If I have a 9V circuit, with:
2 resistors in parallel; followed by:
2 single resistors in series;
and I calculate the effective R of the parallel pair (using 1/R=1/R1+1/R2); can i then add this value to the series resistors to get a total for the circuit? As in:
(1/R=1/R1+1/R2)+R3+R4 = Reff
If so, can I now use Ohm's Law to get the current flowing through the last single resistor?
Thanks
The way you did it in the formula, you added the reciprocal of the parallel resistance to R3 and R4.
Rparallel= (R1*R2/(R1+R2))
then add that to R3 and R4
I get it. My text shows your formula, but recommends 1/R= 1/R1+1/R2 etc as easier! I'm with you. If the numbers match, is my thinking ok?
I forgot to say "take the reciprocal". Does that make more sense?
Yes, you can add the effective resistance of the parallel resistors to the series resistors to get the total resistance of the circuit. However, there is a small mistake in your equation. Instead of adding the resistances directly, you should take their reciprocal values, add them up, and then take the reciprocal of the sum to get the effective resistance.
The correct equation would be:
1/Reff = 1/R1 + 1/R2 + 1/R3 + 1/R4
Once you have calculated the effective resistance (Reff) using this equation, you can then use Ohm's Law (V = I * R) to calculate the current flowing through the last single resistor. Ohm's Law states that the current (I) flowing through a resistor is equal to the voltage (V) across it divided by its resistance (R).
So, if you know the voltage across the last single resistor (which in this case is the same as the voltage across the entire circuit, since the resistors are in series) and you have calculated the effective resistance (Reff), you can rearrange Ohm's Law to solve for the current (I).
I = V / Reff
Note that in this circuit, the voltage across the last single resistor would be the same as the voltage across the parallel resistors and the series resistors, since they are all connected in series.