Power transferred by an alternating current
Section: Alternating Currents | Syllabus: Cambridge AS Level Physics 9702
Power transferred by an alternating current Derivation of Mean Power Key Concept Mean Current: The average value of a sinusoidal current is zero because the positive half-cycle cancels the negative half-cycle.
Mean Power: Since P I^2, power is always positive (heating occurs regardless of current direction). The mean power is not zero . Step-by-Step Derivation Step 1: Write the instantaneous power equation .
The power dissipated in a resistor R at any instant is: P = I^2 R Step 2: Substitute the expression for instantaneous alternating current . Since I = I_0 ω t: P = (I_0 ω t)^2 × R Step 3: Expand the squared term.
P = I_0^2 ^2 ω t × R P = I_0^2 R ^2 ω t Step 4: Identify the maximum (peak) power . When ^2 ω t = 1 (i.e., when ω t = ± 1): P_max = I_0^2 R Step 5: Calculate the mean value of ^2 ω t . The average value of ^2 θ over a complete cycle is 1/2.
(This can be shown graphically: the ^2 curve oscillates symmetrically between 0 and 1, so its mean is 0.5) Step 6: Calculate the mean power . P_mean = I_0^2 R × ^2 ω t P_mean = I_0^2 R × 1/2 P_mean = 1/2 I_0^2 R = 1/2 P_max The mean power in a resistive load is half the maximum power for a sinusoidal alternating current.
Diagram: Power vs Time Graph Show Power varying as a ^2 function (always positive, oscillating between 0 and P_max). Mark the Peak Power (I_0^2 R) and the Mean Power (1/2 I_0^2 R) as a horizontal dashed line.
Root-Mean-Square (r.m.s.) Values The r.m.s. value is defined as: Examiner Definition The r.m.s. value of an alternating current is the value of a steady direct current (d.c.) that produces the same heating effect (mean power) in a given resistor.
Derivation of r.m.s. Current Formula We derive I_rms by equating the mean power of a.c. to the power of an equivalent d.c. Step 1: Write the expression for mean power of a.c. (derived above): P_mean = 1/2 I_0^2 R Step 2: Write the expression for power of a steady d.c.
with current I_rms: P_dc = I_rms^2 R Step 3: By definition, r.m.s. current produces the same heating effect , so equate the two expressions: I_rms^2 R = 1/2 I_0^2 R Step 4: Cancel the resistance R from both sides: I_rms^2 = 1/2 I_0^2 Step 5: Take the square root of both sides: I_rms = √1/2 × I_0 = I_0/√2 Similarly, for voltage: I_rms = I_0/√2 0.707 I_0 V_rms = V_0/√2 0.707 V_0 Using r.m.s.
values, the power equations become identical to the d.c. equations: - P_mean = I_rms^2 R - P_mean = V_rms I_rms - P_mean = V_rms^2R Worked Examples Example 1 Question: An a.c. supply has a peak voltage of 170 V.
Calculate the r.m.s. voltage. Answer: V_rms = 170 / √2 = 120 V. Example 2 Question: A heater has a resistance of 50 and is connected to a 240 V a.c. supply (r.m.s.). Calculate the correct peak power and mean power.
Answer: Mean Power: P = V_rms^2 / R = 240^2 / 50 = 1152 W. Peak Power: P_0 = V_0^2 / R = (240√2)^2 / 50 = (2 × 240^2) / 50 = 2304 W (Double the mean power). Diagram: Sinusoidal Waveform Graph of Potential Difference (V) against Time (t).
Label: Peak Value (V_0), Peak-to-Peak Value (2V_0), Period (T), and one complete cycle. Common Mistake Students often use Peak Voltage (V_0) to calculate Mean Power. Remember: Mean Power uses r.m.s., Peak Power uses Peak values.
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