# Microchip Technology: The Past and the Future by Charles Kerridge By Charles Kerridge

Similar electronics books

Using Robots in Hazardous Environments: Landmine Detection, De-Mining and Other Applications

There were significant contemporary advances in robot structures which may exchange people in project unsafe actions in hard or risky environments. released in organization with the CLAWAR (Climbing and strolling Robots and linked applied sciences organization) (www. clawar. org), this significant ebook studies the improvement of robot platforms for de-mining and different dicy actions comparable to fire-fighting.

Quality by Design for Electronics

This ebook concentrates at the caliber of digital items. Electronics in most cases, together with semiconductor expertise and software program, has develop into the major know-how for huge components of business construction. In approximately all increasing branches of electronics, in particular electronic electronics, is concerned.

Encyclopedia of Electronic Components Volume 2: LEDs, LCDs, Audio, Thyristors, Digital Logic, and Amplification

Need to know the way to use an digital part? This moment booklet of a three-volume set contains key details on electronics elements on your projects--complete with pictures, schematics, and diagrams. you are going to research what each does, the way it works, why it really is worthy, and what variations exist. regardless of how a lot you recognize approximately electronics, you will find attention-grabbing information you have by no means come upon earlier than.

Additional resources for Microchip Technology: The Past and the Future

Sample text

Then 180 + 200 − 600 = −220 (CHECKS) If v3 = −100 V, then v1 = −120 V and v2 = −200 V. 14 [a] Write a KVL equation clockwise aroud the right loop, starting below the 300 Ω resistor: −va + vb = −0 so va = v b Using Ohm’s law, va = 300ia and vb = 75ib Substituting, so 300ia = 75ib ib = 4ia Write a KCL equation at the top middle node, summing the currents leaving: −ig + ia + ib = 0 so ig = ia + ib = ia + 4ia = 5ia Write a KVL equation clockwise around the left loop, starting below the voltage source: −200 V + v40 + va = 0 From Ohm’s law, v40 = 40ig and va = 300ia CHAPTER 2.

6 A. 6 A) = 120 V. 4 A)] = −(200 V)(2 A) = −400 W Thus the voltage source delivers 400 W of power to the circuit. 16 [a] Write a KVL equation clockwise around the right loop: −v60 + v30 + v90 = 0 From Ohm’s law, v60 = (60 Ω)(4 A) = 240 V, v30 = 30io , v90 = 90io Substituting, −240 V + 30io + 90io = 0 Thus io = so 120io = 240 V 240 V = 2A 120 Now write a KCL equatiohn at the top middle node, summing the currents leaving: −ig + 4 A + io = 0 so ig = 4 A + 2 A = 6 A [b] Write a KVL equation clockwise around the left loop: −vo + v60 = 0 so vo = v60 = 240 V [c] Calculate power using p = vi for the source and p = Ri2 for the resistors: psource = −vo ig = −(240 V)(6 A) = −1440 W p60Ω = 42 (60) = 960 W p30Ω = 30i2o = (30)22 = 120 W p90Ω = 90i2o = (90)22 = 360 W Pdev = 1440 W Pabs = 960 + 120 + 360 = 1440 W CHAPTER 2.

5 [a] Yes, independent voltage sources can carry whatever current is required by the connection; independent current source can support any voltage required by the connection. 6 Write the two KCL equations, summing the currents leaving the node: KCL, top node: 25A − 20A − 5A = 0A KCL, bottom node: − 25A + 20A + 5A = 0A Write the three KVL equations, summing the voltages in a clockwise direction: KVL, left loop: KVL, right loop: − v25 + v20 = 0 60V − 100V − v5 − v20 = 0 Problems 2–13 60V − 100V − v5 − v25 = 0 KVL, outer loop: Note that since v5 , v20 , and v25 are not speciﬁed, we can choose values that satisfy the equations.