Question

The heat produced in a \( 4 \, \Omega \) resistor when an electric current of \( 5 \, A \) flows in it for \( 2 \) seconds is

• A. \( 200 \, J \)
• B. \( 40 \, J \)
• C. \( 50 \, J \)
• D. \( 80 \, J \)

Hint:

Joule's law of heating is a fundamental concept in electricity. Remember the formula \( H = I^2 R t \) and the units of each quantity.

Answer 1

The Correct Option is A

Step 1: Recall Joule's law of heating.
Joule's law states that the heat produced in a resistor is directly proportional to the square of the current, the resistance, and the time for which the current flows. Mathematically, it is given by: \[ H = I^2 R t \] where \( H \) is the heat produced (in Joules), \( I \) is the current (in Amperes), \( R \) is the resistance (in Ohms), and \( t \) is the time (in seconds). Step 2: Identify the given values.
From the problem statement, we have:
Resistance, \( R = 4 \, \Omega \)
Electric current, \( I = 5 \, A \)
Time, \( t = 2 \, s \)
Step 3: Substitute the values into Joule's law formula. \[ H = (5 \, A)^2 \times (4 \, \Omega) \times (2 \, s) \] \[ H = (25 \, A^2) \times (4 \, \Omega) \times (2 \, s) \] \[ H = 100 \, \Omega \cdot A^2 \times 2 \, s \] Step 4: Calculate the heat produced.
Since \( 1 \, \Omega \cdot A^2 \cdot s = 1 \, J \) (because \( V = IR \implies \text{Volt} = \text{Ampere} \times \text{Ohm} \) and \( P = VI = I^2 R \implies \text{Watt} = \text{Ampere}^2 \times \text{Ohm} \), and \( E = Pt \implies \text{Joule} = \text{Watt} \times \text{second} \)), we have: \[ H = 200 \, J \] The heat produced in the resistor is \( 200 \, J \).