Simple logic diagram-with switch and resistor

[SyNtHeSiS]

Homework Statement

Why is it that the output is equal to V+ (when the switch is open) in diagram A, if there is no current flowing?

Why is it that A (1/0) is used by convention? I mean the lamp I got at home has a "0" for off and "1" for on, on the switch. This is confusing me.

Homework Equations

High= 1
Low = 0

The Attempt at a Solution

Not sure at all.

 

[RSW00001]

Suggestion: Apply the voltage divider rule assuming the switch to be a resistor which can have one of two values (∞ or 0) depending on whether it's open or closed.

 

[RajChakrabrty]

leave voltage diving rules and others.

very simply,
if upper resistor (called Pull up resistor)-----is present while lower (called Pull down)
resister is not present,
that means that total amount of current coming from power supply V+,
will pass through the o/p node.
thus,
o/p node gets full amount of current coming from power supply V+.


** the reason of your confusion is,
there is no connection made to ground, then how does the current flows through network.
but, there is always a load (or any other network, that is connected to GND obviously.)
so, current flows through that path.

Hope your confusion will be solved.
Best of Luck.

 

[SyNtHeSiS]

Oh thanks. I also want to know if there is only 1 resistor in a circuit, why is it that the total V+ will be dropped across the resistor, even if V=100V, I mean what if that resistor is small and only consumes a bit of V?

 

[Antiphon]

Resistors don't consume V. They flow a current proportional to whatever V is applied to their terminals. They consume power which is V*I.

The voltages in a circuit loop must total to 0.

 

[RajChakrabrty]

Resistors don't consume V. They flow a current proportional to whatever V is applied to their terminals. They consume power which is V*I.

The voltages in a circuit loop must total to 0.

yes.
that's right.
whenever that type of ckt you will see
you think like a o/p network is attached with it.
this type of ckt is made for this purpose
you think, that without any network or resister at o/p ,
how can you measure the voltage.
even if you measyre with a voltmeter or multi-meter, there will also be a resitor to measure it.
and low resistance case is discussed in just above post.

 

[RSW00001]

While resistors don't "consume" voltage they do "drop" voltage. That is why I suggested using the voltage divider rule (substituting a resistance for the switch) and see what happens as you vary the bottom resistance from some finite value down to zero and then up to infinity.

 

[SyNtHeSiS]

Why also is it that current is defined as I = V/R all over a circuit? As in why is it that the current where the circuit has no resistance different? e.g. voltage line between 2 resistors and line between Earth and a resistor.

 

[berkeman]

Why also is it that current is defined as I = V/R all over a circuit? As in why is it that the current where the circuit has no resistance different? e.g. voltage line between 2 resistors and line between Earth and a resistor.

A real wire has a real (but very small) resistance. So the current flowing through it does generate a small voltage drop. You can figure out how much the drop will be, based on wire resistance tables and the current that is flowing.

Wire resistance table: http://en.wikipedia.org/wiki/American_wire_gauge

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