Get Answers to all your Questions

header-bg

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) $R_E=0.125 \, \mathrm{k} \Omega$

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

5 g of Na₂SO₄ was dissolved in x g of H₂O. The change in freezing point was found to be 3.82⁰C. If Na₂SO₄ is 81.5% ionised, the value of x (K

A capacitor is made of two square plates each of side 'a' making a very small angle $\alpha$

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO₃ to give fraction A. The leftover organic phase was extracted with d

Trending Articles/News

anam-khan

In JEE Main 2025 April 3 shift 2, chemistry takes U-turn to be tougher than physics, mathematics

anam-khan

JEE Main 2025 April 3 Exam Analysis 2025: Shift 1 paper easy to moderate; chemistry section ‘easiest’

anam-khan

JEE Mains 2025 admit card out for April 7, 8, 9 exams; download on jeemain.nta.nic.in

Latest Question