NTSE Stage 2 SAT 2019 Exam Question Paper

After simplifying the given circuit, we get,${\mathit{R}}_{\mathit{A}\mathit{B}}=\frac{1}{3}\mathrm{\Omega }$
Then equivalent resistance across the battery,
${\mathit{R}}_{\mathit{e}\mathit{q}}=2+\frac{1}{3}+2=\frac{13}{3}\mathrm{\Omega }$
So current in circuit,$\mathit{I}=\frac{\mathit{V}}{{\mathit{R}}_{\mathit{e}\mathit{q}}}\Rightarrow \frac{1.3}{13}\times 1\mathit{a}\mathit{m}\mathit{p}$
$\mathit{I}=\frac{3}{10}\mathit{a}\mathit{m}\mathit{p}$
Power dissipated across arm AB,
${\mathit{P}}_{\mathit{A}\mathit{B}}={\mathit{I}}^3\times {\mathit{R}}_{\mathit{A}\mathit{B}}={\left(\frac{3}{10}\right)}^2\times \frac{1}{3}$
${\mathit{P}}_{\mathit{A}\mathit{B}}=\frac{3}{100}=0.03$Watt
Total power dissipated in circuit,
${\mathit{P}}_{\mathit{c}\mathit{k}\mathit{t}}={\mathit{I}}^2\times {\mathit{R}}_{\mathit{e}\mathit{q}}={\left(\frac{3}{10}\right)}^2\times \frac{13}{3}$
${\mathit{P}}_{\mathit{c}\mathit{k}\mathit{t}}=\frac{39}{100}=0.39$watt
Ratio of power across A and B to total power = Ratio of work done across A and B to total circuit
$\because \mathit{W}=\mathit{P}\times \mathit{t}$
So,$\frac{{\mathit{P}}_{\mathit{A}\mathit{B}}}{{\mathit{P}}_{\mathit{c}\mathit{k}\mathit{t}}}=\frac{{\mathit{W}}_{\mathit{A}\mathit{B}}}{{\mathit{W}}_{\mathit{c}\mathit{k}\mathit{t}}}=\frac{0.03}{0.39}=\frac{1}{13}$