To determine the d.c. equivalent of the given a.c. current \( I(t) = 2 \sin(\omega t) \), we need to understand the relationship between alternating current (a.c.) and direct current (d.c.) in terms of their effective values. ### Step-by-Step Explanation 1. **Understanding the a.c. Current**: - The given current is \( I(t) = 2 \sin(\omega t) \). This means that the current varies sinusoidally with time, oscillating between +2 A and -2 A. - The amplitude of this sine wave is 2 A. 2. **Calculating the Root Mean Square (RMS) Value**: - For a sinusoidal current, the d.c. equivalent is often represented by the RMS value. The RMS value of a sinusoidal current is calculated using the formula: \[ I_{\text{RMS}} = \frac{I_0}{\sqrt{2}} \] where \( I_0 \) is the peak (or maximum) value of the current. 3. **Applying the Formula**: - In our case, the peak value \( I_0 \) is 2 A. Therefore, we can substitute this value into the RMS formula: \[ I_{\text{RMS}} = \frac{2 \, \text{A}}{\sqrt{2}} = \sqrt{2} \, \text{A} \] 4. **Conclusion**: - The d.c. equivalent of the a.c. current \( 2 \sin(\omega t) \) is \( \sqrt{2} \, \text{A} \). ### Why the Other Options are Incorrect - **Option A: \(\frac{1A}{\sqrt{2}}\)**: - This value is incorrect because it does not represent the RMS value of the given a.c. current. It is actually the RMS value of a current with a peak value of 1 A, which is not relevant here. - **Option C: 2 A**: - This option represents the peak value of the current, not the RMS value. The d.c. equivalent is not simply the peak value; it must be adjusted to account for the nature of the a.c. waveform. - **Option D: 2\(\sqrt{2}\) A**: - This value is incorrect as it suggests a misunderstanding of the RMS calculation. It would imply a peak value of \( 2\sqrt{2} \) A, which is not the case here. ### Summary of Key Points - The d.c. equivalent of an a.c. current is typically represented by its RMS value. - For a sinusoidal current, the RMS value is calculated as \( I_{\text{RMS}} = \frac{I_0}{\sqrt{2}} \). - In this case, the peak current is 2 A, leading to a d.c. equivalent of \( \sqrt{2} \, \text{A} \). - Understanding the difference between peak current and RMS current is crucial for solving problems related to a.c. circuits. ### Final Answer **Correct Option: B. \(\sqrt{2}\) A**