Question

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

• A. 7A
• B. 12.5A
• C. 20A
• D. 25A

Answer 1

The correct option is A 7A

$X_2=\frac{V}{I}=\frac{125}{10};LW=12.5;L=\frac{12.5}{2\pi f}L=\frac{0.125}{\pi }$
$R=\frac{V}{I}=\frac{125}{12.5}=10\Omega$
$I_{rms}=\frac{V_{rms}}{Z}=\frac{V_{rms}}{\sqrt{(}R{)}^2+(LW{)}^2}$
$=\frac{100}{\sqrt{{10}^2+{(\frac{0.125}{\pi }\times 2\pi \times 40)}^2}}=\frac{100}{10\sqrt{2}}$
$=5\sqrt{2}=5\times 1.414$