Question

A resistor dissipates 192 J of energy in 1 s when a current of 4 A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5 s is _________ J.

Hint:

Since \( H \propto I^2 t \), if current doubles (\( \times 2 \)), heat becomes \( 2^2 = 4 \) times for the same duration. Since time also increases from 1 s to 5 s (\( \times 5 \)), the total heat becomes \( 4 \times 5 = 20 \) times the original.
\( 192 \times 20 = 3840 \) J.

Answer 1

Correct Answer: 3840

Step 1: Understanding the Concept:
Thermal energy (heat) dissipated by a resistor is given by Joule's Law of heating, which states that the heat produced is proportional to the square of the current, the resistance, and the time for which the current flows.
Step 2: Key Formula or Approach:
The formula for heat dissipation is:
\[ H = I^2Rt \]
Where:
\( H \) = Heat energy (Joules)
\( I \) = Current (Amperes)
\( R \) = Resistance (Ohms)
\( t \) = Time (Seconds)
Step 3: Detailed Explanation:
Initial Case:
Given: \( H_1 = 192 \) J, \( I_1 = 4 \) A, \( t_1 = 1 \) s.
Using the formula:
\[ 192 = (4)^2 \times R \times 1 \]
\[ 192 = 16R \]
\[ R = \frac{192}{16} = 12\,\Omega \]
Second Case:
Current is doubled, so \( I_2 = 2 \times 4 = 8 \) A.
Time \( t_2 = 5 \) s.
Resistance \( R \) remains the same (\( 12\,\Omega \)).
The new heat dissipated \( H_2 \) is:
\[ H_2 = I_2^2 \times R \times t_2 \]
\[ H_2 = (8)^2 \times 12 \times 5 \]
\[ H_2 = 64 \times 12 \times 5 \]
\[ H_2 = 64 \times 60 \]
\[ H_2 = 3840 \text{ J} \]
Step 4: Final Answer:
The thermal energy dissipated in 5 seconds is 3840 J.