Question

Equivalent resistance between vertex E and F is:

• A. R/5
• B. R/3
• C. 2R/5
• D. R/2

Answer 1

Answer: (B)

R/3

Answer 2

The equivalent resistance between vertices E and F in the given hexagonal pyramid network is calculated by considering the symmetry of the network and applying Kirchhoff's laws. By exploiting the symmetry with respect to the plane passing through the apex O, the center of the base P, and the line connecting the midpoints of EF and BC, we can simplify the network. This symmetry implies that $V_A = V_D$ and $V_B = V_C$, where $V_A$, $V_B$, $V_C$, and $V_D$ are the potentials at vertices A, B, C, and D, respectively.

By applying a potential difference V between E and F ($V_E = V$ and $V_F = 0$) and analyzing the current distribution using Kirchhoff's junction rule at nodes A, B, O, and P, we can express the potentials at these nodes in terms of V. Specifically, we find that:

• $V_A = \frac{4}{15}V$
• $V_B = \frac{1}{3}V$
• $V_O = V_P = \frac{11}{30}V$

Using these potentials, we can calculate the current flowing into the network at vertex E:

$I_E = \frac{V_{EF}}{R} + \frac{V_{ED}}{R} + \frac{V_{EO}}{R} + \frac{V_{EP}}{R} = \frac{V}{R} + \frac{V - V_A}{R} + \frac{V - V_O}{R} + \frac{V - V_P}{R}$

Substituting the values of $V_A$ and $V_O = V_P$:

$I_E = \frac{V}{R} + \frac{V - \frac{4}{15}V}{R} + \frac{2(V - \frac{11}{30}V)}{R} = \frac{V}{R} + \frac{\frac{11}{15}V}{R} + \frac{2 \cdot \frac{19}{30}V}{R} = \frac{V}{R} (1 + \frac{11}{15} + \frac{19}{15}) = \frac{3V}{R}$

Therefore, the equivalent resistance $R_{EF}$ between vertices E and F is:

$R_{EF} = \frac{V}{I_E} = \frac{V}{\frac{3V}{R}} = \frac{R}{3}$

Thus, the equivalent resistance between vertex E and F is $\frac{R}{3}$.