Question

The resistor in which maximum heat will be produced is

• A. $6\Omega$
• B. $2\Omega$
• C. $5\Omega$
• D. $4\Omega$

Answer 1

Answer: (D)

$3\Omega$ , $6\Omega$ and $2\Omega$ resistances are in parallel. So, the potential drop across them will be equal. Of these, three resistances maximum heat will be generated across $2\Omega$ resistance
$\left(H=\frac{V^2}{R}torH\propto \frac{1}{R}\right)$
Similarly, $5\Omega$ and $4\Omega$ resistances are also in parallel so, more heat will be generated across $4\Omega$ resistor. Now the given circuit can be redrawn as :



$\frac{V_1}{V_2}=\frac{1}{\frac{20}{9}}=\frac{9}{20}$
$\therefore V_1=\left(\frac{9}{29}\right)V$
and $V_2=\left(\frac{20}{29}\right)V$
Power developed across $2\Omega$ resistor will be
$P_1=\frac{V_1^2}{2}={\left(\frac{9}{29}\right)}^2\frac{V}{2}$
and power developed across $4\Omega$ 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 $4\Omega$ resistance.