Question

What is the combined resistance of a 6 ohm resistor which is connected in series with a parallel arrangement of two resistors, each resistance of 2 ohm? Thanks

Answer 1

`R_"eq" = 7 Omega`

Explanation:

Looking just at the 2 `2Omega` resistors, their parallel combination is equivalent to ... `darr`

This is easy because the 2 resistors have equal value. In such a case, the equivalent resistance is 1/2 of the individual resistance -- therefore `1Omega`.

The `6Omega` resistor is in series with the previously determined `1Omega`. Therefore `R_"eq" = 6Omega + 1/2*2Omega=7 Omega`

I hope this helps,
Steve

Answer 2

Given `R_1 = 6 Omega, R_2 = 2Omega, and R_3 = 2Omega`

`R = R_1+ 1/(1/R_2+1/R_3)`

`R = 6Omega+ 1/(1/(2Omega)+1/(2Omega))`

`R = 7Omega`