physics voltage question?
The circuit diagram features a lightbulb with a resistance of 1.4Ω. Let ΔVab represent the magnitude of the potential difference between points a and b. I have the following questions:
- What is the value of ΔV12?
- What is the value of ΔV23?
- What is the value of ΔV23 if the lightbulb is unscrewed from its socket?
1 Answers
The 3V source see a resistance of 2Ω+1.4Ω = 3.4Ω.
Therefore, the current leaving the source = 3V/3.4Ω = 0.9A
1. What is the value of ΔV12? 0.9A*2Ω = 1.8V
2. What is the value of ΔV23? 0.9A*1.4 = 1.2V
3. What is the value of ΔV23 if the bulb is unscrewed from its socket? When the bulb is unscrewed, you have a open circuit so current will cease which means V12 = 0V which means V2 = V1. V3 and V4 are really the same point since they are shorted together. so V23 = V14 = 3V
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