Random helpful notes for your experiment:
EQ5 V = I_{1}R_{1} + I_{1}R_{eq} where R_{eq} is the resistance of the of the parallel resistors. Since they're in parallel, this is equal to: 1/R_{eq} = 1/R_{2} + 1/R_{3} Plugging this in above yields, 1 ==> V = I_{1}R_{1} + I_{1}(----------) 1/R_{2} + 1/R_{3} The term in parantheses can be rearranged as: R_{2}R_{3} R_{eq} = ---------- R_{2} + R_{3} Factoring out the I_{1} and dividing out yields: V ==> I_{1} = ---------- R_{1} + R_{2}R_{3} ------- R_{2} + R_{3} Multiplying and dividing R_{1} by R_{2}+R_{3}, gives: V ==> I_{1} = --------------------- R_{1}(R_{2} + R_{3}) + R_{2}R_{3} -------------------- R_{2} + R_{3} Which finally yields: V(R_{2} + R_{3}) ==> I_{1} = --------------------- (EQ5) R_{1}R_{2} + R_{1}R_{3} + R_{2}R_{3} EQ6 I_{1}R_{eq} = V_{eq} = I_{2}R_{2} Using R_{eq} from above gives: I_{1}R_{eq} I_{1}R_{2}R_{3} I_{2} = ------ ==> ------------ R_{2} R_{2}(R_{2} + R_{3}) Cancelling out the R_{2}'s, gives: I_{1}R_{3} ==> --------- (EQ6) R_{2} + R_{3} EQ7 I_{1}R_{eq} = V_{eq} = I_{3}R_{3} Using R_{eq} from above gives: I_{1}R_{eq} I_{1}R_{2}R_{3} I_{3} = ------ ==> ------------ R_{3} R_{3}(R_{2} + R_{3}) Cancelling out the R_{3}'s, gives: I_{1}R_{2} ==> --------- (EQ7) R_{2} + R_{3}
Required Materials: