Find the equivalent resistance between points A and B for the group of resistors shown in Figure 21-29, where R1 = 93 and R2 = 40 . ok i know that [tex]R_{1}[/tex] and 35[tex]\Omega[/tex] are in series and [tex]R_{2}[/tex] is in parallel to the top 2 resistors but I am obviously making errors either in how i set up the problem or in my calculations. I have tried (1/[tex]R_{2}[/tex]) + (1/2R) = 3/2R (as shown in my book) and then invert to show 2/3(R) and the answer i'm getting is wrong.. please help
You should first combine the two resistors in series, and then combine the result with the other resistor in parallel. What are the laws of combination for resistors in series? What about parallel?
for series: [tex]R_{eq} = R_{1}+R_{2}+R_{3}...[/tex] for parallel: [tex]1/R_{eq}= 1/R_{1}+1/R_{2}+1/R_{3}...[/tex] so you're saying: 93[tex]\Omega[/tex]+35[tex]\Omega[/tex]+(1/40[tex]\Omega[/tex]) OR (1/93 +35) + (1/40) ???
Clearly the first statement can't be right since you're adding things that have different units. The second looks ok as long as you mean 1/(93+35). Like I said, first combine the two resistors in series. Take that result and combine it with the other resistor in parallel.