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In a common base (CB) configuration, a transistor has a collector current $\left(\mathrm{I}_{\mathrm{C}}\right)$ of $4 \mathrm{~mA}$ and a collector-emitter voltage $\left(\mathrm{V}_{\mathrm{CE}}\right)$ of $10 \mathrm{~V}$ . If the voltage across the emitter resistor $\left(R_E\right)$ is $0.5 \mathrm{~V}$ , what is the value of $R_E$ ?
Option: 1
$R E=0.125 \, \mathrm{k} \Omega$

Option: 2
$R E=1 \, \mathrm{k} \Omega$

Option: 3
$R E=2.5 \, \mathrm{k} \Omega$

Option: 4
$R E=8 \, \mathrm{k} \Omega$

Answers (1)

best_answer

Given:

$\begin{aligned} & I_C=4 \mathrm{~mA} \\ & V_{C E}=10 \mathrm{~V} \\ & V_{R E}=0.5 \mathrm{~V} \end{aligned}$

Using Ohm's Law, the voltage across the emitter resistor $\left(R_E\right)$ can be calculated as:

$V_{R E}=I_E \cdot R_E$

Since the emitter current $\left(I_E\right)$ can be approximated as the collector current $\left(I_C\right)$ in a common base configuration.

Substituting the given values, we have:

$(0.5) V=(4) m A \cdot\left(R_E\right)$

To solve for $\mathrm{R}_{\mathrm{E}$ , divide both sides of the equation by $\text { (4) } m A$ :

$R_E=\frac{0.5 V}{4 m A}=0.125 k(\Omega)$

Therefore, the correct option is a) $R_E=0.125 \, \mathrm{k} \Omega$

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