Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • In a circuit with a forward-biased diode having a voltage drop of and a safe current li

In a circuit with a forward-biased diode having a voltage drop of 0.5 \mathrm{~V} and a safe current limit of 10 \mathrm{~mA}, a battery with an EMF of 2.5 \mathrm{~V} is used. What is the minimum resistance value required to be connected in series with the diode to ensure that the current does not exceed the safe limit?

Option: 1

100 \, \Omega


Option: 2

150\, \Omega


Option: 3

200 \, \Omega


Option: 4

500 \, \Omega


Answers (1)

best_answer

Given:

Diode forward voltage drop, V_{\text {diode }}=0.5 \mathrm{~V}

Safe current limit, I_{\text {limit }}=10 \mathrm{~mA}=0.01 \mathrm{~A}

Battery EMF, E=2.5 \mathrm{~V}

To calculate the minimum resistance ( \left.R_{\min }\right) , we can use the formula: 

\begin{aligned} & V_R=E-V_{\text {diode }} \\ & R_{\text {min }}=\frac{V_R}{I_{\text {limit }}} \end{aligned}

Substituting the given values:

\begin{aligned} & V_R=2.5 \mathrm{~V}-0.5 \mathrm{~V}=2 \mathrm{~V} \\ & R_{\min }=\frac{2 \mathrm{~V}}{0.01 \mathrm{~A}}=200 \Omega \end{aligned}
Therefore, the correct option is (C) 200 \Omega \text {. }

Posted by

SANGALDEEP SINGH

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question