Find the equivalent resistance between the point a and b of the circuit shown in figure

Explanation

Suppose a current `i` enters the circuit at the point `a`, a part `i_1` goes through the `10 Omega` resistor and the rest `I - i_1` throught he `5 Omega` resistor. By symmerty, the current `i` coming out from the point `b` will be composed of a part `i_1` from the `10 Omega` resistor and `i - i_1` from the `5 Omega` (figure) resistor. Applying Kirchhoff a junction law, we can find the current through the middle `5 Omega` resistor. The current distribution is shown in figure.
We have
`V_a - V_b = (V_a - V_c) + (V_c - V_b)`
`= (10 Omega) i_i + (5 Omega) (i-i_1)`
`= (5 Omega) i + (5 Omega) i_1`. .... (i)
Also, `V_a - V_b = (V_a - V_c) + (V_c - V_d) + (V_d - V_b)`
`= (10 Omega) i_1 + (30 Omega) i_1` ...(ii)
Multiplying (i) by 6 and subtracting (ii) from it, we eliminate `i_1` and get,
`5(V_a - V_b) = (35 Omega) i`
or, `(V_a - V_b)/(i) = 7 Omega`.
Thus the equivalent resistance between the points `a and b` is `7 Omega`.