A student was trying to construct the circuit shown in the figure below marked (a), but ended up constructing the circuit marked (b). Realizing her mistake, she corrected the circuit, but to her surprise, the output voltage (across $R$) did not change.

The value of resistance $R$ is :-

For circuit (a),

$i_R=\left(\frac{10}{{\displaystyle \underset{\mathrm{Current} \mathrm{through} \mathrm{cell}}{\underset{\uparrow }{\frac{300R}{300+R}}}}+300}\right)\times \frac{300}{300+R}$

[Note : $300 \mathrm{\Omega } \& R$ are in parallel which is in series with $100 \& 200 \mathrm{\Omega }$]

$\therefore V_{R_a}=\frac{10\times 300R}{300R+{300}^2+300R}$

[$V_{R_a}$ is potential difference across resistance $R$]

For circuit (b),

$i_R=\left(\frac{10}{{\displaystyle \underset{\mathrm{Current} \mathrm{through} \mathrm{cell}}{\underset{\uparrow }{\frac{(200+R)\left(300\right)}{200+R+300}}}+100}}\right)\times \frac{300}{300+200+R}$

[Note: $R \& 200 \mathrm{\Omega }$ are in series which is in parallel with $300 \mathrm{\Omega }$ & again the combination is in series with $100 \mathrm{\Omega }$]

$\therefore V_{R_b}=\frac{100\times 300R}{300\times 200+300R+100\times 500+100R}$

[$V_{R_b}$ is potential difference across resistance $R$]

According to given situation

$V_{R_a}=V_{R_b}$

$\therefore 300R+9\times {10}^4+300R=6\times {10}^4+400R+5\times {10}^4$

$\Rightarrow 200R=2\times {10}^4\Rightarrow R=100 \mathrm{\Omega }$