Primarygun Posted September 7, 2005 Posted September 7, 2005 Here are two questions concerning the circuit things. 1.In a series circuit with a bulb in the middle of it with 5V. (Please neglect resistance of wire) My teacher said that each electronwould lose all of the energy, i.e.5J per second, after they have passed the bulb. So it has a decline in its velocity after passing the bulb? 2.In the circuit above, a voltmeter is connected. The voltmeter starts a new loop for the circuit, so the energy given by each electron to the bulb will change and then how does a voltmeter work? Thanks for any replies.
insane_alien Posted September 7, 2005 Posted September 7, 2005 1. the electrons still move along the wire i.e. there is still a current. but they have little to no energy available to do work. 2. a voltmeter is practically an open circut so its effect on the voltage across the bulb is negligble. i'm not actually quite sure how a voltmeter works.
mezarashi Posted September 7, 2005 Posted September 7, 2005 To your first question. No, it's not a loss in velocity. Actually the drift velocity of electrons in a copy wire isn't all that great. Consider how many electrons are in a centimeter of wire. How much fast does charge need to flow to get 1A? The loss is the loss of electrical potential. You can imagine yourself on an elevator coming down from the 80th floor. When the elevator passes by the 10th floor, you are at the same speed, but you have lost gravitational potential. For question 2. The voltmeter (atleast in theory) does not affect the circuit due to its very very high input resistance. Meaning no current will go into the voltmeter. Modern voltmeters will probably use a ADC (analog to digital converter) somewhere to detect the charges on internal capacitors.
5614 Posted September 7, 2005 Posted September 7, 2005 The drift velocity of an electron will be approximately 1cm/s, this obviously varies, this is an approximation in more ordinary circumstances.
Primarygun Posted September 8, 2005 Author Posted September 8, 2005 To your first question. No, it's not a loss in velocity. Actually the drift velocity of electrons in a copy wire isn't all that great. Consider how many electrons are in a centimeter of wire. How much fast does charge need to flow to get 1A? The loss is the loss of electrical potential. You can imagine yourself on an elevator coming down from the 80th floor. When the elevator passes by the 10th floor, you are at the same speed, but you have lost gravitational potential. I know what the fact is, thank you. So we get a measure of the velocity of the electrons after they pass through the bulb before getting the conclusion of they lose electric potential instead of velocity , or something else? For question 2. The voltmeter (atleast in theory) does not affect the circuit due to its very very high input resistance. Meaning no current will go into the voltmeter. Modern voltmeters will probably use a ADC (analog to digital converter) somewhere to detect the charges on internal capacitors. so is having high resistance is necessary for a proper voltmeter?
mezarashi Posted September 8, 2005 Posted September 8, 2005 I know what the fact is' date=' thank you.So we get a measure of the velocity of the electrons after they pass through the bulb before getting the conclusion of [b']they lose electric potential instead of velocity [/b], or something else? Yes, we say that there is a voltage drop across the resistor. Voltage is also known as electric potential. so is having high resistance is necessary for a proper voltmeter? Yes. Because ideally you don't want to disturb the circuit. You want to be able to probe any two points on your circuit without changing the circuit dynamics. This is only possible if there is a high input resistance.
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