Animated Solution for Physics - Semiconductors: The current $i$ in the network is

A Lesson in Potential and Biasing

At first glance, a bridge circuit packed with resistors and diodes can look like an intimidating maze. However, complex circuits often simplify beautifully once you understand the core principles of potential difference and diode biasing. Let's break down this network step-by-step.

The first and most crucial step in any DC circuit analysis is identifying the source of electromotive force. Here, we have a $9\text{V}$ battery. By carefully observing the battery symbol, we note that the longer parallel line is on the left. This indicates that the positive terminal is connected to the left side of the circuit, and the negative terminal is on the right.

Because current naturally flows from a region of higher potential to a region of lower potential, we can establish that the overall current will attempt to flow from the left node (let's call it $L$) to the right node ($R$).

Now, we must evaluate the state of the two diodes. An ideal diode acts as a closed switch (short circuit) when forward-biased and an open switch (open circuit) when reverse-biased. A diode is forward-biased only if its 'arrow' points in the direction of the potential drop.

Let's look at the top-left diode. It is oriented to allow current to flow from the top node ($T$) to the left node ($L$). However, we already established that node $L$ is connected to the positive terminal, making it the highest potential point in the bridge. Since the diode points from a lower potential ($T$) to a higher potential ($L$), it is reverse-biased.

Similarly, examine the bottom-right diode. It points from the right node ($R$) to the bottom node ($B$). Node $R$ is connected to the negative terminal, making it the lowest potential point. The diode is pointing from a lower potential ($R$) to a higher potential ($B$), meaning it is also reverse-biased.

Since both diodes are reverse-biased, they act as open circuits. We can effectively erase the top-left and bottom-right branches from our diagram. What remains is a single, continuous path for the current!

The current leaves the positive terminal, travels through the $10\Omega$ resistor on the bottom-left, goes up through the central $5\Omega$ resistor, and finally passes through the $10\Omega$ resistor on the top-right before returning to the negative terminal.

This transforms our complex bridge into a simple series circuit. The equivalent resistance $R_{eq}$ is simply the sum of all the resistors in this loop, including the $5\Omega$ resistor in series with the battery:

With the equivalent resistance found, we can apply Ohm's Law to find the total current $i$ flowing through the network:

Substituting our values:

By systematically checking the conditions of the non-linear components (the diodes), we turned a seemingly difficult problem into a straightforward application of series resistance. Always look for the hidden simplicity!