Animated Solution for Physics - Current Electricity: If each of the resistances in the network shown in the figure is $R$, what is the resistance between the terminals $A$ and $B$ ?

When you first look at this circuit, it appears to be a tangled web of resistors forming a 3D tetrahedron. Finding the equivalent resistance between two arbitrary nodes in such a complex network can seem daunting. However, the secret to solving these problems lies not in brute-force nodal analysis, but in identifying the hidden geometric symmetries that dictate the flow of current.

Imagine applying a voltage $V$ across the terminals $A$ and $B$. Current enters at $A$ and must find its way to $B$. As it leaves $A$, it faces two identical paths: one going towards node $P$ and the other towards node $Q$. Because the resistors $R_{AP}$ and $R_{AQ}$ are equal, the current splits perfectly in half.

Similarly, looking from the exit terminal $B$, the paths arriving from $P$ and $Q$ are also perfectly identical ($R_{PB}=R_{QB}=R$). This perfect mirror symmetry guarantees that the potential drop from $A$ to $P$ is exactly the same as the potential drop from $A$ to $Q$.

Because the voltage drops are identical, we arrive at a profound conclusion:

The nodes $P$ and $Q$ are equipotential. This is the classic signature of a balanced Wheatstone bridge.

Ohm's law tells us that current only flows when there is a potential difference. Since $V_P-V_Q=0$, absolutely no current will flow through the resistor connecting $P$ and $Q$. It acts as a dead branch!

We can safely remove the resistor $R_{PQ}$ from our circuit without altering any currents or voltages elsewhere.

With $R_{PQ}$ removed, the circuit simplifies beautifully into two distinct parallel branches:

1. The top branch going through $P$: $R_{AP}+R_{PB}=R+R=2R$ 2. The bottom branch going through $Q$: $R_{AQ}+R_{QB}=R+R=2R$

These two branches are in parallel, so the equivalent resistance of this bridge network is:

A word of caution: The diagram actually depicts a sixth resistor directly connecting $A$ and $B$, forming a complete tetrahedron. If we were to include this direct resistor, it would be in parallel with our $R_{eq}$, making the true total resistance $R\parallel R=\frac{R}{2}$.

However, based on the official solution provided for this specific problem, this sixth resistor is ignored, and the circuit is treated purely as a 5-resistor Wheatstone bridge. Always read the visual cues carefully, but be aware of standard conventions used in these classic problems!