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Power from Alternating Current | |||||||
Energy, (W), is needed to move a charge, (Q), through a potential difference, (V).
W = Q×V Electric current, (I), is the charge, (Q), flowing per unit time, (t).=> Q = I×t Combining these equations together gives=> W = I×t×V Power, (P) is the energy, (W), per unit time, (t)=> P = I×V For a resistorOhm's Law gives V = I×R=> P = I×V = I2×R = V2/R Introduction to the sine wave formulaFor a sine wave alternating voltage, V = V0sin(ωt), and the power dissipated in a resistor R over one time period, (T) is Using the cos(2θ) formula, this can be rearranged to give Substituting and simplifying using T = 2π/ω gives Averaging the power over one time period (dividing by 2π/ω) gives that the mean power is Examining this formula shows that the quantity V0/√2 is the steady (direct) voltage that would dissipate the same power. This quantity is known as the root mean square (rms) voltage. For a capacitorIntroduction to capacitors in AC circuitsUsing P = I × V and summing over one time period Using the sin(2θ) formula, this can be rearranged to giveTherefore the mean power dissipated by an ideal capacitor over one time period is zero. |