Question 52 of 380

In the given circuit, the current flowing through the resistance $20\Omega$ is $0.3A$, while the ammeter reads $0.9A$. The value of ${R}_{1}$ is _____$\Omega$.

Your answerCheck answer

Show answer and solution▾

Answer: 30

Given, ${i}_{1}=0.3A,{i}_{1}+{i}_{2}+{i}_{3}=0.9A$
So, ${V}_{AB}={i}_{1}\times 20\Omega =0.3\times 20V=6V$
${i}_{2}=\dfrac{6V}{15\Omega }=\dfrac{2}{5}A$
Now,
${i}_{1}+{i}_{2}+{i}_{3}=\dfrac{9}{10}A$
$\Rightarrow \dfrac{3}{10}+\dfrac{2}{5}+{i}_{3}=\dfrac{9}{10}$
$\Rightarrow \dfrac{7}{10}+{i}_{3}=\dfrac{9}{10}$
$\Rightarrow {i}_{3}=0.2A$
So, ${i}_{3}\times {R}_{1}=6V$
$\Rightarrow (0.2){R}_{1}=6$
$\Rightarrow {R}_{1}=\dfrac{6}{0.2}=30\Omega$

JEE Main Physics practice question on Current Electricity (Resistance and Resistivity), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.