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physics voltage question?

The lightbulb in the circuit diagram has a resistance of 1.4Ω . Let ΔVab be the magnitude of the potential difference between point a and point b.1. What is the value of ΔV12?2. What is the value of ΔV23?3. What is the value of ΔV23 if the bulb is unscrewed from its socket?
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Everardo Feil

Feb 20, 2025

The 3V source see a resistance of 2Ω+1.4Ω = 3.4Ω. Therefore, the current leaving the source = 3V/3.4Ω = 0.9A1. What is the value of ΔV12? 0.9A*2Ω = 1.8V2. What is the value of ΔV23? 0.9A*1.4 = 1.2V3. 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... Show More
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