if there is a circuit with a resistor , a 6v battery and a switch, when the switch is opened, there is no current passes through in the circuit, by V=IR, THE voltage across the switch should be 0, but why the voltage across it is 6?
I have think of it and i think maybe air act as a resistor of infinity resisrance so it shares all the voltage of the battery. but as voltage= potential difference = work done to bring a charge from 1 point to another, if the circuit is imcompeted, there will be no work done, the voltage will be 0. there will be contradition.
Your notion of viewing the resistance of the switch as infinite is a good one if you feel the need to write an equation for voltage drop. If that's a bother, think of it instead as many mnay times the resistance of the actual resistor.
As you learn more about circuit analysis, you'll learn about equivelent circuits. The mathematical equivelent of a simple resistive circuit is a voltage source and a series resistor (or a current source and a shunt resistor). As far as the terminals are concerned, you can't tell the difference between the equivlent circuit and the actual one. My memory says they're called Norton (current source and shunt) or Thevelen (sp?) (voltage souce and series) circuits.
The voltage drop across the *resistor* is 0 volts (and hence the full voltage is delivered to the switch contacts).
To the degree that the voltmeter loads the circuit, there will be current flow and a voltage drop across the resistor, which will make the voltage across the switch *lower* (less than the applied 6 volts).
-- Floyd L. Davidson Ukpeagvik (Barrow, Alaska) snipped-for-privacy@barrow.com
No contradiction. As you stated, the work done is a charge moving (current) through a potential difference. Since the switch resistance is infinite, there is no current (no charges moving). So despite the fact that there is a potential difference, no work is being done.
Hi Erica, Yes, the resistance of the switch is infinite so the current is 0. Because the current is zero, the voltage drop across the resistor is 0. Therefore, all of the battery voltage appears across the switch. Another way to look at the voltage being measured across the switch is that it's the voltage drop across the battery and resistor, which is
6-0=6.
Work - I think the best way to look at it is that the switch ceases to exist when it is open. You are measuring the potential of the battery with a resistor on one end with no current flowing, similar to an equivalent circuit for a voltage power source with nothing connected to it. The battery potential is 6 volts but the potential is not being used to perform work via (VI) across a passive device such as a resistor because there is no current flowing thru it to perform actual work. Close the circuit and work will be performed in the amount of VI where Vacross and Ithru are measured at the device in question.
V=IR, so I=V(battery)/R(total), next solve for current... I=Vb/(Rs+Rr), next solve for voltage across the switch.... Vs=IRs=VbRs/(Rs+Rr), next slove for voltage across the resistor... Vr=IRr=VbRr/(Rs+Rr)
As the switch resistance grows, the voltage across it approaches 6 volts as Rs/Rs+Rr approaches unity (infinity/infinity). As the switch resistance grows, the current thru it drops as Vb/(Rs+Rr) approaches zero (1/infinity). Work being performed in the switch, defined as VI, approaches zero as the current approaches zero (1/infinity). Work being performed in the resistor also drops to zero as the current approaches zero (1/infinity). Once the switch is open, all work stops.
In this case, you are only measuring a potential, not work actually being performed and must have a current in order to perform work. The work performed in a resistance is the energy required to forced the electrons to move across it. If the electrons don't move, there is not work being performed. The potential work is in the battery, which you are measuring. I'm sorry but I've been out of school too long to explain it any better.
The way Erica described her circuit, I assumed she was presenting it as exercise in physics and electrical engineering, Gerald. If that's the case, the voltage would be calculated rather than measured with a real meter, and the 6 volt battery would be lossless. In real life, yes a meter used to measure the voltage across the switch would introduce a very small amount of leakage current around the switch.
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