VP = ap Xt=8×0.5 = 4 m/s(向西) Vz = az xt=6×0.5 = 3 m/s(向東)

請問為什麼-1不合 謝謝

3 jy平面上,一光線通過(3,4) ,經y軸反射後與實心圓盤相切,而此實心圓盤外圍方程 式為: C-D7+0y-2)" =2。設原光線打在y軸上的點坐標為4(0,o) ,則實數g= 遍 ! (3.4對y軸的對稱點為(3,4) 。 設過(一3,4)對圓的切線為一4三訪(x+3) 4 疙 監」 2 全 注 讀 (4zz二2)? ==27 +2 人zz2 十世 2 由 一14z77 二1677十2三0 人有訪 或-1(丰合) VOLtA 可知切線為 py-4=-7G+3) 了 及 中全D 人証信2在施 2 6x一42一2避 0及三直線 和信司光還 。則,在圓C 平面上有一個圓C: 二 蘆蝦 平面 關和2 2和下直線好 的距離為正整數值。