Question

The resistor in which maximum heat will be produced is

• A. $2\Omega$
• B. $3\Omega$
• C. $4\Omega$
• D. $6\Omega$

Answer 1

Answer: (C)

Here $2\Omega ,3\Omega$ and $6\Omega$ are in parallel.
So, potential drop across them will be the same.
As heat produced, $H=\frac{V^2}{R}t$
i.e. $H\propto \frac{1}{R^{\prime }}$ so maximum heat will be generated across $2\Omega$ resistance.
Similarly $4\Omega$ and $5\Omega$ are also in parallel, so more heat will be generated across $4\Omega .$
Now the effective circuit will become

Total resistance $=I+\frac{20}{9}=\frac{29}{9}\Omega$
Current $I=\frac{V}{\frac{29}{9}}=\frac{9V}{29}A$
$V_1=\frac{9V}{29}\times 1=\frac{9}{29}V$
and $V_2=\frac{9V}{29}\times \frac{20}{9}=\frac{20V}{29}$
Power spent across $2\Omega$
$P_1=\frac{V_1^2}{2}=\frac{{\left(\frac{9V}{29}\right)}^2}{2}=\frac{40.5V^2}{(29{)}^2}$
Power spend across $4\Omega$
$P_2=\frac{V_2^2}{4}=\frac{\left(\frac{20V}{29}\right)}{4}=\frac{50V^2}{(29{)}^2}$
$\therefore P_2>P_1$. Hence maximum heat is produced in $4\Omega$ resistance.