JEE Main 2019 — Current Electricity Question with Solution

Given reading of an ideal voltmeter for resistance $R_4$  is $V_4 = 5 V$, then current (let $i_1$)is given branch (as resistance $R_3$ in series with resistance $R_4$) is 
$i_1=\frac{V}{R_4}=\frac{5}{500}=0.01 A$
Then for resistance $R_3$, the potential difference will be 

$V_3 = i_1 R_3 =0.01\times 100 =1V$

So, for given branch of resistance potential difference is

$V_3 + V_4 = 1 + 5 =6 V$

From figure for resistance $R_1 , R_3$ and $R_4$ a total applied potential difference is $18 V$ then,

$$\begin{gathered} V_1+ V_3+V_4 = 18 \\ \therefore V_1 =18-\left(V_3+V_4\right) =18-6 =12 V \end{gathered}$$

So, the total current drawn from the source is 

$\therefore$ Current through $R_2$

$i_2=0·03-0·01= 0·02 A$
and potential difference for $R_2$ is 

$V_2 =V_3 + V_4 =6 V$ ( $R_2$ is in parallel arrangement with the resistances $R_3$ and $R_4$)

Now the value of resistance $R_2$ is :
 $$\begin{gathered} V_2 = i_2 R_2 \\ \Rightarrow R_2 = \frac{V_2}{i_2} =\frac{6}{0·02} = 300 \Omega \end{gathered}$$