Question

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

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

Answer 1

The correct option is A $10/\sqrt{2}A$

$X_L=V_{rms}/I_{rms}=125/10=12.5\Omega$
$R=125/12.5=10\Omega$
at 40Hz, $X_L^{\prime \prime }=12.5\frac{40}{50}=10\Omega$
When connected in series, imepdance = $\sqrt{R^2+{X_L^{\prime \prime }}^2}=\sqrt{{10}^2+{10}^2}=10\sqrt{2}\Omega$
Current in series combination = $\frac{100}{10\sqrt{2}}=\frac{10}{\sqrt{2}}A$