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)
And Gerald claims to teach this stuff...
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) email@example.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.
Good luck with the rest of your homework. ;-)
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
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...
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.