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

The correct option is A $6\Omega$

The potential drop across $3\Omega ,6\Omega \text{and}2\Omega$ resistors are same as parallel.

We know Heat $H\alpha \frac{1}{R}$

Thus $2\Omega$ producing more heat.

As $5\Omega and4\Omega$ are in parallel, the $4\Omega$ resistor will produce more heat.

Thus the equivalent circuit

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

as equivalent resistance of $3\Omega ,6\Omega ,2\Omega is1\Omega$

and equivalent resistance of $5\Omega and4\Omega is\left(20/9\right)\Omega$

$V_1=\left(9/29\right)V{V}_2=\left(20/29\right)V$

Power across$2\Omega \text{resistor}{P}_1=V_1^2/2$

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

Power across $4\Omega$ resistor

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

Hence $P_2>P_1$

Therefore $6\Omega$ resistor will generate more heat.