Voltage After a Single Resistor or Between Two?

Thread Starter

speedster239

Joined Nov 24, 2009
12
I understand how voltage drop is calculated when two or more resistors are in series and you're attempting to calculate the voltage drop between a resistor, but what about in these two situations when you're trying to calculate the voltage drop after or between two resistors?

This might seem like a silly question to those of you with tons of experience with electricity, but I'm at a loss to figure this out. I really appreciate your help.



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R!f@@

Joined Apr 2, 2009
10,007
An open circuit has no current, which results in having the source voltage at the open circuit

which basically means that if you open a circuit and measure voltage at that point, u will get the voltage as same as the source voltage.
Does not matter where you open the circuit.
 

R!f@@

Joined Apr 2, 2009
10,007
On the other hand, you cannot measure a voltage on a conductor.
Unless you have a highly sensitive milivoltmeter
 

Ghar

Joined Mar 8, 2010
655
Voltage drops add as you go around a loop.

In your second picture (assuming it's a closed circuit, not as drawn) you have two 6 ohm resistors.
The current in your circuit is 12 V / (6 + 6) = 1 A.

The voltage drop across each resistor is 1A * 6 ohms = 6 V.

The voltage in the middle is the battery voltage 12, minus one drop, 6, giving you 6 V in the middle node.
 

Thread Starter

speedster239

Joined Nov 24, 2009
12
Voltage drops add as you go around a loop.

In your second picture (assuming it's a closed circuit, not as drawn) you have two 6 ohm resistors.
The current in your circuit is 12 V / (6 + 6) = 1 A.

The voltage drop across each resistor is 1A * 6 ohms = 6 V.

The voltage in the middle is the battery voltage 12, minus one drop, 6, giving you 6 V in the middle node.
Yes, both of the schematics are meant to show a closed circuit, sorry.

Why is it only minus one drop? I don't understand that fully. Why aren't both resistors accounted for? Or are they?
 

Ghar

Joined Mar 8, 2010
655
There are 3 voltage 'drops'.

6 across the first resistor, 6 across the second resistor, and -12 across the battery (so it's actually a voltage rise)

The sum of the 3 is 0:
6 + 6 - 12 = 0

This is true around any closed loop.

So if you put the negative of the battery as 0, the positive of the battery is +12.
Go through one resistor, it's 12 - 6 = 6.
Go through both resistors and you're back at the negative of the battery, 12 - 6 - 6 = 0.

This is Kirchoff's voltage law and is one of the basic tools in circuit analysis.

This is the AAC chapter you want:
http://www.allaboutcircuits.com/vol_1/chpt_6/index.html
 

Thread Starter

speedster239

Joined Nov 24, 2009
12
Thank you guys for your help, I figured that last question out. Now I have another one...

I know that you can use a transistor to regulate current by allowing current to flow between the base and emitter terminals, which will in turn allow a larger current to flow between the collector and emitter terminal.

Is there a way that I can wire up such a "switch" in the exact opposite sense. Ie, if I allow electricity to flow between the base and emitter terminals, current can NOT flow between the collector and emitter terminal, but while current isn't flowing between base and emitter it CAN flow between the collector and emitter.

Is there a way to wire this up?

Thanks again,
Vaughan
 

Ghar

Joined Mar 8, 2010
655
Not like how you explained because that's just how a transistor works. However you can have the same functional result by using the transistor in a different way or by using a combination of transistors.
I'm assuming you just want a switch that's off when the input signal is high?
 

Thread Starter

speedster239

Joined Nov 24, 2009
12
Not like how you explained because that's just how a transistor works. However you can have the same functional result by using the transistor in a different way or by using a combination of transistors.
I'm assuming you just want a switch that's off when the input signal is high?
Yes indeed. And when there is no input the current does flow. How would I accomplish such a task?
 
Last edited:

Georacer

Joined Nov 25, 2009
5,182
If you have small current demands, you could use a PNP transistor. Unlike NPN, PNP transistors, allow current to flow from the Emitter to the Collector when the Base is kept at the Low voltage. If you want to stop current from flowing, raise the Base to High voltage. Be careful of the difference in pin layout between PNP and NPN transistors. And remember that the maximum current you can draw through the transistor is a manufacurer parameter you cannot overcome.
 
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