Answer:
[tex]R = \frac{x(x+1)}{2x+1}[/tex] --- total resistance
Step-by-step explanation:
Given
[tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]
Required
Find R when
[tex]R_1 = x[/tex]
[tex]R_2 = x+1[/tex]
So, we have:
[tex]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}[/tex]
Substitute values for both R's
[tex]\frac{1}{R} = \frac{1}{x} + \frac{1}{x+1}[/tex]
Take LCM
[tex]\frac{1}{R} = \frac{x+1+x}{x(x+1)}[/tex]
Collect like terms
[tex]\frac{1}{R} = \frac{x+x+1}{x(x+1)}[/tex]
[tex]\frac{1}{R} = \frac{2x+1}{x(x+1)}[/tex]
Inverse both sides
[tex]R = \frac{x(x+1)}{2x+1}[/tex]
