Physics - Current Electricity Question with Solution | TestHub

$1\Omega$ and $3\Omega$ are in series so, egin{aligned} & R_s=R_1+R_2 \\ & R_s=3+1=4 \Omega \end{aligned} and $R_2=8\Omega$ net $i=$ current in the circuit Current through $R_1$ is $i_1=\frac{i\times R_2}{R_1+R_2}$ $=\frac{i\times 8}{12}=\frac{21}{3}$ Current $R_2$ is $i_2=\frac{1\times R_1}{R_1+R_2}=\frac{i\times 4}{12}$ $=\frac{i\times 8}{12}$ Power dissipated in $3\Omega$ resistance is: $P_1=l_1^2\times 3$ Power dissipated in $8\Omega$ resistance is: $P_2=l_2^2\times 8$ For (i) and (ii) $\begin{array}{rl}& \frac{P_1}{P_2}=\frac{i_1^2\times 3}{i_2^2\times 8}\\ & \Rightarrow \frac{P_1}{P_2}\end{array}$ $\begin{array}{rl}& =\frac{{\left(\frac{2i}{3}\right)}^2\times 3}{{\left(\frac{i}{3}\right)}^2\times 8}=\frac{3}{2}\\ \therefore P_1& =\frac{3}{2}\times P_2=\frac{3}{2}\times 2=3\mathrm{W}\end{array}$ $\begin{array}{rlr}\because & P& =100\mathrm{W},V=125V\\ \therefore & P& =VI\\ \Rightarrow & I& =\frac{P}{V}=\frac{100}{125}A\end{array}$