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Let's work through the problem systematically. The transistor is made of **Germanium**, and the current gain \( \beta = 100 \). We'll go step by step to find the required parameters:

### i. Determine the base current \( I_B \).

We can apply **Kirchhoff’s Voltage Law (KVL)** in the base loop to find \( I_B \). The base-emitter voltage for a Germanium transistor is typically around **0.3V**.

**Base loop:**

\[
V_{BB} = I_B R_B + V_{BE}
\]

Where:
- \( V_{BB} = 5V \)
- \( R_B = 10k\Omega \)
- \( V_{BE} = 0.3V \) (for Germanium)

Substitute the values into the equation:

\[
5 = I_B \cdot 10k + 0.3
\]

Solve for \( I_B \):

\[
I_B = \frac{5 - 0.3}{10k} = \frac{4.7}{10k} = 0.47 \text{ mA}
\]

Thus, the base current \( I_B = 0.47 \text{ mA} \).

---

### ii. Calculate the voltage drop \( V_{CE} \) across the collector-emitter junction.

For \( V_{CE} \), we'll use **Kirchhoff’s Voltage Law (KVL)** in the collector-emitter loop.

\[
V_{CC} = I_C \cdot R_C + V_{CE}
\]

Now, we know:

- \( V_{CC} = 10V \)
- \( R_C = 100\Omega \)
- \( I_C = \beta \cdot I_B \)

Since \( \beta = 100 \) and \( I_B = 0.47 \text{ mA} \):

\[
I_C = 100 \times 0.47 \text{ mA} = 47 \text{ mA}
\]

Now substitute the values into the KVL equation:

\[
10V = 47 \text{ mA} \times 100\Omega + V_{CE}
\]

\[
10V = 4.7V + V_{CE}
\]

So,

\[
V_{CE} = 10V - 4.7V = 5.3V
\]

Thus, the voltage drop \( V_{CE} = 5.3V \).

---

### iii. Determine the total emitter current \( I_E \).

The total emitter current \( I_E \) is related to the base and collector currents by:

\[
I_E = I_B + I_C
\]

Substitute the values of \( I_B \) and \( I_C \):

\[
I_E = 0.47 \text{ mA} + 47 \text{ mA} = 47.47 \text{ mA}
\]

Thus, the emitter current \( I_E = 47.47 \text{ mA} \).

---

### Summary of Results:
- \( I_B = 0.47 \text{ mA} \)
- \( V_{CE} = 5.3 \text{ V} \)
- \( I_E = 47.47 \text{ mA} \)

Would you like more details or have any questions?

---

Here are 5 related questions to extend this topic:
1. How would the results change if the transistor were made of Silicon instead of Germanium?
2. What is the effect of changing \( R_C \) to 200 ohms on the current \( I_C \)?
3. How can you calculate the power dissipated by the transistor in this circuit?
4. What happens to \( I_B \) if \( V_{BB} \) is increased to 6V?
5. How would varying \( \beta \) affect the overall behavior of the circuit?

**Tip:** Always check the type of transistor (Germanium or Silicon) when calculating the base-emitter voltage \( V_{BE} \), as it significantly affects the current calculations.

For the circuit in Fig 2, determine the following parameters. Assume that the transistor is made of Germanium. Assume beta = 100. i. Determine the base current, I_B. ii. Calculate the voltage drop, V_CE, across collector-emitter junction. iii. Determine the total emitter current, I_E.

This problem involves calculating the base current, voltage drop across the collector-emitter junction, and the total emitter current for a circuit using a Germanium transistor. Assumes a current gain of beta = 100, V_BB = 5V, and V_CC = 10V.

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