Suppose I have a +15 and - 15 volt supply with 3 resistors in series connected between the supply. Also the middle resistor is a pot. How do I calculate the current and drop across the each resistor and finally at the center tapped lead of the middle resistor.

Thanks

 

it would be better if u posted the circuit
from what i figure the net voltage will be 30V and entire 30 volts appear across 3 resistors what r the values of resistances?

 

Use I = E/R for the series current. Then E = IR to get the drop across each resistor. The drop across the pot will give you the min and max voltage the wiper can output.

 

Use I = E/R for the series current. Then E = IR to get the drop across each resistor. The drop across the pot will give you the min and max voltage the wiper can output.

So is the total voltage 30? to which I apply ohms law. Here is a picture. I want to calculate for the bridge. I have just never done it with a negative voltage. I do this as a hobby only. Hence my need for help. I just dont know how to go about setting up the problem.

thanks

 

So is the total voltage 30? to which I apply ohms law. Here is a picture. I want to calculate for the bridge. I have just never done it with a negative voltage. I do this as a hobby only. Hence my need for help. I just dont know how to go about setting up the problem.

thanks

Ohms law is valid for all resistors. First calculate the total resistance of the voltage divider. You do that by calculating the equivalent resistance of R203/R205 parallel association. Then you calculate the resistance for the whole series. Then you calculate the current that biase the series or resistors by applying the Ohm's Law, knowing that V = 30V.

Assuming that you have a rock solid voltage divider, that is, the current drawn by the amplifier will be negligible and won't affect the potentials significantly, you can use the results for the current to calculate the voltage drop for each resistor. Notice that R203 and R205 are in parallel, so the current will split for each. Consider the parallel association resistance that you calculated before to calculate the voltage drop across the paralel association (which will be equal for R203 and R205).

Equations to use:
Rseries = R1 + R2 + R3 + ... + Rn
R paralel = 1 / (1 / R1 + 1 / R2 + 1 / R3 + ... + 1 / Rn)
R = V / I

That would be it.

 

Assuming no load on the pot wiper, R203 and R205 in parallel are equivalent to 2372.35 ohm. The total resistance of all the resistors is then 80672.35 ohm. the current flowing is 30 V / 80672.35 ohm = 0.0003718746 A. By multiplying this current by the value of each resistor we find the voltage drop in each one. Beginning from the bottom: 2.3056, 15.6187, 0.8822, 7.8094, 3.3841 V.

So, the voltage as measured from the -15 V rail can range between 17.9244 and 18.8066 V depending on the position of the wiper.

We could redo the calculations taking into account the tolerances of the values of the resistors.

 

Assuming no load on the pot wiper, R203 and R205 in parallel are equivalent to 2372.35 ohm. The total resistance of all the resistors is then 80672.35 ohm. the current flowing is 30 V / 80672.35 ohm = 0.0003718746 A. By multiplying this current by the value of each resistor we find the voltage drop in each one. Beginning from the bottom: 2.3056, 15.6187, 0.8822, 7.8094, 3.3841 V.

So, the voltage as measured from the -15 V rail can range between 17.9244 and 18.8066 V depending on the position of the wiper.

We could redo the calculations taking into account the tolerances of the values of the resistors.

No, the voltage (always relative to ground) varies from 2.924V to 3.8065V. Since you are considering the voltage drops from 15V to -15V, you have to subtract 15V from these values, in order to get the voltage at each node.

You can redo the calculations based on the tolerances, if you have to guarantee that the voltages cannot pass certain limits. If you don't have to guarantee that, the nominal values of the resistors will be fine for your calculations.

 

No, the voltage (always relative to ground) varies from 2.924V to 3.8065V.

And how is this different from what I said?

 

Different reference point is all. Some classes teach to always reference from ground. Other classes teach to always state reference point.

 

Maybe I am missing something but the fact is that my answer was correct and to that cumesoftware replied

No,

which I think can only be interpreted as a negation of what I had posted. Maybe he meant "yes"?

 

He was taught to reference from ground. You specified reference from the negative rail. That's why you got different numbers. "No" means his numbers did not agree with yours. No big deal, really.