Animated Solution for Physics - Current Electricity: The figure shows a circuit having eight resistances of $1\Omega$ each, labelled $R_1$ to $R_8$, and two ideal batteries with voltages ${\epsilon }_1=12\text{ V}$ and ${\epsilon }_2=6\text{ V}$. Which of the following statement(s) is(are) correct?

When you first lay eyes on this circuit, it looks like a classic trap designed to drain your time during an exam. With eight resistors, two batteries, and multiple interconnected loops, setting up standard Kirchhoff's Loop equations would result in a massive, error-prone system of linear equations.

However, in advanced physics problems, brute force is rarely the intended path. The key to unlocking this problem lies in taking a step back and observing the geometry of the connections. Physics is deeply intertwined with symmetry, and recognizing it can turn a nightmare into a walk in the park.

Let's carefully trace the connections to the top node ($T$) and the bottom node ($B$). Notice that node $T$ is connected to the left node ($L$), the right node ($R$), and the center node ($C$) through identical $1\Omega$ resistors ($R_6$, $R_7$, and $R_2$).

Now, look at node $B$. It is also connected to the exact same three nodes ($L$, $R$, and $C$) through identical $1\Omega$ resistors ($R_5$, $R_8$, and $R_4$). Because the upper half and the lower half of the circuit are perfect mirror images of each other, the electrical potential at the top node must be exactly equal to the electrical potential at the bottom node.

Since there is no potential difference between them, we can virtually "fold" the circuit in half, merging nodes $T$ and $B$ into a single equivalent node, which we will call $M$. When we do this, the corresponding resistors from the top and bottom halves end up in parallel. For example, the $1\Omega$ resistor from $L$ to $T$ is now in parallel with the $1\Omega$ resistor from $L$ to $B$.

Applying this to all three pairs, our complex web collapses into a beautifully simple star network centered at $M$, with $0.5\Omega$ legs reaching out to nodes $L$, $R$, and $C$.

Smart Reference Node

Now we have a central node $M$ connected to the active branches containing the batteries. To make our Nodal Analysis (KCL) as clean as possible, we can choose any node as our reference potential. Instead of choosing ground ($0\text{ V}$) arbitrarily, let's make a strategic choice: let's set the potential of the center node $C$ to $12\text{ V}$.

Why $12\text{ V}$? Look at the right branch connecting $C$ to $R$. It contains a $12\text{ V}$ battery. If we start at $C$ ($12\text{ V}$) and cross the battery, the potential drops by exactly $12\text{ V}$, landing us at a perfect $0\text{ V}$ just before the resistor $R_1$. This means the entire right branch is simply a connection from node $M$ to $0\text{ V}$ through a total resistance of $1.5\Omega$ (the $0.5\Omega$ leg plus the $1\Omega$ of $R_1$).

Similarly, the left branch connects $C$ to $L$ with a $6\text{ V}$ battery. Starting at $C$ ($12\text{ V}$) and moving left, the potential increases by $6\text{ V}$, putting the node before $R_3$ at $18\text{ V}$. Thus, the left branch connects node $M$ to $18\text{ V}$ through $1.5\Omega$.

Our entire circuit is now reduced to a single unknown node $M$ (let's call its voltage $V_0$) connected to three known potentials: $18\text{ V}$, $12\text{ V}$, and $0\text{ V}$. We apply Kirchhoff's Current Law at node $M$:

To solve this elegantly, multiply the entire equation by $1.5$:

With the central voltage known, calculating the currents is trivial. The current through $R_1$ is the current in the right branch:

The current through $R_3$ is the current in the left branch:

To find the currents in the folded resistors, we calculate the total branch current and split it. The current flowing from $C$ to $M$ is $\frac{12-10.8}{0.5}=2.4\text{ A}$. Because of symmetry, this splits equally between $R_2$ and $R_4$, giving $I_{R2}=1.2\text{ A}$.

Similarly, the $4.8\text{ A}$ flowing from the left splits equally at node $L$, giving $I_{R5}=2.4\text{ A}$. Evaluating all the options, we find that every single statement provided is perfectly correct!