iGCSE Physics (0625) 4.2.5 Resistance-Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Detailed solution:
To calculate the energy transferred in an electrical circuit, we use the formula $E = I \times V \times t$.
From the question, the current $I = 2.0 \text{ A}$, the potential difference $V = 12 \text{ V}$, and the time $t = 30 \text{ s}$.
Substituting these values into the equation: $E = 2.0 \text{ A} \times 12 \text{ V} \times 30 \text{ s}$.
First, calculate the power: $P = I \times V = 2.0 \times 12 = 24 \text{ W}$.
Then, multiply the power by the time: $E = 24 \text{ W} \times 30 \text{ s} = 720 \text{ J}$.
Thus, the total energy transferred in the resistor is $720 \text{ J}$, which corresponds to option D.
▶️ Answer/Explanation
Detailed solution:
Electrical power $P$ is defined as the rate at which energy is transferred, given by $P = IV$.
Energy transferred $E$ is the product of power and the time $t$ for which the current flows, expressed as $E = P \times t$.
By substituting the expression for power into the energy equation, we derive $E = (IV) \times t$.
Therefore, the total energy transferred by the resistor is $E = IVt$.
This matches option D, while the other options represent incorrect algebraic arrangements of these variables.
▶️ Answer/Explanation
Detailed solution:
First, convert the time from hours to seconds: $t = 1.0 \times 3600 = 3600\text{ s}$.
Using the electrical energy formula $E = IVt$, where $I = 2.5\text{ A}$ and $V = 12\text{ V}$.
Substitute the values: $E = 12\text{ V} \times 2.5\text{ A} \times 3600\text{ s}$.
This gives $E = 108\,000\text{ J}$.
Options B and D are incorrect because Watts ($\text{W}$) is the unit for power, not energy.
Option A is incorrect as it fails to account for the conversion of hours into seconds.