translation ; we know the values of
R1 = 1 ohm
R2= 2 ohm
R3= 6 ohm
R4= 2 ohm
R5= 4 ohm
R6= 4 ohm
we can also see the voltmeter reading a potential difference value of 8 volts , and the ampere meter giving a reading of 2 amps at the indicated locations
The Attempt at a Solution;
so , as you can see with the following attachements i ended up with value for the missing resitance of 8 ohm and a f.e.m ( electro force ) oh 54.
by plugging those in we get a potential difference of 8V and a 2 amp signal .
but i got those value by testing a real circuit, and cant get the same numbers through formulas. by symetry , shouldnt the right side have the same potential difference of 8 volts as the left side ?
pretty sure my resistance value of 8 ohm looks ok .
am i wrong in my calculations ? if my handwriting is too bad i can rewrite the equations here . we know the current coming in at a junction has to be the same as the current coming out.
since there are only resistance and only 1 battery , im sure i already used the formula for the answer in one of my worksheet. . i know since i ended up with a value of 8R., gave me a hint that the missing R value was 8 .