Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

In the circuit shown, when key $\mathrm{K_1}$ is closed, then the ammeter reads $\mathrm{I_0}$ whether $\mathrm{K_2}$ is open or closed. But when $\mathrm{K_1}$ is open the ammeter reads $\mathrm{I_0/2}$ , when $\mathrm{K_2}$ is closed:

Assuming that ammeter resistance is much less than $\mathrm{R_2}$ , the values of $\mathrm{r}$ and $\mathrm{R_1}$ is ohms are:
Option: 1
$100,50$

Option: 2
$50,100$

Option: 3
$0,100$

Option: 4
$0,50$

Answers (1)

best_answer

When $\mathrm{K}_1$ is closed, $\mathrm{R}_1$ is short-circuited

When $\mathrm{K_2}$ is open, $\mathrm{i_0=\frac{E}{r+R_2}=\frac{E}{r+100}}$

When $\mathrm{K_2}$ is closed, $\mathrm{i_0=\frac{1}{2}\left[\frac{E}{r+50}\right]}$

From these two equations, we get $\mathrm{r=0}$

When $\mathrm{K_1}$ is open and $\mathrm{K_2}$ is closed. $\mathrm{\frac{i_0}{2}=\frac{E}{2\left(R_1+50\right)}}$

From eqs. (i) and (iii), we have $\mathrm{R}_1=50 \Omega$

Hence option 4 is correct

Posted by

Pankaj

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

JEE Main 2026 session 2 application correction begins on jeemain.nta.nic.in; edit details by tomorrow

anam-khan

JEE Main 2026 Registration LIVE: Session 2 exam from April 2 to 9; shift timings, admit card, exam pattern

anam-khan

JEE Main 2026 session 2 registration ends today for BTech, BArch; apply at jeemain.nta.nic.in

Latest Question