Physics: Circuitry question?

Assuming resistor A = resistor B.

The Voltage across A is halved when B is added in series with A, but A's resistance stays the same.

Power in Watts of A in original circuit = [(1^2)V]/(R of A)

Power in Watts of A after resistor B is added = [(.5^2)V] /(R of A)

Power ratio = [(.5^2)V]/(R of A)] / [(1^2)V]/(R of A)] = .25/1 = 1/4

Assuming the SIZE of resistors A and B are the SAME in both circuits,

then

when two of the same size resistors exist instead of only one, the current that flows = I = V/2R, which is one half the original current of the 1st circuit. However, the POWER used by one of these two identical resistors is P = I²R so,

The value of the current (I) has been halved, but that value is SQUARED in the formula for P, so IT (meaning the power used by a resistor in this circuit = (1/2)² = 1/4 <= *

*Notice: If BOTH resistors are taken into account, the TOTAL power used by the 2nd circuit is 1/2 the power of that used by the 1st circuit. :>)

If A and B are equal, circuit current halves, but circuit resistance doubles.

So e.g. if circuit resistance was 100 ohms and supply voltage 100V., (I = E/R) = (100/100) = 1A. current.

The power dissipated by the resistor will be (EI) = 100 x 1 = 100W.

Now double the resistance, = 200 ohms. The current will halve, = 0.5A.

THE VOLTAGE ACROSS EACH RESISTOR is (100/2) = 50V.!

Power dissipated by each = (EI) = (0.5A x 50V) = 25W.

There you are, you are asked for the dissipation only of 1 of the resistors, R1, not the total dissipation.

Hard to say. Do you know anything about the relative values of the 2 resistors? For example if A=B, then the current goes to 1/2 of what it was. And power is given by

P=I^2*R

P=I²R. If Ra=Rb, then I is reduced to ½ value when Rb is added. So, I is now I(½). Thus P=[I(½)²R(2)]. Each resister has ¼P so total P(Ra+Rb)=½P(Ra).

You are right it is halved if A =B