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Q.
The resistor in which maximum heat will be produced is
NTA AbhyasNTA Abhyas 2020Current Electricity
Solution:
$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
$\left(H = \frac{V^{2}}{R} t or H \propto \frac{1}{R}\right)$
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 :
$\frac{V_{1}}{V_{2}}=\frac{1}{\frac{20}{9}}=\frac{9}{2 0}$
$\therefore V_{1}=\left(\frac{9}{2 9}\right)V$
and $V_{2}=\left(\frac{2 0}{2 9}\right)V$
Power developed across $2\Omega $ resistor will be
$P_{1}=\frac{V_{1}^{2}}{2}=\left(\frac{9}{2 9}\right)^{2}\frac{V}{2}$
and power developed across $4\Omega $ resistor is
$P_{2}=\frac{V_{2}^{2}}{2}=\left(\frac{2 0}{2 9}\right)^{2}\frac{V}{4}>P_{1}$
$\therefore $ Maximum heat is generated across $4\Omega $ resistance.
