Physics - Current Electricity Question with Solution | TestHub

From null current $\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=\frac{1}{4}.\dots \dots \left(i\right)$

Resistance of the bridge portion can not exceed $1\mathrm{k}\Omega$. Therefore entire circuit resistance will lie in the range $1\mathrm{M}\Omega$ to $1.001\mathrm{M}\Omega$ and can be taken approximately $1\mathrm{M}\Omega$ Let battery current in ${\mathrm{I}}_0$ When D is at A

$\left({\mathrm{I}}_0-2\mu \mathrm{A}\right)(1\mathrm{k}\Omega )=(2\mu \mathrm{A}){\mathrm{R}}_2..\left(2\right)$ When D is at B

$\left({\mathrm{I}}_0-3\mu \mathrm{A}\right)(1\mathrm{k}\Omega )=(3\mu \mathrm{A}){\mathrm{R}}_1...\left(3\right)$ from (2)/(3) we have$\frac{{\mathrm{I}}_0-(2\mu \mathrm{A})}{{\mathrm{I}}_0-(3\mu \mathrm{A})}=\frac{(2\mu \mathrm{A}){\mathrm{R}}_2}{(3\mu \mathrm{A}){\mathrm{R}}_1}..\dots \left(4\right)$ From (1) \& (4) $\frac{{\mathrm{I}}_0-(2\mu \mathrm{A})}{{\mathrm{I}}_0-(3\mu \mathrm{A})}=\frac{(2\mu \mathrm{A})}{(3\mu \mathrm{A})}\times \frac{4}{1}$ $\frac{{\mathrm{I}}_0-(2\mu \mathrm{A})}{{\mathrm{I}}_0-(3\mu \mathrm{A})}=\frac{8}{3}\Rightarrow {\mathrm{I}}_0\approx 3.6\mu \mathrm{A}$ As entire circuit has resistance $\approx 1\mathrm{M}\Omega$, the battery voltage ${\mathrm{V}}_0\approx 3.6\mu \mathrm{A}\times 1\mathrm{M}\Omega =3.6\mathrm{V}$

Now from equation (3) $(3.6\mu \mathrm{A}-3\mu \mathrm{A})\times 103\Omega =(3\times 10-6\mathrm{A}){\mathrm{R}}_1$ $\Rightarrow {\mathrm{R}}_1\approx 0.2\mathrm{k}\Omega$