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Step-by-Step Solution
Step 1: Express the Current in Terms of Time
Given the applied voltage:
$ v(t) = 220\,\sin(100\,\pi\,t) \text{ volt} $.
For a purely resistive circuit of resistance $ R = 50\,\Omega $,
the current is
$ I(t) = \dfrac{v(t)}{R} = \dfrac{220}{50}\,\sin(100\,\pi\,t). $
Hence,
$ I_{\max} = \dfrac{220}{50} \,\text{A} .$
Step 2: Determine the Time Period
The angular frequency is $ \omega = 100\,\pi\,\text{rad/s} $.
The time period $ T $ is given by
$ T = \dfrac{2\,\pi}{\omega} = \dfrac{2\,\pi}{100\,\pi} = \dfrac{1}{50}\,\text{s}. $
Step 3: Find the Time at Which Current is at Maximum
For a sine wave, the current reaches its first maximum at one-fourth of the time period. Thus,
$ t_1 = \dfrac{T}{4} = \dfrac{1}{50} \times \dfrac{1}{4} = \dfrac{1}{200}\,\text{s}. $
Step 4: Find the Time at Which Current is Half of Its Maximum Value
We want $ I(t_2) = \dfrac{I_{\max}}{2} $. Using
$ I(t) = I_{\max}\sin(100\,\pi\,t), $
we set
$ \sin(100\,\pi\,t_2) = \dfrac{1}{2}. $
One suitable angle for this is
$ 100\,\pi\,t_2 = \dfrac{5\,\pi}{6}, $
which gives
$ t_2 = \dfrac{5\,\pi}{6} \times \dfrac{1}{100\,\pi} = \dfrac{5}{600} = \dfrac{1}{120}\,\text{s}. $
Step 5: Calculate the Time Difference
The time taken for the current to rise from half of the peak value to the peak value is
$ \Delta t = t_2 - t_1 = \dfrac{1}{120} - \dfrac{1}{200}. $
Finding a common denominator:
$ \dfrac{1}{120} = \dfrac{5}{600}, \quad \dfrac{1}{200} = \dfrac{3}{600}. $
Hence,
$ \Delta t = \bigl(\dfrac{5}{600} - \dfrac{3}{600}\bigr)\,\text{s} = \dfrac{2}{600}\,\text{s} = \dfrac{1}{300}\,\text{s}. $
In milliseconds,
$ \dfrac{1}{300}\,\text{s} = 3.3 \,\text{ms}. $
Final Answer
The time taken is 3.3 ms.
