Animated Solution for Physics - Current Electricity: The effective resistance between points $P$ and $Q$ of the electrical circuit shown in the figure is

When you first look at this circuit, it might seem like a tangled web of resistors. With multiple $2R$ and $r$ resistors connected between points $P$ and $Q$, your first instinct might be to start writing out Kirchhoff's Voltage and Current Laws.

However, before diving into complex algebra, it is always wise to take a step back and look for patterns. The most powerful pattern in circuit analysis is symmetry.

Notice the horizontal axis passing straight through the input point $P$ and the output point $Q$. If you look closely, the top half of the circuit is a perfect mirror image of the bottom half.

Because of this perfect horizontal symmetry, the electrical potential at any node on the top half must be exactly equal to the potential at the corresponding mirrored node on the bottom half. Specifically, the potential at the top node $A$ must be exactly equal to the potential at the bottom node $B$ ($V_A=V_B$).

Since nodes $A$ and $B$ are at the same potential, we can conceptually "fold" the circuit along this horizontal axis. When we do this, the corresponding top and bottom resistors are connected in parallel.

For example, the two $2R$ resistors connected to point $P$ are now in parallel, giving an equivalent resistance of:

Applying this folding technique to all symmetric pairs, the circuit simplifies dramatically.

After folding the circuit, look at the structure we are left with. It is a classic Wheatstone bridge!

We have resistors $R$ and $R$ on the top arms, and resistors $r$ and $r$ on the bottom arms. Between the central nodes, we have a vertical resistor $R$.

To solve a Wheatstone bridge, we must always check if it is balanced. The condition for a balanced bridge is that the ratio of the resistances in the adjacent arms must be equal.

Let's check our ratios:

Since $\frac{R}{r}=\frac{R}{r}$, the bridge is perfectly balanced!

In a balanced Wheatstone bridge, the potential difference across the central arm is zero. Consequently, no current flows through the middle resistor $R$. It is essentially a dead wire, and we can completely remove it from our calculations.

With the middle resistor gone, the circuit simplifies beautifully into two parallel branches.

The top branch consists of two resistors $R$ and $R$ in series:

The bottom branch consists of two resistors $r$ and $r$ in series:

Finally, we just have these two branches ($2R$ and $2r$) in parallel. The equivalent resistance between points $P$ and $Q$ is their product divided by their sum:

And there we have it! By leveraging symmetry and recognizing a balanced Wheatstone bridge, we turned a complex network into a straightforward calculation.