In connection with the circuit drawn below, the value of current flowing through 2k ohm resistor is

Q: Why doesn't the $10\text{V}$ battery push $10\text{V}$ across the $2\text{k}\Omega$ resistor?

The $10\text{V}$ is the total voltage supplied by the battery. However, there is a $1\text{k}\Omega$ resistor in series with the rest of the circuit. This series resistor will have a voltage drop across it, so the remaining voltage for the parallel branches will be less than $10\text{V}$. Furthermore, the Zener diode clamps this remaining voltage to exactly $5\text{V}$.

Q: How do we know for sure that the Zener diode is in the breakdown region?

To check this, we imagine removing the Zener diode and calculating the 'unloaded' voltage across that branch. The circuit becomes a simple voltage divider with a $1\text{k}\Omega$ and a $2\text{k}\Omega$ resistor. The voltage across the $2\text{k}\Omega$ resistor would be $10\text{V} \times \frac{2\text{k}\Omega}{1\text{k}\Omega + 2\text{k}\Omega} = 6.67\text{V}$. Since $6.67\text{V}$ is greater than the Zener's breakdown voltage of $5\text{V}$, the diode will definitely break down and clamp the voltage at $5\text{V}$.

Q: What would happen if the battery voltage was only $6\text{V}$ instead of $10\text{V}$?

If the battery was $6\text{V}$, the unloaded voltage across the parallel branch would be $6\text{V} \times \frac{2}{3} = 4\text{V}$. Because $4\text{V}$ is less than the $5\text{V}$ breakdown voltage, the Zener diode would not break down. It would act like an open switch, and the current would simply be determined by the $1\text{k}\Omega$ and $2\text{k}\Omega$ resistors in series.

Q: Why did we change the answer from $2.5 \times 10^{-3}$ to $25 \times 10^{-4}$?

This is a classic exam trap! The question specifically asks for the value to fill in the blank for the expression $....... \times 10^{-4}\text{ A}$. To find the correct integer to put in that blank, we must mathematically manipulate our scientific notation to match their requested power of ten.

Q: Does the $1\text{k}\Omega$ resistor affect the current flowing through the $2\text{k}\Omega$ resistor at all?

As long as the Zener diode is in the breakdown region, it acts as a perfect voltage regulator. It forces the voltage across the $2\text{k}\Omega$ resistor to be exactly $5\text{V}$. Therefore, the $1\text{k}\Omega$ resistor does not affect the current through the $2\text{k}\Omega$ resistor. However, the $1\text{k}\Omega$ resistor *does* determine the total current drawn from the battery, and consequently, how much current flows through the Zener diode itself.

Question

In connection with the circuit drawn below, the value of current flowing through $2	ext{k}\Omega$ resistor is ....... $	imes 10^{-4}	ext{ A}$.

Sigma Percentile · Accepted Answer

### Unlocking the Power of the Zener Diode

Welcome to a fascinating journey into the world of semiconductor electronics! Today, we are going to dissect a classic circuit problem involving a Zener diode. At first glance, circuits with multiple branches and specialized components can look intimidating. But don't worry—once you understand the fundamental personality of a Zener diode, these problems become incredibly satisfying to solve.

Imagine you are an electron leaving the positive terminal of the $10	ext{V}$ battery. Your journey begins by traveling through the $1	ext{k}\Omega$ series resistor. After passing through this resistor, you reach a junction where the path splits into two parallel branches. One path leads through a $2	ext{k}\Omega$ resistor, and the other path leads through a Zener diode.

### The Clamping Action of the Zener Diode

Let's focus on the Zener diode. Notice its orientation in the circuit diagram. The cathode (the side with the bar) is facing the positive terminal of the battery. This means the Zener diode is connected in **reverse bias**.

In a normal diode, reverse bias means it acts like a closed door, blocking all current. However, a Zener diode is special. It is heavily doped and designed to operate safely in the breakdown region. When the reverse voltage across it reaches a specific threshold—known as the **Zener breakdown voltage** ($V_Z$)—it suddenly opens the door just enough to maintain that exact voltage across its terminals.

In our circuit, the Zener diode has a breakdown voltage of $5	ext{V}$. Because the unloaded voltage of this branch (if the Zener were removed) would be $6.67	ext{V}$, the Zener diode is forced into breakdown. It acts as a perfect voltage regulator, clamping the voltage across that entire parallel section to exactly $5	ext{V}$.

### The Parallel Connection

Here is the crucial logical bridge: The $2	ext{k}\Omega$ resistor is connected perfectly in parallel with the Zener diode. In any parallel circuit, the potential difference across all branches must be identical.

Since the Zener diode has locked the voltage at $5	ext{V}$, the voltage across the $2	ext{k}\Omega$ resistor is also strictly clamped at $5	ext{V}$.

$$V_{2	ext{k}\Omega} = V_{Z} = 5	ext{V}$$

This is a beautiful simplification! We don't even need to worry about the $1	ext{k}\Omega$ resistor or the $10	ext{V}$ battery to find the current through the $2	ext{k}\Omega$ resistor. The Zener diode has isolated it from the rest of the circuit's fluctuations.

### Executing the Calculation

Now that we know the resistance ($R = 2	ext{k}\Omega = 2 	imes 10^3 \Omega$) and the voltage across it ($V = 5	ext{V}$), finding the current is a straightforward application of Ohm's Law.

$$I = \frac{V}{R}$$

Let's substitute our known values into the equation:

$$i = \frac{5}{2 	imes 10^3}$$

$$i = 2.5 	imes 10^{-3}	ext{ A}$$

We have our current! But wait, we must always read the question carefully. The examiner has set a small trap regarding the format of the answer.

### The Final Formatting Step

The question asks for the value to fill in the blank for the expression $....... 	imes 10^{-4}	ext{ A}$. Our current answer is in the power of $10^{-3}$. We need to mathematically manipulate our scientific notation to match their requested power of ten.

To change $10^{-3}$ to $10^{-4}$, we multiply the exponent part by $10^{-1}$ and the decimal part by $10^1$.

$$i = 2.5 	imes 10^{-3}	ext{ A} = 25 	imes 10^{-4}	ext{ A}$$

By comparing this to the requested format $x 	imes 10^{-4}	ext{ A}$, we can clearly see that our integer value is **25**.

And there you have it! By understanding the clamping behavior of the Zener diode and recognizing the parallel circuit structure, we turned a complex-looking problem into a simple application of Ohm's Law.

Animated Solution for Physics - Semiconductors: In connection with the circuit drawn below, the value of current flowing through $2 k Ω$ resistor is ....... $\times 1 0^{- 4} A$ .

Enter Numerical Value:

Visualized Solution

Circuit Analysis

Zener Breakdown Region

Voltage across $2 k Ω$ Resistor

Calculating Current

Final Current Value

Further Exploration

The Sigma Insight: P-N Junction Diode

Unlocking the Power of the Zener Diode

The Clamping Action of the Zener Diode

The Parallel Connection

Executing the Calculation

The Final Formatting Step

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Animated Solution for Physics - Semiconductors: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In connection with the circuit drawn below, the value of current flowing through 2kΩ resistor is ....... ×10−4 A.