Animated Solution for Physics - Semiconductors: The circuit contains two diodes each with a forward resistance of $50\Omega$ and with infinite reverse resistance. If the battery voltage is $6\text{ V}$, the current through the $120\Omega$ resistance is ......... $\text{mA}$.

Welcome to this fascinating journey into the world of semiconductor circuits! Today, we are going to unravel a problem that beautifully combines the principles of diode biasing with classic circuit analysis.

Imagine you are an electron ready to flow out of a $6\text{ V}$ battery. You travel up the wire and suddenly face a fork in the road: two parallel branches, each guarded by a diode ($D_1$ and $D_2$) and a resistor. Your mission is to find out how much total current eventually makes it through the $120\Omega$ resistor at the bottom. Let's break this down step by step.

Before we rush into any mathematical calculations, we need to understand the state of our "guards"—the diodes. A diode is like a one-way valve for electricity. It allows current to flow freely when it is forward-biased but completely blocks it when it is reverse-biased.

How do we know which is which? We look at the battery. The longer line on the battery symbol represents the positive terminal, which is at a higher potential. The current wants to flow out of this positive terminal and travel through the circuit.

Let's trace the path from the positive terminal to the top branch containing diode $D_1$. The positive terminal is connected to the flat triangle side of $D_1$, which is its p-side. When the p-side is at a higher potential, the diode is forward-biased. The problem tells us that in this state, the diode isn't a perfect conductor; it has a forward resistance of $50\Omega$. So, we can mentally replace $D_1$ with a $50\Omega$ resistor.

Now, let's look at the middle branch with diode $D_2$. The positive terminal of the battery is connected to the straight line side of $D_2$, which is its n-side. This means $D_2$ is reverse-biased. According to the problem, it has infinite reverse resistance. In practical terms, this means $D_2$ acts as an open circuit. It's a dead end! Absolutely no current will flow through this middle branch.

With the middle branch completely out of the picture, our seemingly complex parallel circuit simplifies into a beautiful, single series loop.

The current will flow from the battery, travel exclusively through the top branch (passing through the $50\Omega$ resistance of $D_1$ and the $130\Omega$ resistor), and then continue through the $120\Omega$ resistor at the bottom before returning to the battery.

Since all the active components are now in series, we can find the total current using Ohm's Law:

The total voltage $V$ is simply our battery voltage, $6\text{ V}$.

The equivalent resistance $R_{eq}$ is the sum of all the resistances in our single loop. We must be careful not to forget the internal resistance of the forward-biased diode!

Let's do the math. Adding the resistances together:

Now, we substitute this back into Ohm's Law:

Simplifying this fraction, we divide both the numerator and the denominator by 6:

We have our current, but the question specifically asks for the answer in milliamperes ($\text{mA}$). To convert from amperes to milliamperes, we multiply by $1000$:

And there we have it! The current flowing through the $120\Omega$ resistance is exactly $20\text{ mA}$.

As a fun thought experiment, what if we flipped the battery around? $D_1$ would become the open circuit, and $D_2$ would conduct. The physics remains the same, but the path changes. Keep exploring, and happy learning!