Physics - Magnetism Question with Solution | TestHub

Current through arms of resistances $P$ and $Q$ in series $$i_1=\frac{i\times 330}{330+110}=\frac{3}{4}i$$ Here $i=$ total current Similarly, current through arms of resistances $R$ and $S$ in series $$i_2=\frac{i\times 110}{330+110}=\frac{1}{4}i$$ Heat developed per second $=i^{2}R$ $\therefore$ Ratio of heat developed per sec $$\begin{array}{l}{H}_P:H_Q:H_R:H_S\\ ={\left(\frac{3}{4}i\right)}^2\times 100:{\left(\frac{3}{4}i\right)}^2\times 10:\\ {\left(\frac{1}{4}i\right)}^2\times 300:{\left(\frac{1}{4}i\right)}^2\times 30\\ =30:3:10:1\end{array}$$