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 Oct 26th 2017, 05:58 AM #1 Junior Member   Join Date: Sep 2017 Posts: 5 Circuits Thank you for helping me with two homework questions: Two attachments are attached to my question. The one with the two light bulbs and the open switch is about the potential difference between point A and B. I thought it was 0V, because the circuit isn't closed, so there is no potential. However the answer is 12 V. Can someone explain? The other picture (three lightbulbs and an open switch) is about what happens to the brightness of the lights when the switch closes. Obviously light C will start burning, leaving light B to burn less bright, because both of them now share the current that was first fully going to B. However, I thought that nothing would change for light A, because it is still on the same wire and nothing happens to its current or voltage. However, the answer says light A will burn more brightly. Why? Thank you so much for your help. cheers, Max Attached Thumbnails
 Oct 26th 2017, 06:44 AM #2 Senior Member   Join Date: Apr 2017 Posts: 177 1. .... the potential between A and B is the same as the potential across the battery , the only difference is , one light is in the circuit , which will not affect the potential ... 2. when the switch is on , lighting up C , will the current drawn from the battery change ??? It must increase, because you've reduced the overall resistance the battery sees , and that increased current is flowing through A
 Oct 26th 2017, 11:28 AM #3 Senior Member     Join Date: Jun 2016 Location: England Posts: 377 The key to the first question is the word "potential" It is the voltage that the Battery could potentially supply. The actual voltage supplied by a battery will depend on the current being drawn, which in turn depends on the resistance of the circuit. For a high resistance circuit drawing a low current the battery has the potential of delivering 12V. Your second example is definitely counter-intuitive (which is no doubt why your tutor included it as one of the questions) The key is to work out the way the resistance of the circuit changes. introducing the second bulb actually reduces the overall resistance of the circuit. as oz93666 says, this means extra current is drawn from the battery so lamp A gets brighter. __________________ ~\o/~

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