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KCET · Physics · Alternating Current

An ideal choke draws a current of \(8 \mathrm{~A}\) when connected to an AC supply of \(100 \mathrm{~V}, 50 \mathrm{~Hz}\). A pure resistor draws a current of \(10 \mathrm{~A}\) when connected to the same source. The ideal choke and the resistor are connected in series and then connected to the AC source of \(150 \mathrm{~V}, 40 \mathrm{~Hz}\). The current in the circuit becomes

  1. A \(\frac{15}{\sqrt{2}} \mathrm{~A}\)
  2. B \(8 \mathrm{~A}\)
  3. C \(18 \mathrm{~A}\)
  4. D \(10 \mathrm{~A}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{15}{\sqrt{2}} \mathrm{~A}\)

Step-by-step Solution

Detailed explanation

Resistance, \(\mathrm{R}=\frac{100}{10}=10 \Omega\)
Inductive reactance, \(\mathrm{X}_{\mathrm{L}}=2 \pi \mathrm{fL}\)
\(\frac{100}{8}=2 \pi \times 50 \times \mathrm{L} \)
\(\Rightarrow \mathrm{L}=\frac{1}{8 \pi} \mathrm{H} \)
\(\mathrm{X}_{\mathrm{L}}^{\prime}=2 \pi \mathrm{f}^{\prime} \mathrm{L}=2 \pi \times 40 \times \frac{1}{8 \pi}=10 \Omega\)
Impedance of the circuit is \(\mathrm{Z}=\sqrt{\mathrm{R}^{2}+\mathrm{X}_{\mathrm{L}}^{\prime 2}}\)
\(=\sqrt{(10)^{2}+(10)^{2}}=10 \sqrt{2} \Omega\)
Current in the circuit is \(\mathrm{i}=\frac{\mathrm{V}}{\mathrm{Z}}=\frac{150}{10 \sqrt{2}}=\frac{15}{\sqrt{2}} \mathrm{~A}\)