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Alternating Current — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsAlternating Current5 Marks
Question
An electrical device draws 2kW power from AC mains $(\text{voltage }223\text{V}\text{(rms)}=\sqrt{50.000}\text{V})$. The current differs (lags) in phase by $\phi\Big(\tan\phi=\frac{-3}{4}\Big)$ as compared to voltage. Find,
  1. R,
  2. $X_C - X_L,$
  3. $I_M$. Another device has twice the values for $R, X_C$ and $X_L$. How are the answers affected?
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Answer

Power $\text{P}=\frac{\text{V}^2}{\text{Z}}$Given, P = 2kW = 2000W,$ V_{rms} = V = 223V$
$\text{Z}=\frac{\text{V}^2}{\text{P}}=\frac{223\times223}{2\times10^3}=25$
Now, impedance $\text{Z}=25\Omega$
Impedance $\text{Z}=\sqrt{\text{R}^2+(\text{X}_\text{C}-\text{X}_\text{L})^2}$
$\Rightarrow\ \sqrt{\text{R}^2+(\text{X}_\text{C}-\text{X}_\text{L})^2}=25$
$\Rightarrow\ \text{R}^2+(\text{X}_\text{C}-\text{X}_\text{L})^2=625$
As, $\tan\phi=\frac{\text{X}_\text{C}-\text{X}_\text{L}}{\text{R}}=-\frac{3}{4}$
$\therefore\ 625=\text{R}^2+\Big(-\frac{3}{4}\text{R}\Big)^2=\frac{25}{16}$
$\text{R}^2=400$
$\Rightarrow\ \text{R}=20\Omega$
  1. Resistance $\text{R}=200\Omega.$
  2. $\text{X}_\text{L}-\text{X}_\text{C}=\frac{3\text{R}}{4}=\frac{3}{4}\times20=15\Omega.$
  3. Main current $\text{I}_\text{M}=\text{I}\sqrt{2}=\frac{\text{V}}{\text{Z}}\sqrt{2}=\frac{223}{25}\times\sqrt{2}=12.6\text{A}$.

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