luc said:
I have a circuit with 3 resistors.
R1=5500 ohms, R2= 2200 ohms, R3=?
The only other thing that is known is the voltage over R3, which is 4.5V.
How can one derive from this, the current running in this circuit and R3.
What do you have to look for first and why?
First off, if this is a series circuit (as you posted below) then you
are still missing some data. All we can do at this point is lump R1 and R2
and use that as a single 7700 ohm resistor.
So, we have a single known resistance (7700 ohms), a voltage drop across
an unknown resistor (R3), and no idea what the source voltage is. From
that, we can determine only that a) the circuit has at least 7700 ohms of
resistance, b) the supply voltage is at least 4.5 volts, and c) we do not
know enough to figure out anything else.
If R3 is an open circuit, then the supply voltage would indeed be 4.5
volts, and you could not draw more than (4.5V / 7700 ohms) or 584.4
microamps through it. And clearly, R3 cannot be a short or the voltage drop
across it would not be 4.5 volts.
The best you can do is to create a graph showing the relationship
between R3 and the supply voltage, and eliminate many values as being
impossible at either end of the graph. Values of the supply voltage can
only range from 4.5 to infinity at a first approximation.
And, on the resistance axis of the plot, the values of R3 can only range
from some non-zero value to infinity (open circuit). So you have two
definite limits to the graph in any case. Do it on log graph paper, using
sample value of R3 and Ohm's formula to determine the required source
voltage in each test case. The yield will be a curve that will show you for
each assumed value of R3 what the resulting supply voltage will be.
Once you have that value, you can easily determine the current through
the circuit through the formula (Vsupply / (7700+R3)). In fact, just assume
the 7700 ohms as being in series to start with and your graph will yield
your current for any given supply voltage and usable value of R3.
Cheers!
Chip Shults