The potential difference across the collector of a transistor, used in common emitter mode is 1.5 V, with the collector resistance of `3 kOmega`, the emitter current is [`beta= 50`]

Options

A. 0.70mA
B. 0.51 mA
C. 1.1 mA
D. 1.9 mA

Explanation

To solve the problem step by step, we will follow the logical sequence of calculations based on the information provided.

### Step 1: Identify the given values
- Potential difference across the collector (Vc) = 1.5 V
- Collector resistance (Rc) = 3 kΩ = 3 × 10³ Ω
- Current amplification factor (β) = 50

### Step 2: Calculate the collector current (Ic)
The formula for the collector current (Ic) in terms of the potential difference and collector resistance is given by:
\[ Ic = \frac{Vc}{Rc} \]

Substituting the values:
\[ Ic = \frac{1.5 \, \text{V}}{3 \times 10^3 \, \Omega} \]
\[ Ic = \frac{1.5}{3000} \]
\[ Ic = 5 \times 10^{-4} \, \text{A} \]

### Step 3: Calculate the base current (Ib)
The relationship between the collector current (Ic) and the base current (Ib) is given by:
\[ β = \frac{Ic}{Ib} \]
Rearranging this gives:
\[ Ib = \frac{Ic}{β} \]

Substituting the values:
\[ Ib = \frac{5 \times 10^{-4} \, \text{A}}{50} \]
\[ Ib = 1 \times 10^{-5} \, \text{A} \]

### Step 4: Calculate the emitter current (Ie)
The emitter current (Ie) is the sum of the collector current (Ic) and the base current (Ib):
\[ Ie = Ic + Ib \]

Substituting the values:
\[ Ie = 5 \times 10^{-4} \, \text{A} + 1 \times 10^{-5} \, \text{A} \]
\[ Ie = 5 \times 10^{-4} + 0.1 \times 10^{-4} \]
\[ Ie = 6 \times 10^{-4} \, \text{A} \]
\[ Ie = 0.51 \, \text{mA} \]

### Final Answer
The value of the emitter current (Ie) is:
\[ Ie = 0.51 \, \text{mA} \]

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