Question

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

• A. 10A
• B. 8A
• C. $5\sqrt{2}$amp
• D. $10\sqrt{2}$amp

Answer 1

Answer: (C)

$5\sqrt{2}$amp

Answer 2

X_L = $\frac{100}{8}$, R = $\frac{100}{10} = 10\Omega$ ; L 100π = $\frac{100}{8}$ or

L = $\frac{1}{8\pi}$H

Z = $\sqrt{\left( \frac{1}{8\pi} \times 2\pi \times 40 \right)^{2} + 10^{2}} = 10\sqrt{2}$

I = $\frac{E}{Z} = \frac{100}{10\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}A$

Hence (3) is correct.