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Determine if the diode (ideal) in Fig. 9 (i) is forward biased or reverse biased.
Take 1kilo ohm instead 2, of Take 3 kilo ohm instead of 8  4 kohm will be 4     Take 1kilo ohm instead of 6 and  18 volt instead of 10 volt

Let's solve the given circuit analysis problem with the modifications you've requested.

### Modified Circuit Components
1. Replace the **2 kΩ** resistor with **1 kΩ**.
2. Replace the **8 kΩ** resistor with **3 kΩ**.
3. Keep the **4 kΩ** resistor unchanged.
4. Replace the **6 kΩ** resistor with **1 kΩ**.
5. The power supply changes from **10 V** to **18 V**.

We need to determine if the diode in the left circuit (Fig. 9 (i)) is forward biased or reverse biased.

### Step 1: Analyze the Circuit Configuration
In the left circuit:
- The circuit has a diode, and its polarity will determine if the diode is forward or reverse biased.
- A forward-biased diode has a higher voltage at its anode (left) than at its cathode (right), and it allows current to flow.
- A reverse-biased diode has a higher voltage at its cathode, blocking current flow.

### Step 2: Voltage at \( V_1 \) and \( V_2 \)
We need to find the voltages at \( V_1 \) and \( V_2 \) to check the biasing of the diode.

- For \( V_1 \) (voltage at the left side):
  \[
  V_1 = 18V \times \frac{R_{\text{lower left}}}{R_{\text{upper left}} + R_{\text{lower left}}}
  \]
  The resistances are \( R_{\text{upper left}} = 1 k\Omega \) and \( R_{\text{lower left}} = 3 k\Omega \).

- For \( V_2 \) (voltage at the right side):
  \[
  V_2 = 18V \times \frac{R_{\text{lower right}}}{R_{\text{upper right}} + R_{\text{lower right}}}
  \]
  The resistances are \( R_{\text{upper right}} = 4 k\Omega \) and \( R_{\text{lower right}} = 1 k\Omega \).

### Step 3: Calculate the Voltages and Compare
I will now compute these voltages and determine if the diode is forward or reverse biased.The calculated voltages are:
- \( V_1 = 13.5 \, V \)
- \( V_2 = 3.6 \, V \)

### Step 4: Determine Diode Biasing
- The anode of the diode (left side) is at \( V_1 = 13.5 \, V \).
- The cathode of the diode (right side) is at \( V_2 = 3.6 \, V \).

Since \( V_1 > V_2 \), the diode is **forward biased** because the voltage at the anode is higher than the voltage at the cathode. Therefore, current can flow through the diode.

Would you like further details on this, or have any additional questions?

Here are 5 related questions you can explore:
1. How does changing the supply voltage affect the diode's biasing?
2. What would happen if the diode were reverse biased?
3. How do we model non-ideal diodes in similar circuits?
4. Can we calculate the current through the circuit now?
5. What is the power dissipated by each resistor in this circuit?

**Tip:** The voltage divider rule is a powerful tool for quickly finding voltages across resistors in series.

Determine if the diode (ideal) in Fig. 9 (i) is forward biased or reverse biased. Take 1 kilo ohm instead of 2, 3 kilo ohm instead of 8, 4 kilo ohm instead of 4, 1 kilo ohm instead of 6, and 18V instead of 10V.

Explore how to determine if a diode is forward or reverse biased in a modified circuit using the voltage divider rule. Learn about circuit analysis concepts, key formulas, and diode biasing rules. This problem involves replacing resistors with specific values and changing the voltage from 10V to 18V.

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