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Okay, so suppose you have 3 light bulbs connected in series: /Battery/-----------(light bulb 1)------------(light bulb 2)-----------(light bulb 3)--------connects back to battery. If light bulb 2 is unscrewed, will light bulb 1 and light bulb 3 also go out? My guess is that light bulb 3 will go out but not light bulb 1. Thanks anyway!
And got the following answer:
There is some logic behind your reasoning but it has led you to the wrong conclusion. With bulb 2 unscrewed and with the positive terminal of the battery connected to light bulb 1 you probably reasoned that bulb 1 would still have a positive applied Voltage and bulb 3 would not. That reasoning is correct. With bulb 2 unscrewed and with the negative test lead of a Voltmeter connected to the negative terminal of the battery and the positive test lead connected to either terminal of bulb 1 the full battery Voltage would be measured whereas zero Volts would be measured at either terminal of bulb 3 under the same circumstances. But bulb 2`s filament makes (bulb 2 screwed in) or breaks (bulb 2 screwed out) the circuit`s current path for both bulb 1 and bulb 2 very similar to the way an on-off switch would do. Of course with zero current through the light bulb filaments the light bulb(s) will not light. Therefore your conclusion should be that if either of the three bulbs in a series connected circuit such as the one described is unscrewed then current will cease to flow and all three light bulbs will go out.