Animated Solution for Physics - Semiconductors: The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?

Imagine you are looking at a water pipe system, but instead of water, we have electrical current, and instead of valves, we have diodes. In this problem, we are presented with a $12\text{V}$ battery connected to a network containing a $4\Omega$ resistor and two parallel branches.

The left branch contains diode ${\Delta }_1$ and a $3\Omega$ resistor, while the right branch contains diode ${\Delta }_2$ and a $2\Omega$ resistor. Our ultimate goal is to find the total current flowing out of the battery. But before we can calculate anything, we need to understand how these diodes are behaving.

Diodes are like one-way streets for electrical current. An ideal diode has a very simple rulebook: - If it is forward-biased (current tries to flow in the direction of the triangle), it acts as a perfect wire with zero resistance ($R=0$). - If it is reverse-biased (current tries to flow against the triangle), it acts as a broken wire or an open circuit with infinite resistance ($R=\infty$).

To determine the bias, we look at the potential difference. The long line of the battery symbol represents the positive terminal (higher potential), and the short line represents the negative terminal (lower potential).

Let's trace the potential from the battery's positive terminal along the top wire. This entire top wire is at a higher potential compared to the bottom wire.

Now, focus on diode ${\Delta }_1$. Its triangle (the p-side or anode) is pointing upwards, meaning it is connected to the lower potential bottom wire. Its straight line (the n-side or cathode) is connected to the higher potential top wire. Because the n-side is at a higher potential than the p-side, diode ${\Delta }_1$ is reverse-biased. It will completely block any current from flowing through its branch!

Next, let's examine diode ${\Delta }_2$. Its triangle is pointing downwards, directly connected to the higher potential top wire. Its n-side is connected towards the lower potential bottom wire. Since the p-side is at a higher potential than the n-side, diode ${\Delta }_2$ is forward-biased. It will act like a simple wire and allow current to pass freely.

Because diode ${\Delta }_1$ is reverse-biased, its entire branch acts as an open circuit. We can completely ignore the $3\Omega$ resistor because absolutely zero current will flow through it.

Suddenly, our complex parallel circuit isn't so complex anymore! The current will leave the battery, flow through the $4\Omega$ resistor, bypass the first branch entirely, flow down through diode ${\Delta }_2$ and the $2\Omega$ resistor, and return to the battery.

We are left with a simple series circuit containing just the $4\Omega$ and $2\Omega$ resistors.

In a series circuit, the equivalent resistance $R_{eq}$ is simply the sum of the individual resistances:

Now, we can bring in our trusty friend, Ohm's Law, to find the total current $I$:

Substitute our known values into the equation:

And there we have it! The total current flowing in the circuit is exactly $2.00\text{A}$. Always remember to check your diode polarities before diving into the math—it can turn a tricky parallel circuit into a walk in the park!