Energy radiated by a transmitter – Rankers Physics

Power: Practice Problem & Solution

The energy that will be ideally radiated by a $100\text{ kW}$ transmitter in $1\text{ hour}$ is: (2022)
$1 \times 10^5\text{ J}$
$36 \times 10^7\text{ J}$
$36 \times 10^4\text{ J}$
$36 \times 10^5\text{ J}$

Solution Explained:

To solve this problem, we apply the core principles of Power. Understanding the underlying formula is key to arriving at the correct answer below:

Energy is given by $E = P \times t$. Here, power $P = 100\text{ kW} = 10^5\text{ W}$ and time $t = 1\text{ hour} = 3600\text{ s}$. Thus, $E = 10^5 \times 3600 = 3.6 \times 10^8\text{ J} = 36 \times 10^7\text{ J}$.

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