Question

An AC circuit contains a resistor of $10\Omega$ and an inductor of $2.0\text{H}$, joined in series. If an AC voltage source of $120\text{V},60\text{Hz}$ is applied across this circuit, the peak AC will be -

• A. $0.52\text{A}$
• B. $0.42\text{A}$
• C. $0.22\text{A}$
• D. $0.32\text{A}$

Answer 1

The correct option is C $0.22\text{A}$

Given:

$R=10\Omega$

Now,

$X_L=\omega L=2\pi fL=2\pi \times 60\times 2=240\pi \Omega \approx 754\Omega$

So, impedance,

$Z=\sqrt{R^2+X_L^2}$

Here, $X_L>>R$

$\Rightarrow Z\approx X_L=754\Omega$

Further,

$i_{rms}=\frac{V_{rms}}{Z}=\frac{120}{754}\approx 0.16\text{A}$

Therefore, peak AC,

$i_0=\sqrt{2}{i}_{rms}=1.41\times 0.16=0.22\text{A}$

Hence, option $\left(C\right)$ is the correct answer.