Question

If an AC ammeter reads $IA$ in an AC circuit, then the peak value of the current is

• A. $\frac{I}{\sqrt{2}}A$
• B. $IA$
• C. $I\sqrt{2}A$
• D. $2IA$

Answer 1

The correct option is C

$I\sqrt{2}A$

Step 1: Given data

If an AC ammeter reads $IA$ in an AC circuit,

Step 2: Formula used

In an AC circuit, current is given by the equation

$I=I_0\sin \omega t$

rms value of sine function is at $45°$

Therefore,

$$\begin{gathered} I=I_0\sin \omega t \\ \Rightarrow I_{rms}=I_0\sin 45° \\ \Rightarrow I_{rms}=\frac{I_0}{\sqrt{2}} \end{gathered}$$

where $I_0$ is the peak value of current, $\omega$ is the angular frequency and $t$ is time.

Step 3: Calculation

We know, the value of current we read in an AC ammeter in an AC circuit is the rms value.

So, it is given $I_{rms}=IA$

$$\begin{gathered} I_{rms}=\frac{I_0}{\sqrt{2}} \\ {I}_0=I\sqrt{2}A \end{gathered}$$

The peak value of current is $I\sqrt{2}A$

Therefore, option C is correct