Series Resistor Circuit Voltage (2 Power Sources)

ebolbol

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Calculate the voltage at point B in the given circuit.

I'm trying to polish up on circuitry in my rusty brain and this was a sample quiz question I failed on. I did great with single power source series circuits, but this got me baffled. The correct answer was +10.48 V

Could anybody explain how you would arrive at that answer?

OHM's law didn't fail me, I failed it and I'm very disappointed at my self.

This is probably simple stuff for most of you.
 
Both power sources share a common negative, so the voltage difference is only 3 volts.
That 3 volts is dropped across the total of all 4 resistors with a total value of 150k ohms.
Point B is very close to mid point of resistor values R1 plus R2 equal 74k ohms,
and R3 plus R4 equal 76k ohms , so at point B it equals 1.48volts above 9 volts.
Result being 10.48 volts at point B above earth
 

Both power sources share a common negative, so the voltage difference is only 3 volts.
That 3 volts is dropped across the total of all 4 resistors with a total value of 150k ohms.
Point B is very close to mid point of resistor values R1 plus R2 equal 74k ohms,
and R3 plus R4 equal 76k ohms , so at point B it equals 1.48volts above 9 volts.
Result being 10.48 volts at point B above earth

Thanks to both of you. I did a ton of Googling and couldn't find what you actually posted via the link.

The explanation makes sense. Thanks again.
 
I get 1.52 v at B referenced to ground
the 9V should be subtracted from the 12V source before you calculate the total resistance. the voltage on the circuit is only 3 V there is jut no way to get 10 v at point B
 
I get 1.52 v at B referenced to ground
the 9V should be subtracted from the 12V source before you calculate the total resistance. the voltage on the circuit is only 3 V there is jut no way to get 10 v at point B

3 volts is between voltage source plus terminals.

On point b there is 3V/150 ohm * (56ohm + 20ohm)= 1,52 volts to 12V source's terminal.
To ground voltage is 12V-1,52 = 10,48V
 
Alternative approach using superposition:

VB due to 12v supply = 12*74/(76+74)

VB due to 9V supply = 9*76/(76+74)

VB due to both =(888+684)/150 = 10.48V
 
you guys keep forgetting the first rule of electricity
the sum of all voltage in a circuit = 0
the sum of all current in a circuit = 0
following that the voltage at B with reference to ground (Neg of both sources ) is 1.52 V
 
the voltage at B with reference to ground (Neg of both sources ) is 1.52 V
How can the voltage at B, with respect to common, be less than either of the source voltages?
 
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what is the source voltage, in then case it is 12V - 9V = 3 V
the 9v cancels out 9v's of the 12 so the source voltage is 3V
most people forget to add the neg side of the source
a single battery would be +12 bat -12 sum 0 it doesn't work any other way
 
Gary, I'm afraid you're incorrect on this one.

If your measurement reference for zero was the positive of the 9V battery, then you'd be correct. The total load voltage is indeed only 3 volts, but it is offset from zero by the 9V that is common to both sources.

It's an easy one to test... go grab two different voltage batteries and hook up a couple big value resistors in the arrangement shown and measure voltage with respect to the common negative.
 
that the first ting I was taught when I first starts out over 35 years ago the basic rules of electricity don't change . you just haven't learned how to add the voltages correctly
I have trouble shot many system just with that knowledge and it has never missed.
but you have to add all voltage that both pos and the neg on each device. and you notice I said the voltage is 1.52 referenced to the ground or neg side of both sources If you reference to the pos side of one of the sources the voltage will be different you must always know where your reference is when measuring voltage.
 
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think about this what if both sources were 12V what would the voltage be and what would the current be
current is simple 0 but you could still have a voltage measured at point B also don't forget the meter loading effect he meter will draw some current itself.
 
Doug that would depend on how you measure it the meter will have a load effect on the circuit it maybe small but it will effect it.
on high impedance meter I would think it would be 12V at any point
assuming that both sources are 12V and again with a reference to ground
 

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