e) If the maximum current \( (100 \mathrm{~mA}) \) is given by the formula \( I=I_{\text {max }} \sin \omega t \), calculate the amount of current generated after; (i) 0.0025 seconds \( = \) (ii) 0.0125 seconds \( = \)
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To calculate the current generated after specific time intervals using the formula \( I = I_{\text{max}} \sin(\omega t) \), we first need to understand the sine function's behavior. Assuming \( \omega = 2\pi f \) where \( f = 60 \text{Hz} \) (a common frequency for AC circuits), we get \( \omega = 120\pi \ \text{rad/s} \). Now, substituting \( I_{\text{max}} = 100 \text{mA} \): (i) For \( t = 0.0025 \text{s} \): \[ I = 0.1 \sin(120\pi \times 0.0025) = 0.1 \sin(0.9425) \approx 0.1 \times 0.806 = 0.0806 \text{A} \text{ or } 80.6 \text{mA} \] (ii) For \( t = 0.0125 \text{s} \): \[ I = 0.1 \sin(120\pi \times 0.0125) = 0.1 \sin(4.7125) \approx 0.1 \times (-1) = -0.1 \text{A} \text{ or } -100 \text{mA} \] (Note: The negative sign here indicates the current direction in an AC circuit at that specific time is reversed.)