Physics - Current Electricity Question with Solution | TestHub

$3 \mathrm{\Omega }$,$6 \mathrm{\Omega }$and$2 \mathrm{\Omega }$resistances are in parallel. So, the potential drop across them will be equal. Of these, three resistances maximum heat will be generated across$2 \mathrm{\Omega }$resistance

$\left(H=\frac{V^2}{R}t or H\propto \frac{1}{R}\right)$

Similarly,$5 \mathrm{\Omega }$and$4 \mathrm{\Omega }$resistances are also in parallel so, more heat will be generated across$4 \mathrm{\Omega }$resistor. Now the given circuit can be redrawn as :





$\frac{V_1}{V_2}=\frac{1}{{\displaystyle \frac{20}{9}}}=\frac{9}{20}$

$\therefore V_1=\left(\frac{9}{29}\right)V$

and$V_2=\left(\frac{20}{29}\right)V$

Power developed across$2 \mathrm{\Omega }$resistor will be

$P_1=\frac{V_1^2}{2}={\left(\frac{9}{29}\right)}^2\frac{V}{2}$

and power developed across$4 \mathrm{\Omega }$resistor is

$P_2=\frac{V_2^2}{2}={\left(\frac{20}{29}\right)}^2\frac{V}{4}>P_1$

$\therefore$Maximum heat is generated across$4 \mathrm{\Omega }$resistance.