Animated Solution for Physics - Semiconductor Electronics: In the figure, potential difference between $A$ and $B$ is

Imagine a bustling, high-tech city with a complex network of elevated highways. Cars, representing electrical charge, are constantly flowing, trying to get from the high-energy downtown core to the low-energy, quiet suburbs.

In our electrical circuit, the battery is that downtown core. It acts as a powerful pump, pushing electrons with a mighty electromotive force of $30\text{ V}$.

The resistors in the circuit are like narrow, winding, unpaved roads. They naturally slow the cars down, creating traffic and dissipating energy as heat.

But there is a very special, highly advanced toll booth on one of these roads: the p-n junction diode.

This diode is the ultimate, uncompromising gatekeeper of the circuit. It operates on one simple, unbreakable rule: it only lets traffic flow in one specific direction.

If you approach this toll booth from the correct side, the gates open wide, and you can breeze right through. But if you approach from the wrong side, the gates slam shut, and the road becomes a dead end.

Our absolute first mission in any circuit containing a diode is to interrogate the gatekeeper. We must ask: "Are you open for business, or are you closed?"

Let's carefully trace the intended path of the current. It leaves the positive terminal of our $30\text{ V}$ battery, eager to explore the network.

It travels smoothly through the first $10\text{ k}\Omega$ resistor and arrives at a major crossroads, which we have labeled as node $A$.

Here, the current faces a choice. It can go straight down through a $10\text{ k}\Omega$ resistor, or it can turn right and face the mysterious diode.

Look very closely at the diode's symbol on the schematic. The flat side of the triangle represents the p-side (anode), and the straight vertical line represents the n-side (cathode).

Because our current is arriving directly from the positive terminal of the battery, it hits the flat p-side first.

This means the p-side is at a significantly higher electrical potential than the n-side.

The gatekeeper smiles. The conditions are met. The diode is forward-biased.

Now, in the messy, imperfect real world, passing through this toll booth always costs a little bit of energy. For a standard silicon diode, this "toll fee" is usually about $0.7\text{ V}$.

But in the pristine, mathematically perfect world of ideal physics problems, we operate under different rules. Unless a specific cut-in voltage is explicitly mentioned in the problem, we assume the diode is ideal.

An ideal forward-biased diode is a perfect, frictionless host. It offers absolutely zero resistance to the flow of current.

It acts exactly like a plain, simple, perfectly conducting copper wire.

So, in our minds—and on our scratchpad—we can completely erase the diode symbol and replace it with a solid, unbroken line.

With the diode now acting as a perfect wire, the entire circuit suddenly looks much friendlier and far less intimidating.

The portion of the current that turned right at node $A$ passes effortlessly through the "wire" and goes down through the rightmost $10\text{ k}\Omega$ resistor to reach the bottom return wire, which we call node $B$.

Simultaneously, the portion of the current that went straight down from node $A$ also passes through its own $10\text{ k}\Omega$ resistor to reach that exact same node $B$.

Both of these $10\text{ k}\Omega$ resistors are connected directly between the exact same two points: node $A$ and node $B$.

This is the absolute textbook definition of a parallel combination.

When two identical resistors are placed in parallel, the mathematical calculation is beautifully simple and elegant. The equivalent resistance is exactly half of their individual value.

Our complex, intimidating crossroads has now been mathematically simplified into a single, highly manageable $5\text{ k}\Omega$ resistor.

Now we can step back and see the big picture of the circuit.

The entire network has been reduced to just the first $10\text{ k}\Omega$ resistor in series with our newly calculated $5\text{ k}\Omega$ equivalent resistor.

Because they are in series, the total equivalent resistance of the entire circuit is simply their algebraic sum.

With the total resistance finally in hand, we can call upon the legendary Ohm's Law to find the total current flowing out of the battery.

This $2\text{ mA}$ current marches out of the battery and immediately encounters the first $10\text{ k}\Omega$ resistor.

Pushing through this resistor requires a significant amount of energy, which manifests as a voltage drop across the component. We use Ohm's Law one more time to calculate the exact magnitude of this drop.

Let's think about the conservation of energy. The current started its journey with $30\text{ V}$ of electrical potential energy.

It just spent $20\text{ V}$ of that energy fighting its way through the very first resistor.

By the time it finally reaches node $A$, how much potential energy is left in the tank? We simply subtract the drop from the original source voltage.

And there it is! The potential difference between node $A$ and node $B$ is exactly $10\text{ V}$.

Before we completely wrap up this problem, let's flex our physics muscles with a quick thought experiment.

What if a careless engineer accidentally installed the $30\text{ V}$ battery upside down?

The current would try to flow in the opposite direction. It would travel up from the bottom wire and approach the diode from the straight-line n-side.

The gatekeeper would immediately frown, and the gates would slam shut.

The diode would become reverse-biased, acting as a complete open circuit with effectively infinite resistance.

Absolutely no current would be able to flow through the rightmost branch. That entire section of the circuit would become dead weight.

The entire active circuit would just be the first resistor and the middle resistor connected in a simple series loop.

The total resistance would jump to $20\text{ k}\Omega$, the total current would drop to $1.5\text{ mA}$, and the voltage $V_{AB}$ would change to $15\text{ V}$.

A single flipped battery completely changes the entire mathematical landscape of the circuit. This is exactly why identifying the state of the diode is always your most crucial first step!