Question

The $H$ amount of thermal energy is developed by a resistor in $10 s$ when a current of $4 A$ is passed through it If the current is increased to $16 A$, the themal energy developed by the resistor in $10 s$ will be:

• A. $\frac{ H }{4}$
• B. H
• C. $4 H$
• D. $16 H$

Hint:

The thermal energy developed in a resistor is proportional to the square of the current.

Answer 1

The Correct Option is D

\(H∝ i^2\) for t = constant
\(\frac{H}{H'} = (\frac{4}{16})^2\)
\(H'= 16\,H\)
So, The Correct answer is option (D): 16H

Answer 2

The thermal energy developed by the resistor is given by the formula: \[ E = I^2 R t \] where \( E \) is the thermal energy, \( I \) is the current, \( R \) is the resistance, and \( t \) is the time. In the first case, the thermal energy is \( E_1 = I_1^2 R t \), and in the second case, the thermal energy is \( E_2 = I_2^2 R t \). Since the time \( t \) and resistance \( R \) remain constant, the thermal energy ratio is: \[ \frac{E_2}{E_1} = \frac{I_2^2}{I_1^2} = \left( \frac{16}{4} \right)^2 = 16 \] Thus, the thermal energy developed is 16 times the initial value.