Question

An ideal choke takes a current of $10Amperes$ when connected to an ac supply of $125volt$ and $50Hz$. A pure resistor under the same condition takes a current of $12.5Ampere$. If the two are connected to an ac supply of $100\sqrt{2}$ volt and $40Hz,$ then the current in a series combination of the above resistor and inductor is:

• A. $10A$
• B. $12.5A$
• C. $20A$
• D. $25A$

Answer 1

The correct option is A $10A$

In ideal choke coil all voltage drop occur at inductor. Resistance is neglected in case of ideal choke coil. so there is only inductance.
Let inductance of choke coil $=L$
given value of AC voltage $\left(V\right)=125volt$, frequency $\left(f\right)=50Hz$
Current in inductor $=\frac{V}{\omega L}$
$\omega L=$Resistance due to inductor,
$\omega$ = $2\pi f$
Current in inductor $=\frac{V}{\omega L}=\frac{V}{2\pi fL}=\frac{125}{2\pi \times 50\times L}$
Value of current is given in question$=10A$.
$\frac{125}{2\pi \times 50\times L}=10⟹L=\frac{125}{1000\pi }H$
Let the value of resistance is $R$
So in pure resistance case
Current $=\frac{V}{R}$ , current given $=12.5A$
$\frac{125}{R}=12.5⟹R=10\Omega$
Now this $R\&L$ is connected in series to another AC source of $V=100\sqrt{2}$ volt,
frequency $\left(f\right)=40Hz$
so impedance of combination $\left(Z\right)=\sqrt{R^2+{\omega L}^2}=\sqrt{R^2+{2\pi fL}^2}\Omega$
$Z=\sqrt{{10}^2+{\left(2\pi \times 40\times \frac{125}{1000\pi }\right)}^2}$ , putting value of $R,L,$ and new $f$.
$Z=10\sqrt{2}\Omega$
Current in new combination $=\frac{V}{Z}=\frac{100\sqrt{2}}{10\sqrt{2}}=10A$
So, A is correct Answer.