Power transferred by an alternating current

Section: Alternating Currents | Syllabus: Cambridge A 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 sin ω t:

P = (I_0 sin ω t)^2 × R

Step 3: Expand the squared term.

P = I_0^2 sin^2 ω t × R

P = I_0^2 R sin^2 ω t

Step 4: Identify the maximum (peak) power.

When sin^2 ω t = 1 (i.e., when sin ω t = ± 1):

P_max = I_0^2 R

Step 5: Calculate the mean value of sin^2 ω t.

The average value of sin^2 θ over a complete cycle is (1/2).

(This can be shown graphically: the sin^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 × sin^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 sin^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 Ω 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.

Interactive revision notes, videos and practice questions load below.

Power transferred by an alternating current