Animated Solution for Physics - Semiconductors: The circuit shown in the figure contains two diodes each with a forward resistance of $50\Omega$ and with infinite backward resistance. If the battery voltage is $6\text{V}$, the current through the $100\Omega$ resistance (in ampere) is

Have you ever stood at a busy intersection and watched the traffic lights dictate the flow of cars? In the microscopic world of electronics, diodes play the exact same role. They are the ultimate traffic cops, ensuring that electrons only travel in one permitted direction. Today, we are going to dive deep into a fascinating circuit problem that tests our understanding of these microscopic traffic controllers. This isn't just about plugging numbers into a formula; it's about visualizing the journey of current as it navigates through a network of components. So, buckle up, and let's embark on this electrifying adventure!

When we first glance at the circuit diagram, it might look like a complex maze. We see a rectangular network with multiple branches, resistors, and two diodes labeled $D_1$ and $D_2$. At the very foundation of this network lies our power source: a $6\text{ V}$ battery.

Before we do any math, we must understand the landscape of electrical potential. Think of voltage as the electrical pressure pushing the electrons. The battery symbol is our compass here. It consists of a long parallel line and a short, thicker line. In the universal language of circuit diagrams, the long line represents the positive terminal, which is the high-potential side. The short line represents the negative terminal, the low-potential side.

In our specific diagram, the long line of the battery is on the left. This is a crucial observation! It means that the entire left vertical wire of our circuit is connected to the positive terminal. Consequently, this left wire is at a higher electrical potential compared to the right vertical wire, which is connected to the negative terminal. Imagine the left wire as the top of a hill and the right wire as the valley below. Current, much like water, naturally wants to flow from the high potential (the hilltop) to the low potential (the valley).

Now that we have established the high and low potential regions, let's evaluate the paths available for our current. The current travels up the left wire and reaches two parallel branches. It has a choice to make. Will it take the top branch, the middle branch, or both? To answer this, we must interrogate the traffic cops: the diodes.

Let's look at the top branch containing diode $D_1$. The symbol for a diode is an arrow pointing against a vertical line. The flat side of the triangle (the p-side or anode) is on the left, and the vertical line (the n-side or cathode) is on the right. Because the p-side of $D_1$ is connected to the high-potential left wire, the diode is forward-biased. In traffic terms, the light is green! The diode allows current to pass through. However, the problem states that a forward-biased diode isn't a perfect conductor; it has an internal forward resistance of $50\Omega$. So, $D_1$ acts just like a $50\Omega$ resistor.

Next, let's examine the middle branch containing diode $D_2$. Notice how the symbol is flipped? The vertical line (the n-side) is on the left, facing the high-potential wire, while the flat triangle (the p-side) is on the right. When the n-side is at a higher potential than the p-side, the diode is reverse-biased. The traffic light is red! The problem explicitly states that the diodes have infinite backward resistance. This means $D_2$ acts as an open switch. It completely blocks the flow of electrons.

Because diode $D_2$ is reverse-biased and offers infinite resistance, absolutely zero current will flow through the middle branch. For all practical purposes, that middle branch doesn't even exist in our active circuit. We can mentally erase it from the diagram.

What are we left with? The complex multi-branch network beautifully simplifies into a single, continuous outer loop. The current leaves the battery, travels up the left wire, goes entirely through the top branch (passing through $D_1$ and the $150\Omega$ resistor), travels down the right wire, and finally passes through the $100\Omega$ resistor at the bottom to return to the battery.

Since there is only one path for the current to take, all the components in this active loop are in series. In a series circuit, the total equivalent resistance is simply the sum of the individual resistances.

Let's tally them up: 1. The forward resistance of diode $D_1$, which is $50\Omega$. 2. The resistor in the top branch, which is $150\Omega$. 3. The resistor in the bottom branch, which is $100\Omega$.

Our master equation for the equivalent resistance ($R_{\text{eq}}$) becomes:

We have successfully reduced the entire circuit to a single $6\text{ V}$ battery connected to a single $300\Omega$ equivalent resistor. Now, we bring in the most famous law in electronics: Ohm's Law.

Ohm's Law states that the current ($I$) is equal to the voltage ($V$) divided by the resistance ($R$).

Substituting our known values into the equation:

To make the division easier, we can simplify the fraction. Dividing both the numerator and the denominator by 6 gives us:

And $1$ divided by $50$ is exactly $0.02$.

Because this is a series circuit, the same current of $0.02\text{A}$ flows through every component in the loop. Therefore, the current flowing through the $100\Omega$ resistance is exactly $0.02\text{A}$.

Looking at our options, this matches perfectly with option (b).

This problem is a brilliant demonstration of how conceptual understanding simplifies mathematical execution. By correctly identifying the biasing of the diodes, we transformed a daunting parallel circuit into a trivial series loop. Always remember to let the physics guide your math!