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In Step 4, what is the potential difference across D1 after the jumper wire is connected? O Approximately 2.6V because my LED is still lit and the jumper wire is not properly connected. OL because I am not using the correct meter settings. 0.0 mV, because the LED is not lit and the component leads are connected to each other by the jumper wire. O Approximately 5V because I am somehow measuring across the whole path instead of just D1.

In Step 4, what is the potential difference across D1 after the jumper wire is connected? O Approximately 2.6V because my LED is still lit and the jumper wire is not properly connected. OL because I am not using the correct meter settings. 0.0 mV, because the LED is not lit and the component leads are connected to each other by the jumper wire. O Approximately 5V because I am somehow measuring across the whole path instead of just D1.

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, what is the potential difference across D1 after the jumper wire is connected
In Step 4, what is the potential difference across D1 after the jumper wire is connected? Approximately 2.6V because my LED is still lit and the jumper wire is not properly connected. OL because I am not using the correct meter settings. 0.0 mV, because the LED is not lit and the component leads are connected to each other by the jumper wire. Approximately 5V because I am somehow measuring across the whole path instead of just D1.
Transcribed Image Text:In Step 4, what is the potential difference across D1 after the jumper wire is connected? Approximately 2.6V because my LED is still lit and the jumper wire is not properly connected. OL because I am not using the correct meter settings. 0.0 mV, because the LED is not lit and the component leads are connected to each other by the jumper wire. Approximately 5V because I am somehow measuring across the whole path instead of just D1.
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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