In a Series-Parallel circuit, how do you find the resistance and the current if only the voltage is given? 5 resistors in series and 1 resistor in parallel.
R1= 3 ohm Voltage= 3V Current 1A
R2= 5 ohm Voltage =1.65 current 0.33A
R3= ??? voltage= 4 volts (parellel)
R4= 4 ohm Voltage 1.32 Current 0.33
R5= 1 ohm voltage 1 volt current 1A
R6=3ohm voltage 0.99 current 0.33
If all these resistors that are in series are really in series then the current through them all will be the same. If R1 and R2 are in series you can't have 1A through one and .33A through the other.
Eddy Current is a little dizzy over Milli Volt.
For resistors in parallel
to add more than 2 resistors in parallel
do above calc for R1 and R2
Then do same calc for above result and R3
then simply add total series and parallel resistor values together.
All current that goes into a circuit must leave the same circuit.
(since you have 3 resistors that have .33a they must be in series.)
R2 + R4 + R6 so 5ohms + 4ohms +3ohms = 12ohms
Rt1 = 12
It1 = .33amps
Et1 = Rt1 X It1 or 12 X .33 = 3.96v (I will round that to 4v)
two resistors have 1 amp of current flowing they must also be in series.
Rt2 = R1 + R5 or 3ohms + 1 ohm = 4ohms
E2t = It2 X Rt2 or 4ohms X 1 amp = 4volts
Since both have a total of 4 volts across the series branches
both of these series circuits are in parallel
Then add the R3 with a 4volt drop also to the circuit, and since its in parallel, it uses E = I X R
or 4V = 1amp X 1ohm, Since the 4volts is across the resistance of R3 it must be 4ohms and a current of 1 amp.
ohms law proves this R =E / I or 4v / 1 amp = 4 volts
you have a series parallel circuit (that is actually three series circuits, that are in parallel
Just for fun:
Adding the currents and using ohms law, we see that
Rt = Et / It or 4v / 2.33 amps = 1.73 ohms of Total resistance.
This is a close approximation of the total resistance of the circuit.
There are other methods that may work also, but, I find this the most logical, without going into higher math, (which I abhor)
Hope this comes close to the answer your looking for.
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