Question

Each resistance in the resistor is R. The equivalent resistance between P and Q is

• A. R/6
• B. 5R/6
• C. R/3
• D. 2R/3

Answer 1

Answer: (B)

5R/6

Answer 2

The problem asks for the equivalent resistance between points P and Q in a resistor network formed by the edges of a cube, where each edge has resistance R.

We can solve this problem using the concept of symmetry and equipotential points.

1. Identify Equipotential Points:

   Let current enter at P and leave at Q.

   • Layer 0: Point P.
   • Layer 1: The three vertices directly connected to P (let's call them A, B, C) are symmetrically placed with respect to the P-Q axis. Therefore, they are at the same potential. Let this potential be $V_1$.
   • Layer 2: The three vertices directly connected to Q (let's call them D, E, F) are also symmetrically placed with respect to the P-Q axis. They are also at the same potential, say $V_2$.
   • Layer 3: Point Q.

2. Model the Circuit as Series Combinations of Parallel Resistors:

   The entire cube network can be simplified into three stages connected in series:

   • Stage 1: From P to Layer 1 (A, B, C):

     There are three resistors (PA, PB, PC) connecting P to the three equipotential points in Layer 1. These three resistors are effectively in parallel between P and the equipotential surface at $V_1$.

     The equivalent resistance for this stage is $R_{P \to L1} = \frac{R}{3}$.

   • Stage 2: From Layer 1 (A, B, C) to Layer 2 (D, E, F):

     Each of the three nodes in Layer 1 (A, B, C) is connected to two nodes in Layer 2 (D, E, F). For example, A is connected to D and E. There are a total of $3 \times 2 = 6$ resistors connecting Layer 1 to Layer 2. Since all nodes in Layer 1 are at $V_1$ and all nodes in Layer 2 are at $V_2$, these 6 resistors are effectively in parallel between the equipotential surfaces $V_1$ and $V_2$.

     The equivalent resistance for this stage is $R_{L1 \to L2} = \frac{R}{6}$.

   • Stage 3: From Layer 2 (D, E, F) to Q:

     The three nodes in Layer 2 (D, E, F) are connected to Q. These three resistors (DQ, EQ, FQ) are effectively in parallel between the equipotential surface at $V_2$ and Q.

     The equivalent resistance for this stage is $R_{L2 \to Q} = \frac{R}{3}$.

3. Calculate Total Equivalent Resistance:

   Since these three stages are in series, the total equivalent resistance between P and Q is the sum of the equivalent resistances of each stage:

   $R_{PQ} = R_{P \to L1} + R_{L1 \to L2} + R_{L2 \to Q}$

   $R_{PQ} = \frac{R}{3} + \frac{R}{6} + \frac{R}{3}$

   To sum these fractions, find a common denominator, which is 6:

   $R_{PQ} = \frac{2R}{6} + \frac{R}{6} + \frac{2R}{6}$

   $R_{PQ} = \frac{2R + R + 2R}{6}$

   $R_{PQ} = \frac{5R}{6}$

The equivalent resistance between P and Q is $\frac{5R}{6}$.