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GUJCET · Physics · Alternating Current
The peak value of an alternating current is 5 A and its frequency is 60 Hz . Find its rms value and time taken to reach the peak value of current starting from zero ?
- A \(3.536 \mathrm{~A}, 4.167 \mathrm{~ms}\)
- B \(3.536 \mathrm{~A}, 15 \mathrm{~ms}\)
- C \(6.07 \mathrm{~A}, 10 \mathrm{~ms}\)
- D \(2.536 \mathrm{~A}, 4.167 \mathrm{~ms}\)
Answer & Solution
Correct Answer
(A) \(3.536 \mathrm{~A}, 4.167 \mathrm{~ms}\)
Step-by-step Solution
Detailed explanation
(A) \(3.536 \mathrm{~A}, 4.167 \mathrm{~ms}\)
rms value of current
\(
\begin{aligned}
& \mathrm{I}=\frac{i_{m}}{\sqrt{2}} \\
& \mathrm{I}=\frac{5}{1.414} \\
& \mathrm{I}=3.536 \mathrm{~A}
\end{aligned}
\)
Time taken to reach the peak value of current
\(
\begin{aligned}
t & =\frac{\mathrm{T}}{4} \\
t & =\frac{1}{4 v} \\
\therefore t & =\frac{1}{4 \times 60} \\
\therefore t & \approx 4.167 \mathrm{~ms}
\end{aligned}
\)
rms value of current
\(
\begin{aligned}
& \mathrm{I}=\frac{i_{m}}{\sqrt{2}} \\
& \mathrm{I}=\frac{5}{1.414} \\
& \mathrm{I}=3.536 \mathrm{~A}
\end{aligned}
\)
Time taken to reach the peak value of current
\(
\begin{aligned}
t & =\frac{\mathrm{T}}{4} \\
t & =\frac{1}{4 v} \\
\therefore t & =\frac{1}{4 \times 60} \\
\therefore t & \approx 4.167 \mathrm{~ms}
\end{aligned}
\)
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