Question

An ideal inductor takes a current of $10A$ when connected to a $125$ $V$, $50$ $Hz$ AC supply. A pure resistor across the same source takes $12.5$ $A$. If the two are connected in series across a $100\sqrt{2}V$, $40$ $Hz$ supply, the current through the circuit will be

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

Answer 1

The correct option is A $10A$

Inductor

$X_L=\frac{125}{10}=12.5\Omega$

Resistor

$X_R=\frac{125}{12.5}=10\Omega$

$X_L^{\prime }=12.5\left(\frac{{\omega }_2}{{\omega }_1}\right)\Omega =12.5\left(\frac{40}{50}\right)\Omega =10\Omega$

both are connected in series

$z=\sqrt{{10}^2+(10{)}^2}$

$=10\sqrt{2}\Omega$

$Vrms=100\sqrt{2}V$

$Irms=\frac{100\sqrt{2}}{10\sqrt{2}}=10A$