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(D) The power $P$ dissipated by a heating coil is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Given th

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$55\,\Omega$

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(D) The power $P$ dissipated by a heating coil is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Given that the voltage $V = 220\, V$ is constant for all coils,the power $P$ is inversely proportional to the resistance $R$ $(P \propto \frac{1}{R})$.To obtain the maximum power,the resistance $R$ must be the minimum value among the given options.Comparing the resistances $55\,\Omega$,$110\,\Omega$,$220\,\Omega$,and $440\,\Omega$,the minimum resistance is $55\,\Omega$.Therefore,the heater with $55\,\Omega$ resistance will produce the maximum power.

Heating coils rated at $220\, V$ are available with resistances of $55\,\Omega$,$110\,\Omega$,$220\,\Omega$,and $440\,\Omega$. Which resistance value will produce the maximum power?

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