Question

An ideal inductor takes a current of 10 A 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 $100\sqrt{2}V$ 40Hz supply , the current through the circuit will be

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

Answer 1

Answer: (A)

For 50 Hz and 125 V supply
$X_L=\omega L=\frac{V}{i_L}\Rightarrow L=\frac{1}{8\pi },R=\frac{V}{i_R}=10\Omega$
For 40 Hz, 100 $\sqrt{2V}$ supply
$i=\frac{V}{\sqrt{R^2+X^{2_L}}}=\frac{V}{\sqrt{R^2+4{\pi }^2{f}^2{L}^2}}$