# beginner

Discussion in 'Electronic Basics' started by luc, Feb 20, 2004.

1. ### lucGuest

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?

2. ### Rheilly PhoullGuest

More input !!
Are the resistors in series or parallel or other combination. Perhaps you
should have paid more attention to your tutor with regard to ohms law etc ??

3. ### lucGuest

This is a simple series circuit.

4. ### michael turnerGuest

Hmmmm...A homework question...You're supposed to work it out yourself.

6. ### Greg NeillGuest

Even more input !!
Do you know the total voltage across the resistor string?

7. ### Sir Charles W. Shults IIIGuest

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
your current for any given supply voltage and usable value of R3.

Cheers!

Chip Shults

8. ### John PopelishGuest

If you have a circuit made up of only 3 resistors, your meter is
broken if you read 4.5 volts across any two nodes of the circuit. If
you actually have 4.5 volts across R3, there has to also be a power
source somewhere in the circuit. How the 3 resistors are connected to
that power source affects the voltages and currents.

For instance, if all 3 resistors are in parallel across the power
source, then they all share the 4.5 volts across R3, and you can
easily calculate the currents through R1 and 2, because you know their
resistances. But you can't calculate the current through the unknown
resistance, R3. If all resistors are in series across the power
source, You know that all share the same current, but without knowing
either the value of R3 or the source voltage, you can't know what that
current is. Etc.

9. ### Bill BowdenGuest

You have to look for another known value, there is
not enough information. Or, you can just plug in something
for R3 and work out the rest. Suppose R3 is 4.5K. That would
make the current equal to 1 mA which would drop 5.5 volts on
R1 and 2.2 volts on R2 so the total supply voltage would
be 2.2 + 5.5 + 4.5 = 12.2 volts. So, you can see there are
many answers to the problem depending on the value of R3.

-Bill

10. ### Robert C MonsenGuest

There are at least two unknowns in your equation, the total voltage across
the string, and R3. Unless you can eliminate one of them somehow, you can't
figure out what either are without another independent equation.

Regards,
Bob Monsen