Get Answers to all your Questions

header-bg qa

A dry cell delivering 2 A has terminal voltage 1.14V. What is the internal resistance of the cell if its open–circuit voltage is 1.59 V?

Option: 1

5.09\; \Omega 


Option: 2

6.09\; \Omega


Option: 3

7.09\; \Omega


Option: 4

0.09\; \Omega


Answers (1)

best_answer

The open–circuit voltage is simply the emf of the cell, so V = E – ir with

\begin{aligned} & V=1.41 \mathrm{~V}, i=2 \mathrm{~A}, \\ & E=1.59 \mathrm{~V} .1 .41=1.159-2 r, \quad \text { and } \quad r=0.09 \Omega \end{aligned}

Posted by

Devendra Khairwa

View full answer

JEE Main high-scoring chapters and topics

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