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In Step 4, was the potential at the low side of D1, 'pulled' higher or lower, relative to ground by the jumper

In Step 4, was the potential at the low side of D1, 'pulled' higher or lower, relative to ground by the jumper

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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In step 4, was the potential at the lower side of D1, pulled higher or lower relative ground by the jumper wire ?
In Step 4, was the potential at the low side of D1, 'pulled' higher or lower, relative to ground, by the jumper wire? Higher. This is a pull-up connection. The potential was not pulled higher, or lower, because it remained unchanged. The potential was not pulled, it was pushed.I still did not read the title of this lab assignment. Lower. This is a pull-down connection.
Transcribed Image Text:In Step 4, was the potential at the low side of D1, 'pulled' higher or lower, relative to ground, by the jumper wire? Higher. This is a pull-up connection. The potential was not pulled higher, or lower, because it remained unchanged. The potential was not pulled, it was pushed.I still did not read the title of this lab assignment. Lower. This is a pull-down connection.
STEP 4 Remove the jumper wire you added in Step 3 and connect it in parallel with D1 instead, as shown here: D1 R1 D2 Red 330 Q Blue PSB 5 V This time D1 will be shorted and go out, while D2 becomes brighter. With the jumper connecting the low side of D1 directly to the positive breadboard rail, what will be the new potential of that junction point, relative to ground? • Was the potential of that junction point pulled up (higher) or down (lower), relative to ground by the jumper wire?
Transcribed Image Text:STEP 4 Remove the jumper wire you added in Step 3 and connect it in parallel with D1 instead, as shown here: D1 R1 D2 Red 330 Q Blue PSB 5 V This time D1 will be shorted and go out, while D2 becomes brighter. With the jumper connecting the low side of D1 directly to the positive breadboard rail, what will be the new potential of that junction point, relative to ground? • Was the potential of that junction point pulled up (higher) or down (lower), relative to ground by the jumper wire?
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