Given reading of an ideal voltmeter for resistance R4 is V4=5V, then current (let i1)is given branch (as resistance R3 in series with resistance R4) is
i1=R4V=5005=0.01A
Then for resistance R3, the potential difference will be
V3=i1R3=0.01×100=1V
So, for given branch of resistance potential difference is
V3+V4=1+5=6V
From figure for resistance R1,R3 and R4 a total applied potential difference is 18V then,
V1+V3+V4=18∴V1=18−(V3+V4)=18−6=12V
So, the total current drawn from the source is
∴ Current through R2
i2=0⋅03−0⋅01=0⋅02A
and potential difference for R2 is
V2=V3+V4=6V ( R2 is in parallel arrangement with the resistances R3 and R4)
Now the value of resistance R2 is :
V2=i2R2⇒R2=i2V2=0⋅026=300Ω
