Question

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

• A. 3840.00
• B. 3840
• C. 3840.0

Answer 1

Answer: (A), (B), (C)

Energy dissipated in the resistor is $E=i^{2}Rt$

Given that resistor dissipates $E_1=192\text{J}$ of energy in $t_1=1\text{s}$ when a current of $i_1=4\text{A}$ is passed through it.

On substitution, we get
$R=\frac{192}{16}=12\Omega$

Now, current is doubled and time taken is $5\text{s}$
Hence $i_2=8\text{A},t_2=5\text{s}$

Energy dissipated in this case is $E_2=i_2^{2}R{t}_2=64\left(12\right)\left(5\right)=3840\text{J}$