Reasoning: Electric power is the rate at which electrical energy is converted or consumed.

Reasoning: $\mathbf{P = IV = (12)(2) = 24\ W}$

Reasoning: Power is the product of potential difference and current.

Reasoning: $\mathbf{P =}\mathbf{I}^{\mathbf{2\ }}\mathbf{R}$ $\mathbf{\Longrightarrow}$ Doubling current gives $\mathbf{P =}\mathbf{(2I)}^{\mathbf{2\ }}\mathbf{R =}\mathbf{4I}^{\mathbf{2\ }}\mathbf{R}$.

Reasoning: $\mathbf{R =}$ $\frac{\mathbf{P}}{\mathbf{I}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{50}}{\mathbf{25}}\mathbf{= 2\ \Omega}$

Reasoning: This relation contains only voltage and resistance.

Reasoning: $\mathbf{P =}$ $\frac{\mathbf{V}^{\mathbf{2}}}{\mathbf{R}}$
At constant voltage, increasing resistance decreases power.

Reasoning: $\mathbf{P =}$ $\frac{\mathbf{V}^{\mathbf{2}}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{(20)}^{\mathbf{2}}}{\mathbf{10}}\mathbf{=}\frac{\mathbf{400}}{\mathbf{10}}\mathbf{= 40\ W}$

Reasoning: According to $\mathbf{P =}\mathbf{I}^{\mathbf{2\ }}\mathbf{R}$, larger resistance produces more heat for the same current.

Reasoning: $\mathbf{P =}$ $\frac{\mathbf{V}^{\mathbf{2}}}{\mathbf{R}}$ $\mathbf{\Longrightarrow}$ Tripling voltage increases power by $\mathbf{3}^{\mathbf{2}}\mathbf{= 9}$.

Reasoning: $\mathbf{I}\mathbf{=}$ $\frac{\mathbf{P}}{\mathbf{V}}\mathbf{=}\frac{\mathbf{60}}{\mathbf{120}}\mathbf{= 0.5\ A}$

Reasoning: Power measures energy conversion per unit time.

Reasoning: $\mathbf{P = (2I}\mathbf{)}^{\mathbf{2\ }}\mathbf{(2R) = 8}\mathbf{\ I}^{\mathbf{2\ }}\mathbf{R}$

Reasoning: Power varies as $\mathbf{V}^{\mathbf{2}}$ . $\mathbf{(20\ /\ 10)}^{\mathbf{2}}\mathbf{= 4}$
Because power dissipation scales with the square of the voltage, doubling the voltage from $\mathbf{10\ V}$ to $\mathbf{20\ V}$ quadruples the total power dissipated.
⮚ Step-by-Step Calculation:
Since both resistors have the exact same resistance $\mathbf{(R)}$, we can compare their power outputs directly based on their voltage.
● First Resistor $\mathbf{(}\mathbf{P}_{\mathbf{1}}\mathbf{)}$: Connected to $\mathbf{10\ V}$ $\mathbf{\Longrightarrow}$ $\mathbf{P}_{\mathbf{1}}\mathbf{=}$ $\frac{\mathbf{10}^{\mathbf{2}}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{100}}{\mathbf{R}}$
● Second Resistor $\mathbf{(}\mathbf{P}_{\mathbf{2}}\mathbf{)}$: Connected to $\mathbf{20\ V}$ $\mathbf{\Longrightarrow}$ $\mathbf{P}_{\mathbf{2}}\mathbf{=}$ $\frac{\mathbf{20}^{\mathbf{2}}}{\mathbf{R}}\mathbf{=}\frac{\mathbf{400}}{\mathbf{R}}$
⮚ Summary Ratio:
Dividing the power of the second resistor by the first gives:
$\frac{\mathbf{P}_{\mathbf{2}}}{\mathbf{P}_{\mathbf{1}}}\mathbf{=}$ $\frac{\frac{\mathbf{400}}{\mathbf{R}}}{\frac{\mathbf{100}}{\mathbf{R}}}\mathbf{=}$ $\frac{\mathbf{400}}{\mathbf{100}}\mathbf{= 4}$
Because power dissipation scales with the square of the voltage, doubling the voltage from $\mathbf{10V}$ to $\mathbf{20\ V}$ quadruples the total power dissipated.

Reasoning: $\mathbf{I}\mathbf{=}$ $\frac{\mathbf{P}}{\mathbf{V}}\mathbf{=}\frac{\mathbf{100}}{\mathbf{200}}\mathbf{= 0.5\ A}$

Reasoning: Power is measured in watts $\mathbf{(W)}$.

Reasoning: $\mathbf{P \propto}\mathbf{V}^{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ Doubling voltage makes power four times larger.

Reasoning: Substituting $\mathbf{V = IR}$ into $\mathbf{P = IV\ }$gives $\mathbf{P =}\mathbf{I}^{\mathbf{2\ }}\mathbf{R}$.

Reasoning: $\mathbf{P =}\mathbf{I}^{\mathbf{2\ }}\mathbf{R =}\mathbf{(3)}^{\mathbf{2\ }}\mathbf{(8) = 72\ W}$

Reasoning: $\mathbf{P =}$ $\frac{\mathbf{V}^{\mathbf{2}}}{\mathbf{R}}$ $\mathbf{\Longrightarrow}$ Increasing resistance reduces power.

Reasoning: $\mathbf{V =}$ $\frac{\mathbf{P}}{\mathbf{I}}\mathbf{=}\frac{\mathbf{200}}{\mathbf{10}}\mathbf{= 20\ V}$

Reasoning: $\mathbf{P =}\mathbf{I}^{\mathbf{2\ }}\mathbf{R}$

Reasoning: $\mathbf{R =}$ $\frac{\mathbf{P}}{\mathbf{I}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{16}}{\mathbf{4}}\mathbf{= 4\ \Omega}$

Reasoning: All three formulas represent electric power and are related through Ohm’s law.

Reasoning: $\mathbf{R =}$ $\frac{\mathbf{V}^{\mathbf{2}}}{\mathbf{P}}\mathbf{=}\frac{\mathbf{900}}{\mathbf{90}}\mathbf{= 10\ \Omega}$

Reasoning: $\mathbf{P = IV = (220)(5) = 1100\ W}$

Reasoning: $\mathbf{P =}\mathbf{I}^{\mathbf{2\ }}\mathbf{R}$ $\mathbf{\Longrightarrow}$ Halving current gives $\mathbf{(1\ /\ 2)}^{\mathbf{2}}\mathbf{\ P = P\ /\ 4}$.
The new power is $\mathbf{P\ /\ 4}$ (one-fourth of the original power). This is because power is proportional to the square of the current $\mathbf{(P =}\mathbf{I}^{\mathbf{2\ }}\mathbf{R)}$.
⮚ Power Formula Breakdown:
● Original Power $\mathbf{(P)}$: Equals $\mathbf{I}^{\mathbf{2\ }}\mathbf{R}$, where $\mathbf{I}$ is the original current and $\mathbf{R}$ is the resistance.
● New Current: Becomes $\left( \frac{\mathbf{I}}{\mathbf{2}}\mathbf{I} \right)$ (half of the original).
● New Calculation: $\left( \frac{\mathbf{1}}{\mathbf{2}}\mathbf{I} \right)^{\mathbf{2}}\mathbf{\times R =}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{I}^{\mathbf{2\ }}\mathbf{R =}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{P}$

Reasoning: From $\mathbf{P =}\mathbf{I}^{\mathbf{2\ }}\mathbf{R}$, resistance must remain constant.

Reasoning: $\mathbf{R =}$ $\frac{\mathbf{V}^{\mathbf{2}}}{\mathbf{P}}\mathbf{=}\frac{\mathbf{3600}}{\mathbf{180}}\mathbf{= 20\ \Omega}$

Reasoning: For an ohmic resistor, all three formulas must yield the same power when consistent values of $\mathbf{(V)}$, $\mathbf{(I)}$ and $\mathbf{(R)}$ are used.