Question

If the voltage of a source in an AC circuit is represented by the equation, $E=220\sqrt{2}\sin \left(314t\right)$. Calculate the peak value of the current if the net resistance of the circuit is $220\Omega$.

Take $\sqrt{2}=1.4$

• A. $1.8\text{A}$
• B. $1.6\text{A}$
• C. $1.4\text{A}$
• D. $1.2\text{A}$

Answer 1

The correct option is C $1.4\text{A}$

Given:

Voltage, $E=220\sqrt{2}\sin \left(314t\right)$

Comparing with $E=E_0\sin \omega t$, we get, peak value of voltage,

$E_0=220\sqrt{2}\text{V}$

So, peak value of current,

$i_0=\frac{E_0}{R}=\frac{220\sqrt{2}}{220}=1.4\text{A}$