Evaluating the equivalent resistance

simulate this circuit – Schematic created using CircuitLab

I want to evaluate the equivalent resistance. For that, I noticed R1, R2 and R3 are parallel connected resistors, which yields

$$\dfrac{1}{R} = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} = 1 \implies R = 1 \Omega$$

R4 and R5 are also in parallel

$$\dfrac{1}{R} = \dfrac{1}{6} + \dfrac{1}{4} = \dfrac{5}{12} \implies R = \dfrac{12}{5} \Omega$$

Now those groups are in series

$$R_{equiv} = 1+\dfrac {12}5 = \dfrac{17}5$$

I believe that my final answer is wrong. Could you point out where I've gone wrong so far?

Regards

• You need to double-check your schematic. It doesn't make any sense as drawn, and R2 is not in parallel with R1. – Elliot Alderson Oct 13 '18 at 22:17
• The resistance between the left and right ends of your circuit is 3 Ohm (=R2). Other parts do nothing but take some room. Maybe your schematic should be fixed, as already suspected in another comment. – user287001 Oct 13 '18 at 22:22
• Or it's a trick homework question and the cat is now out of the bag. – TimWescott Oct 13 '18 at 23:42
• My diagram seems okay. – Mark Oct 14 '18 at 14:52