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  • 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

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

Option: 1

\mathrm{10 \mathrm{~A}}


Option: 2

\mathrm{8 \mathrm{~A}}


Option: 3

\mathrm{5 \sqrt{2} \mathrm{~A}}


Option: 4

\mathrm{10 \sqrt{2} \mathrm{~A}}


Answers (1)

best_answer

\begin{aligned} & \mathrm{X}_{\mathrm{L}}=\frac{100}{8}, \mathrm{R}=\frac{100}{10}=10 \Omega ; \mathrm{L} \times 100 \pi=\frac{100}{8} \text { or } \mathrm{L}=\frac{1}{8 \pi} \mathrm{H} \\ & \mathrm{Z}=\sqrt{\left(\frac{1}{8 \pi} \times 2 \pi \times 40\right)^2+10^2}=10 \sqrt{2} \\ & \mathrm{I}=\frac{\mathrm{E}}{\mathrm{Z}}=\frac{100}{10 \sqrt{2}}=\frac{10}{\sqrt{2}}=5 \sqrt{2} \mathrm{~A} \end{aligned}

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