Parallel circuit with Two Voltage Sources

[pompeysam123]

Hi guys and girls, first post so be gentle please

I have a circuit with two votage sources and three resistors set out like this:

R1 330 R3 100 R2 270
+ +
15V 10
- -

: Using KVL I am pretty happy with my current calculations of I1=30.516431mA, I2=18.779342mA and I3 (I1+I2)= 49.0295773mA. But I'm pulling my hair out trying to calculate the voltage over each resistor.
Using Ohm's Law I make V1=10.0704223v V2=5.07042234v and V3= 4.9295773v is this Correct?? how come the voltage doesn't add up to 25V? am I going mad???
I also have to calculate the power in the load dissapated, What does this mean?
Apologies if this is an incredibly boring and easy easy question but im pulling my hair out!!! Thanks in advance.

 

[JimB]

The problem is that it is impossible to determine how things are connected.

The simple ASCII drawing has the spaces removed by the forum software.
Could you make a simple drawing using something like MS PAINT and attach it as a JPG or PNG file?

JimB

 

[MrAl]

Hi,

You can also do it by wrapping it in code tags like this:

   +--R1--+--R2--+
   |      |      |
  15v     R3     10v
   |      |      |
   +------+------+

There's no law that says that the sum of voltages in a circuit have to add up to the total of all the voltage sources added together, is there? The law i know of is: "The sum of voltage drops around a series circuit is equal to zero". Your circuit is not a series circuit so this doesnt apply.

To find the voltage drop across each resistor you simply use Ohm's Law:
V=I*R
and knowing I and R for each resistor you should be able to do this no problem. The polarity is according to which side of the resistor has the highest voltage and that's the positive side. For example, the left side of R1 is positive and the right side negative.

 

[Mosaic]

The voltage drop across R3 is common to both legs of the circuit.

Use simultaneous equations.

15 = V1 +v3
10= V3 +v2
(15-10)= V1 +v2

Solving reveals V1 = 5V
V3 = 10V
v2 = 0V

Now u can work out the currents using V=IR.
Now u can work out the Power using P=VI.

Spice analysis claims otherwise...seems there's a polarity issue with V2, depending on how u look at it its -5V to the left side of the circuit and +5V to the right side.

That causes V3 to be 5V, v2 is either +5 or -5 depending on how u look at it and v1 is 10V.

 

[MrAl]

The voltage drop across R3 is 4.92957746 with the top positive, so that makes the drop across R1 equal to 10.704335 with the left side positive and the right side negative, and the drop across R2 is 5.0704225 with the left side negative and the right side positive. It's that simple.