# Question #30e8f

Jan 27, 2017

Option D is correct

#### Explanation:

In a heating device rate of energy dissipation or Power (P) of the device is related with the applied voltage (V) across the heating resistor coil of resistance (R) and current (I) passing through the resistor as follows

$P = {V}^{2} / R \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

$P = {I}^{2} R \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(2\right)$

$P = I \times V \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(3\right)$

Testing option A-
( Tripling the potential difference (V) keeping the resistance (R) the same .) The relation (1) suggests that $P \propto {V}^{2}$ when R is constant . So tripling the voltage keeping resistance same will enhance the rate of dissipation 9 times

Testing option B-
( Tripling the current (I) keeping the resistance (R) the same .)
The relation (2) suggests that $P \propto {I}^{2}$ when R is constant . So tripling the voltage keeping resistance same will enhance the rate of dissipation 9 times

Testing option C-
( Tripling the resistance (R) keeping the potential difference (V) the same .)
The relation (1) suggests that $P \propto \frac{1}{R}$ when V is constant . So tripling the resistance (R) keeping pd same will decrease the rate of dissipation *to $\frac{1}{3}$ times of its previous value.

Testing option D-
( Tripling the resistance (I) keeping the current I) the same .)
The relation (2) suggests that $P \propto R$ when I is constant . So tripling the resistance keeping current(I) same will enhance the rate of dissipation 3 times

Testing option E-
( Tripling. both the potential difference(V) and current (I) .)
The relation (3) suggests that $P \propto V \mathmr{and} P \propto I$ So tripling both the voltage(V) and current (I) will enhance the rate of dissipation 9 times

So only option D is acceptable