Ac Waveform Sum 0F3773

1. **State the problem:** We have two AC waveforms: $$V_1=20\sin(2800\pi t)$$ and $$V_2=170\sin\left(2800\pi t-\frac{\pi}{6}\right)$$ We want to find the resultant waveform $V_3$ which is the sum of $V_1$ and $V_2$. 2. **Formula and approach:** The sum of two sine waves of the same frequency but different amplitudes and phase shifts can be expressed as a single sine wave: $$V_3 = A\sin(\omega t + \phi)$$ where $$A = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos(\theta)}$$ $$\phi = \tan^{-1}\left(\frac{A_2 \sin(\theta)}{A_1 + A_2 \cos(\theta)}\right)$$ Here, $A_1=20$, $A_2=170$, and $\theta = -\frac{\pi}{6}$ (phase difference). 3. **Calculate amplitude $A$:** $$A = \sqrt{20^2 + 170^2 + 2 \times 20 \times 170 \times \cos\left(-\frac{\pi}{6}\right)}$$ Calculate each term: $$20^2 = 400$$ $$170^2 = 28900$$ $$\cos\left(-\frac{\pi}{6}\right) = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2} \approx 0.8660$$ So, $$A = \sqrt{400 + 28900 + 2 \times 20 \times 170 \times 0.8660}$$ Calculate the product: $$2 \times 20 \times 170 = 6800$$ $$6800 \times 0.8660 = 5888.8$$ Sum inside the root: $$400 + 28900 + 5888.8 = 35188.8$$ Therefore, $$A = \sqrt{35188.8} \approx 187.56$$ 4. **Calculate phase $\phi$:** $$\phi = \tan^{-1}\left(\frac{170 \times \sin\left(-\frac{\pi}{6}\right)}{20 + 170 \times \cos\left(-\frac{\pi}{6}\right)}\right)$$ Calculate numerator: $$170 \times \sin\left(-\frac{\pi}{6}\right) = 170 \times -\frac{1}{2} = -85$$ Calculate denominator: $$20 + 170 \times 0.8660 = 20 + 147.22 = 167.22$$ So, $$\phi = \tan^{-1}\left(\frac{-85}{167.22}\right) = \tan^{-1}(-0.5085) \approx -0.47\, \text{radians}$$ 5. **Write the resultant waveform:** $$V_3 = 187.56 \sin(2800\pi t - 0.47)$$ Rounded to 2 decimal places: $$V_3 = 187.56 \sin(2800\pi t - 0.47)$$ **Final answer:** $$\boxed{V_3 = 187.56 \sin(2800\pi t - 0.47)}$$