Animated Solution for Physics - Current Electricity: In the given circuit, currents in different branches and value of one resistor are shown. Then, potential at point $B$ with respect to the point $A$ is

Welcome to an electrifying journey through the world of circuits! Today, we are going to tackle a fascinating problem that will test your mastery of Kirchhoff's Laws.

Imagine you are an electron navigating through this complex network of wires, batteries, and resistors. Our mission is to find the potential at point $B$ with respect to point $A$, which mathematically translates to finding the value of $V_B-V_A$.

Let's break this down step-by-step and uncover the hidden secrets of this circuit.

To find the potential difference between any two points in a circuit, we can take a virtual walk along any continuous path connecting them. As we walk, we simply add up all the potential gains and drops we encounter.

The most direct and logical path from node $A$ to node $B$ is through nodes $C$ and $D$. So, our chosen route is $A\to C\to D\to B$.

However, before we can embark on this journey, we need to know the current flowing through every single component on our path. Looking at the diagram, we know the currents in branches $AC$ and $CF$, but the current in the vertical branch $CD$ is a mystery. We need to solve this mystery first!

Kirchhoff's Current Law

To find the missing current in branch $CD$, we turn to our trusty tool: Kirchhoff's Current Law (KCL). This law states that the total current entering a junction must exactly equal the total current leaving it.

Let's focus our attention on node $C$. We can clearly see a current of $1\text{ A}$ entering node $C$ from the left (from node $A$). We also see a current of $2\text{ A}$ leaving node $C$ to the right (towards node $F$).

Let's assume there is a current $i_3$ flowing downwards from node $D$ into node $C$. Applying KCL at node $C$, we can write:

Solving this simple equation, we find:

The positive result confirms that our assumed downward direction was absolutely correct. Now we are fully equipped to walk our path!

Kirchhoff's Voltage Law

Now comes the fun part. We will apply Kirchhoff's Voltage Law (KVL) along our chosen path $A\to C\to D\to B$. We will start at node $A$ with an initial potential of $V_A$ and carefully track the changes.

Step 1: From A to C As we move from $A$ to $C$, we cross a $1\text{ V}$ battery. Notice the symbol: we are moving from the shorter, thicker line (negative terminal) to the longer, thinner line (positive terminal). This represents a step up in potential.

Step 2: From C to D Next, we move upwards from $C$ to $D$. Here, we encounter a $2\Omega$ resistor. Crucially, we are moving against the direction of our $1\text{ A}$ current $i_3$. Moving against the current is like walking uphill; it results in a potential gain. The gain is given by Ohm's Law ($I\times R$).

Step 3: From D to B Finally, we move from $D$ to $B$. We cross a $2\text{ V}$ battery, but this time we move from the positive terminal to the negative terminal. This is a step down, meaning a potential drop.

We have successfully navigated the path! Now, let's combine all our equations to find the final relationship between $V_B$ and $V_A$.

Substituting the expressions we found:

Notice how beautifully the $+2$ and $-2$ cancel each other out!

To find the potential at $B$ with respect to $A$, we simply rearrange the equation:

And there we have it! The potential difference is exactly $+1\text{ V}$. By carefully applying Kirchhoff's laws and respecting the sign conventions, even the most intimidating circuits become perfectly solvable puzzles. Keep practicing, and you'll master this in no time!