why the voltage across a resistor with two different input DC source is the avg of the two different input DC voltage

May i know how and why does the voltage across the 1ohm resistor is the average of the two input DC source which is shown in the given model below,

voltag across the 1ohm resistor = (200-100)/2

Answer 1

[David Goodmanson]

Hi RAM,

The 1 ohm resistor is quite a bit larger than the two .01 ohm resistors so in a first approximation the 1 ohm resistor can be neglected. (The 1 ohm resistor is basically like the input impedance of a voltmeter, which should be large). So, neglecting the 1 ohm resistor, arbitrarily calling g the ground point and taking votages with respect to g, then Va = 100, Vb = 300, and Vc (being in between the equal valued resistors) is Vb/2 = 150. So the voltage measurement Vc-Va = 50.

b -- R(.01) --- 
|             |
V2(200)       |
|             |  
a             c 
|             |
V1(100)       |
|             |
g -- R(.01) ---

More formally, the current around the loop is

I = (V1+V2)/(2R)

Going around the top of the loop,

voltage Vc-Va = V2-IR which is (V2-V1)/2

.Going around the bottom of the loop

voltage Vc-Va = -V1+IR which is also (V2-V1)/2

When the 1 ohm resistor is inserted, the voltage changes slightly from 50V as in the jpg.

In first approximation (without the 1 ohm resistor) the current is 15kA and the power dissipated in the circuit is 4.5 MW, but in a model anything is possible.

[RAM hanuman]

Hi David,

Thanks for the explanation.