MHT CET · Physics · Alternating Current
An alternating current of frequency \(50 \mathrm{~Hz}\) has the peak value as \(14 \cdot 14 \mathrm{~A}\). The time
taken by the alternating current in reaching from zero to maximum value and r.m.s. value of current will be respectively
- A \(0.025 \mathrm{~s}, 5 \mathrm{~A}\)
- B \(0 \cdot 005 \mathrm{~s}, 5 \mathrm{~A}\)
- C \(0 \cdot 005 \mathrm{~s}, 10 \mathrm{~A}\)
- D \(0 \cdot 025 \mathrm{~s}, 10 \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(0 \cdot 005 \mathrm{~s}, 10 \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
(D)
\(\mathrm{f}=50 \mathrm{~Hz}, \mathrm{~T}=\frac{1}{50} \mathrm{~s}=0.02 \mathrm{~s}\)
Time taken to reach from zero to maximum value is
\(\begin{array}{l}
t=\frac{T}{4}=\frac{0.02}{4}=0.005 \mathrm{~s} \\
I_{m s}=\frac{I_{0}}{\sqrt{2}}=\frac{14.14}{1.414}=10 \mathrm{~A}
\end{array}\)
\(\mathrm{f}=50 \mathrm{~Hz}, \mathrm{~T}=\frac{1}{50} \mathrm{~s}=0.02 \mathrm{~s}\)
Time taken to reach from zero to maximum value is
\(\begin{array}{l}
t=\frac{T}{4}=\frac{0.02}{4}=0.005 \mathrm{~s} \\
I_{m s}=\frac{I_{0}}{\sqrt{2}}=\frac{14.14}{1.414}=10 \mathrm{~A}
\end{array}\)
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