# output voltages problem

Discussion in 'Homework Help' started by metelskiy, Oct 24, 2010.

1. ### metelskiy Thread Starter Member

Oct 22, 2010
66
3
Hi. I have a HW problem that i need help with:
A 12V battery output is divided down to obtain two output voltages. Three 3.3kΩ resistors are used to provide the two outputs with only one output at a time loaded with 10kΩ. Determine the output voltages in both cases.
Can someone please help me to understand this problem? I can't imagine how is the circuit looks like. Thanks.
Here is what i drew regarding to the problem:

Are they basically asking me to find output voltages for A and B ?

2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
What the exercise wants to say is that you will have only one load (the 10kΩ resistor) at a time, hooked up on your circuit. The load will be mounted either between terminal A and Ground or terminal B and Ground.

Your job is to determine the voltage that the load will get each time.

3. ### renotenz New Member

Oct 25, 2010
11
0
Yup, just use voltage divider to do the calculations

4. ### metelskiy Thread Starter Member

Oct 22, 2010
66
3
I need help with lab report which has to do with voltage devider. First, our original design requirements were:

The desired voltage at point A is +6V. The current flowing through R1 and R2 must be 1mA ± 5%. The goal is to implement the circuit with only two standard 5% tolerance carbon composition resistors.
So it was done with result using R1=3.9k and R2=5.6k and 1mA through it (V at point A was very close to 6V).

Then in next lab we had New Design Requirements:
Four hundred units containing the voltage divider you designed last week are now in the field. Field service has determined that the voltage at point A should have been +5V instead of +6V. You have been asked to recommend the simplest field modification that can be made at the customer site. Your supervisor suggested as solution the addition of a parallel resistor across R2.
So we spent almost whole lab figuring out R3 to put in parallel with R2. We used 5V=(3.9k/[(5.6^-1)+(R3^-1)]^-1)*5V which ended up R3=11kΩ.
Now i have to write lab report about part 2 where we had to put R3 in parallel with R2. In my understanding, since we have power source 10V and we need +5V at point A (between two resistances in series) aren't R1 and R2,3 should be same value in order to split V in half? Can someone please explain me how did that 11k resistor in parallel with 5.6kΩ effected the voltage to go down from 6V to 5V? Wouldn't voltage drop through R1 (3.9k) be 4V? Thanks.

5. ### hobbyist Distinguished Member

Aug 10, 2008
802
76
Yeh I see where you are getting confused, that 4 volt drop across R1, ONLY happens when you have R2 as 5.6K ohms.

Any time you change the value of a resistor in a voltage divider, then the current will either increase or decrease, depending on the kind of resistor value change, in your case the resistors in parrallel, would increase the overall current, as less resistance is now present in the entire circuit divider.

So now you need to recalculate the voltage drop across R1, because a new current value flows through R1, different then when it did with R2 alone.

So just remember ;
any change in resistant values in a voltage divider, will always change the amount of current flow through the divider, which inherently changes all voltage drops across all resistors in that same divider.

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