Three alternating voltage sources V_1 = 3 sin | vedclass

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Question

(A) The net voltage $V_{net}$ across the resistor is the sum of the three sources: $V_{net} = V_1 + V_2 + V_3$.Representing these as phasors:$V_1 = 3

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$3$

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(A) The net voltage $V_{net}$ across the resistor is the sum of the three sources: $V_{net} = V_1 + V_2 + V_3$.Representing these as phasors:$V_1 = 3 \angle 0^\circ = 3 + j0$$V_2 = 5 \angle 30^\circ = 5(\cos 30^\circ + j \sin 30^\circ) = 5(\frac{\sqrt{3}}{2} + j \frac{1}{2}) = \frac{5\sqrt{3}}{2} + j 2.5$$V_3 = 5 \angle -127^\circ = 5(\cos(-127^\circ) + j \sin(-127^\circ)) \approx 5(-0.6 - j 0.8) = -3 - j 4$Summing the components:$V_{real} = 3 + \frac{5\sqrt{3}}{2} - 3 = \frac{5\sqrt{3}}{2} \approx 4.33$$V_{imag} = 0 + 2.5 - 4 = -1.5$Peak voltage $V_{max} = \sqrt{V_{real}^2 + V_{imag}^2} = \sqrt{(\frac{5\sqrt{3}}{2})^2 + (-1.5)^2} = \sqrt{\frac{75}{4} + 2.25} = \sqrt{18.75 + 2.25} = \sqrt{21} 	ext{ V}$.Peak current $I_{max} = \frac{V_{max}}{R} = \frac{\sqrt{21}}{\sqrt{7/3}} = \sqrt{\frac{21 	imes 3}{7}} = \sqrt{9} = 3 	ext{ A}$.

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