Problem 2: For this problem we will derive the resistor addition rules for the two basic cases in Fig. 2. This problem is tricky only because it is almost too obvious. We must be careful of what we are trying to derive and keep it separate from what we are taking as fact.

i ) First is the equivalent resistance of two known resistors R1 and R2 in series. What we are trying to do is relate the potential of the battery V to the current output of the battery I by using ohms law with some equivalent resistance Rs. First, write the known voltage V in terms of the unknown voltage across each resistor V1 = R1I and V2 = R2I. (We technically dont know I but we can write it in as a placeholder

ii ) Now Ohm’s law says that Rs = V /I, thus simply divide your result from part i by the current I to yield the equivalent resistance of two resistors in series.

iii )Parts iii and iv do the same for the parallel case in Fig. 2 B. First write the current coming from the battery, I, in terms of the current through each resistor, I1 = V /R1 and I2 = V /R2.

iv) Once again, the equivalent resistance is the voltage across the resistors V divided by the current from the battery I. Therefore divide V by your result from part iii to retrieve the equivalent resistance for the parallel case.