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

2 Answers
May 15, 2018

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

May 15, 2018

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