Animated Solution for Physics - Current Electricity: For the resistance network shown in the figure, choose the correct option(s).

Have you ever looked at a circuit diagram and felt a wave of panic? You are not alone. When we see multiple loops, crossing wires, and a bunch of resistors, our first instinct is often to dive straight into Kirchhoff's laws, setting up a nightmare of simultaneous equations.

But what if I told you there is a secret weapon? A hidden pattern that can make all those equations vanish into thin air?

Welcome to the magic of symmetry.

Take a close look at the resistance network in our problem. We have a $12\text{V}$ battery feeding into a junction that splits into an upper and a lower branch. The upper branch is lined with $2\Omega$ resistors, while the lower branch is lined with $4\Omega$ resistors. Connecting these branches are two vertical $1\Omega$ resistors.

If we were to use standard nodal analysis without thinking, we would have to assign variables to nodes P, Q, S, and T, and solve a system of four equations. That sounds like a lot of work, right?

But let's pause and look for a pattern.

Notice the ratio of the resistors in the upper branch to those in the lower branch. For the first segment, it is $2\Omega /4\Omega =1/2$. For the second segment, it is $2\Omega /4\Omega =1/2$. For the third segment, it is $2\Omega /4\Omega =1/2$.

This constant ratio is a massive clue! It strongly suggests that this circuit is a variation of a balanced Wheatstone bridge.

To test this hypothesis, let's make a bold assumption: What if no current flows through the vertical $1\Omega$ resistors?

If no current flows through the vertical resistors, we can temporarily ignore them. This simplifies our circuit into two independent series branches connected in parallel to the $12\text{V}$ battery.

Let's calculate the potentials at our key nodes to see if this assumption holds up. We will set the negative terminal of the battery to $0\text{ V}$, making the positive terminal $12\text{ V}$.

The Upper Branch: The total resistance is $2\Omega +2\Omega +2\Omega =6\Omega$. The current flowing through the top, which we will call $I_{top}$, is simply $12\text{ V}/6\Omega =2\text{ A}$.

Now, let's track the potential as we move along the top wire. Starting at $12\text{ V}$, we cross the first $2\Omega$ resistor. The voltage drop is $2\text{ A}\times 2\Omega =4\text{ V}$. So, the potential at node P is $12\text{ V}-4\text{ V}=8\text{ V}$.

Continuing to node S, we cross another $2\Omega$ resistor. Another $4\text{ V}$ drop leaves us with a potential of $4\text{ V}$ at node S.

The Lower Branch: The total resistance here is $4\Omega +4\Omega +4\Omega =12\Omega$. The current, $I_{bot}$, is $12\text{ V}/12\Omega =1\text{ A}$.

Let's track the potential along the bottom wire. Starting at $12\text{ V}$, we cross the first $4\Omega$ resistor. The voltage drop is $1\text{ A}\times 4\Omega =4\text{ V}$. So, the potential at node Q is $12\text{ V}-4\text{ V}=8\text{ V}$.

Moving to node T, we cross another $4\Omega$ resistor. Another $4\text{ V}$ drop leaves us with a potential of $4\text{ V}$ at node T.

Look at the potentials we just calculated! $V_P=8\text{ V}$ and $V_Q=8\text{ V}$. $V_S=4\text{ V}$ and $V_T=4\text{ V}$.

Because $V_P=V_Q$, there is absolutely zero potential difference across the resistor connecting P and Q. According to Ohm's law, if there is no potential difference, there is no current. Our assumption was perfectly correct! The current through PQ is indeed zero.

Similarly, because $V_S=V_T$, no current flows through the resistor connecting S and T.

Now that we have completely decoded the circuit, evaluating the options is a breeze.

- Option A: The current through PQ is zero. We just proved this! - Option B: The total current $I_1$ is the sum of the currents in the upper and lower branches. $I_1=I_{top}+I_{bot}=2\text{ A}+1\text{ A}=3\text{ A}$. This is correct. - Option C: The potential at S is $4\text{ V}$, and the potential at Q is $8\text{ V}$. Since $4\text{ V}<8\text{ V}$, this statement is also correct. - Option D: The current $I_2$ is simply the current flowing through the upper branch, which we found to be $2\text{ A}$. This is correct too!

So, all four options are correct.

The next time you face a daunting circuit, don't rush into writing equations. Take a step back, look for symmetry, and see if you can find a hidden bridge. It might just save you a page of calculations!