Question

An $AC$ circuit consists of a $220\Omega$ resistance and a $0.7\text{H}$ choke in series. Find the power absorbed from $220\text{V}$ and $50\text{Hz}$ source connected in series with the circuit elements

• A. $110.08\text{W}$
• B. $600.07\text{W}$
• C. $809.02\text{W}$
• D. $1600\text{W}$

Answer 1

The correct option is A $110.08\text{W}$

Given:

$R=220\Omega ,L=0.7\text{H},V_{rms}=220\text{V}$

$\omega =50\text{Hz}$

The impedance of the circuit is,

$Z=\sqrt{R^2+{\omega }^2{L}^2}=\sqrt{R^2+(2\pi fL{)}^2}$

$=\sqrt{(220{)}^2+(2\times 3.14\times 50\times 0.7{)}^2}$

$\Rightarrow Z=311\Omega$

Now, $\cos \varphi =\frac{R}{Z}=\frac{220}{311}=0.707$

Also, $i_{rms}=\frac{V_{rms}}{Z}=\frac{220}{311}=0.707\text{A}$

$\therefore$ The power absorbed in the circuit

$P=V_{rms}{i}_{rms}\cos \varphi$

$=\left(220\right)\left(0.707\right)\left(0.707\right)$

$=110.08\text{W}$

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