Question

An ideal choke takes a current of $8$ ampere when connected to an AC supply of $100$ volt and $50Hz$. A pure resistor under the same condition takes a current of $10$ ampere. If the two are connected to an AC supply of $150$ volt and $40Hz$, then the current in a series combination of the above resistor and inductor is

• A. A10$ amp
• B. $8$ amp
• C. $18$ amp
• D. $\frac{15}{\sqrt{2}}$ amp

Answer 1

The correct option is D $\frac{15}{\sqrt{2}}$ amp

$i=\frac{V}{L\omega },forthechoke$
$\Rightarrow L\omega =\frac{V}{i}=\frac{100}{8}=12.5$
$\Rightarrow L=\frac{12.5}{100\pi }$
$i=\frac{V}{R},fortheresistor$
$\Rightarrow R=\frac{V}{i}=\frac{100}{10}=10$
$whenbothareconnected,$
$i=\frac{V}{\sqrt{R^2+{\left(L\omega \right)}^2}}$
$=\frac{150}{\sqrt{{10}^2+{(\frac{12.5}{100\pi }\times 80\pi )}^2}}$
$=\frac{150}{\sqrt{{10}^2+{10}^2}}=\frac{15}{\sqrt{2}}amp$