Equivalent Resistance (3)

If we place a battery across the points $O$ and $A$ note that the path of current that will start flowing from junction $O$ will be the same for the junctions $B$ and $D$ , this means that those points will be at the same potential.

• To prove this point will be left as an exercise to the reader.

Hint: Assume a certain potential $V_1$ , $V_2$ and $V_3$ for each junction $B$ , $C$ and $D$ and $V$ and $0$ for junctions $O$ and $A$ . Now assume a current distribution and using Ohm's law, $V = IR$ , write each $V_i$ in terms of $I_i$ and $R = 1$ . Solving for $V_1$ , $V_2$ and $V_3$ by this method will give $V_2 = \dfrac 54 V$ and $V_1 = V_2 = \dfrac 34 V$ , concluding that $B$ and $D$ are at the same potential.

Rearranging the diagram (taking $B$ and $D$ as a single junction), we will get:

• Two resistors in parallel across $C$ and $D$ $\implies \dfrac{R}{2} : (R_1)$ .
• Two resistors in parallel across $D$ and $O$ $\implies \dfrac{R}{2} : (R_2)$ .
• Two resistors in parallel across $D$ and $A$ $\implies \dfrac{R}{2} : (R_3)$ .
• One resistor across $C$ and $O$ $\implies R : (R_4)$ .
• One resistor across $O$ and $A$ $\implies R : (R_5)$ .

Let's solve this step by step. We have

• Two resistors in series across $D$ and $O$ $\implies R_1 + R_4 = \dfrac{R}{2} + R = \dfrac{3R}{2} : (R_6)$ .

Now

• Two resistors in parallel across $D$ and $O$ $\implies \left(R_6^{-1} + R_2^{-1} \right)^{-1} = \left( \dfrac{2}{3R} + \dfrac{2}{R} \right)^{-1} = \dfrac{3R}{8} : (R_7)$ .

Now

• Two resistors in series across $O$ and $A$ $\implies R_7 + R_3 = \dfrac{3R}{8} + \dfrac{R}{2} = \dfrac{7R}{8} : (R_8)$ .

Now

• Two resistors in parallel across $O$ and $A$ $\implies \left(R_8^{-1} + R_5^{-1} \right)^{-1} = \left( \dfrac{8}{7R} + \dfrac{1}{R} \right)^{-1} = \dfrac{7R}{15} : (R_{\text{eq}})$ .

Hence equivalent resistance across $O$ and $A$ is $\dfrac{7R}{15} = \dfrac{7 \times 1}{15} = \boxed{0.467}$ .