Question

An ideal choke takes a current of $10\text{A}$ when connected to an ac supply of $125\text{V}$ and $50\text{Hz}$. A pure resistor under the same conditions takes a current of $12.5\text{A}$. If the two are connected to an $\text{AC}$ supply of $100\text{V}$ and $40\text{Hz}$, then the current in a series combination of the above resistor and inductor is

• A. $5\sqrt{2}A$
• B. $12.5A$
• C. $20A$
• D. $10A$

Answer 1

The correct option is A $5\sqrt{2}A$

From given the circuits are
$V=I{X}_L$
$X_L=\frac{V}{I}=\frac{125V}{10A}=12.5\Omega$
$X_L=\omega L=2\pi f\left(L\right)$
$2\pi \times 50\times L=12.5$
$L=\frac{12.5}{2\pi \times 50}$
$R=\frac{125}{12.5}=10\Omega$

$i=\frac{100}{Z}$
$Z=\sqrt{X_L^{\prime 2}+R^2}$
$\begin{array}{cc}{X}_L^{\prime }=& \omega L=2\pi \times 40\times \frac{12.5}{2\pi \times 50}\\ & X_L^{\prime }=10\Omega \\ Z& =\sqrt{X_L^{\prime 2}+R^2}\\ & =\sqrt{100+100}\\ & =10\sqrt{2}\Omega \end{array}$
Therefore,
$i=\frac{100}{10\sqrt{2}}A=5\sqrt{2}A$.