Question

What is the approximate peak value of an alternating current producing four times the heat produced per second by a steady current of $2A$ in a resistor?

• A. $2.8\text{A}$
• B. $4.0\text{A}$
• C. $5.6\text{A}$
• D. $8.0A$

Answer 1

The correct option is C

$5.6\text{A}$

Step 1: Given data

Steady current, $I_{steady}=2\text{A}$

Step 2: To find

The peak value of an alternating current.

Step 3: Calculate the peak value of an alternating current

$I_{rms}=2I_{steady}$..............(i)

Also,

$I_{rms}=\frac{I_{max}}{\sqrt{2}}$...................(ii)

Here,

$I_{max}$ is the peak value.

$I_{rms}$ is the alternating current.

From (i) and (ii)

$\frac{I_{max}}{\sqrt{2}}=2I_{steady}$
$$\begin{gathered} I_{max}=2\sqrt{2}{I}_{steady} \\ =2\sqrt{2}\times 2\text{A} \\ =5.6A \end{gathered}$$

Hence, option(C) is correct.