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GUJCET · Physics · Alternating Current
A pure inductor of 50.0 mH is connected to a source of 220 V . Then the rms current in the circuit will be __________ . The frequency of the source is 50 Hz .
- A 28 A
- B 7 A
- C 14 A
- D 21 A
Answer & Solution
Correct Answer
(C) 14 A
Step-by-step Solution
Detailed explanation
(C) 14 A
\(\mathrm{V}=220 \mathrm{~V}\)
\(\mathrm{L}=50 \mathrm{mH}\)
\(v=50 \mathrm{~Hz}\)
\(1=\frac{V}{Z}\)
\(1=\frac{V}{\omega L}\)
\(I=\frac{V}{2 \pi \nu L}\)
\(I=\frac{220}{2 \times 3.14 \times 50 \times 50 \times 10^{-3}}\)
\(I=14 \mathrm{~A}\)
\(\mathrm{V}=220 \mathrm{~V}\)
\(\mathrm{L}=50 \mathrm{mH}\)
\(v=50 \mathrm{~Hz}\)
\(1=\frac{V}{Z}\)
\(1=\frac{V}{\omega L}\)
\(I=\frac{V}{2 \pi \nu L}\)
\(I=\frac{220}{2 \times 3.14 \times 50 \times 50 \times 10^{-3}}\)
\(I=14 \mathrm{~A}\)
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