# Simple logic diagram-with switch and resistor

## 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.

High= 1
Low = 0

## The Attempt at a Solution

Not sure at all.

#### Attachments

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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.

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.

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?

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.

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.

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.

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
Mentor
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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