What will be the internal energy when at 2`Omega` resistant 5 ampere current is flown for 30 mts?

Options

A. 9 Joule
B. `9xx10^2` Joule
C. `9xx10^4` Joule
D. `9xx10^3` Joule

Explanation

To find the internal energy when a current flows through a resistor, we can use the formula for electrical energy:

**Step 1: Identify the given values.**
- Resistance (R) = 2 ohms
- Current (I) = 5 amperes
- Time (t) = 30 minutes

**Step 2: Convert time from minutes to seconds.**
Since the formula requires time in seconds, we convert 30 minutes to seconds:
\[ t = 30 \text{ minutes} \times 60 \text{ seconds/minute} = 1800 \text{ seconds} \]

**Step 3: Use the formula for electrical energy.**
The formula to calculate the energy (E) released due to the flow of current through a resistor is:
\[ E = I^2 \cdot R \cdot t \]

**Step 4: Substitute the values into the formula.**
Now substitute the values we have:
\[ E = (5 \text{ A})^2 \cdot (2 \text{ ohms}) \cdot (1800 \text{ s}) \]

**Step 5: Calculate the energy.**
First, calculate \( (5 \text{ A})^2 \):
\[ (5 \text{ A})^2 = 25 \text{ A}^2 \]
Now substitute this back into the equation:
\[ E = 25 \cdot 2 \cdot 1800 \]
\[ E = 50 \cdot 1800 \]
\[ E = 90000 \text{ Joules} \]

**Step 6: Express the final answer in scientific notation.**
The final energy can be expressed as:
\[ E = 9 \times 10^4 \text{ Joules} \]

Thus, the internal energy when a 5 ampere current flows through a 2 ohm resistor for 30 minutes is \( 9 \times 10^4 \) Joules.

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