In the given circuit the internal resistance of the 18V cell is negligible. If R_1=400 Ω , R_3=100 Ω and R_4=500 Ω and the reading of an ideal…

Given reading of an ideal voltmeter for resistance ${R}_{4}$R4​ is ${V}_{4}=5V$V4​=5V, then current (let ${i}_{1}$i1​)is given branch (as resistance ${R}_{3}$R3​ in series with resistance ${R}_{4}$R4​) is

$${i}_{1}=\frac{V}{{R}_{4}}=\frac{5}{500}=0.01A$$i1​=R4​V​=5005​=0.01A

Then for resistance ${R}_{3}$R3​, the potential difference will be

$${V}_{3}={i}_{1}{R}_{3}=0.01\times 100=1V$$V3​=i1​R3​=0.01×100=1V

So, for given branch of resistance potential difference is

$${V}_{3}+{V}_{4}=1+5=6V$$V3​+V4​=1+5=6V

From figure for resistance ${R}_{1},{R}_{3}$R1​,R3​ and ${R}_{4}$R4​ a total applied potential difference is $18V$18V then,

$${V}_{1}+{V}_{3}+{V}_{4}=18 \therefore {V}_{1}=18-({V}_{3}+{V}_{4})=18-6=12V$$V1​+V3​+V4​=18∴V1​=18−(V3​+V4​)=18−6=12V

So, the total current drawn from the source is

$\therefore$∴ Current through ${R}_{2}$R2​

$${i}_{2}=0\cdot 03-0\cdot 01=0\cdot 02A$$i2​=0⋅03−0⋅01=0⋅02A

and potential difference for ${R}_{2}$R2​ is

${V}_{2}={V}_{3}+{V}_{4}=6V$V2​=V3​+V4​=6V ( ${R}_{2}$R2​ is in parallel arrangement with the resistances ${R}_{3}$R3​ and ${R}_{4}$R4​)

Now the value of resistance ${R}_{2}$R2​ is :

$${V}_{2}={i}_{2}{R}_{2} \Rightarrow {R}_{2}=\frac{{V}_{2}}{{i}_{2}}=\frac{6}{0\cdot 02}=300\Omega$$V2​=i2​R2​⇒R2​=i2​V2​​=0⋅026​=300Ω