Question

An ideal choke takes a current of 8A when connected to an ac 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 ac supply of 100V and 40 Hz, 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

$X_{L} = \frac{V_{rms}}{i_{rms}} = \frac{100}{8} = 2\pi \times 50 \times L$ ⇒

$L = \frac{1}{8\pi}Henry$ and $R = \frac{100}{10} = 10\Omega$

So impedance of the series RC circuit at a frequency of 40 Hz is $Z = \sqrt{\left( \frac{1}{8\pi} \times 2\pi \times 40 \right)^{2} + 10^{2}} = 10\sqrt{2}$

Hence current in the RC circuit now

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