11.2.2 Root-Mean-Square Current & Voltage

• Direct current sources provide a constant voltage and current over time, making it easy to measure
• In situations involving alternating voltage and current, the average values of voltage and current will always be zero
  • This can make it difficult to measure

The mean value for alternating current and voltage is always zero

• The use of root mean square values gets around this problem
  • First remove all the negative signs by simply squaring the peak current, or voltage
  • Find the average of the squared value
  • And finally, take the square root

• Root-mean-square (rms) values of current, or voltage, are a useful way of comparing a.c current, or voltage, to its equivalent direct current (d.c), or voltage
  • The rms values represent the direct current, or voltage, values that will produce the same heating effect, or power dissipation, as the alternating current, or voltage

• The rms value of an alternating current is defined as:

The square root of the mean of squares of all the values of the current in one cycle

• An alternate definition is:

The equivalent direct current that produces the same power

• The rms current I_rms is defined by the equation:

• Where:

• • I₀ = peak current (A)

• The rms value of an alternating voltage is defined as:

The square root of the mean of squares of all the values of the voltage in one cycle

• An alternate definition is:

The equivalent dc voltage that produces the same power

• The rms voltage V_rms is defined by the equation:

• Where:
  • V₀ = peak voltage (V)

• Rms current is equal to 0.707 × I₀, which is about 70% of the peak current I₀
  • This is also the case for rms voltage

• The rms value is therefore defined as:

The steady direct current, or voltage, that delivers the same average power in a resistor as the alternating current, or voltage

• A resistive load is any electrical component with resistance eg. a lamp

V_rms and peak voltage. The rms voltage is about 70% of the peak voltage

Worked Example

Step 1: Write down the V_rms equation 

Step 2: Rearrange for the peak voltage, V₀

V₀ = √2 × V_rms

Step 3: Substitute in the values

V₀ = √2 × 230

Step 4: Calculate the peak-to-peak voltage

• • The peak-to-peak voltage is the peak voltage (V₀) × 2
  • Peak-to-peak voltage = (√2 × 230) × 2 = 650.538 = 651 V (3 s.f)

Exam Tip

• The average power of a supply is the product of the rms current and voltage:

Average power = I_rms × V_rms

Worked Example

Step 1: Write down the V_rms equation

Step 2: Substitute in the values and calculate V_rms

= 230 V

Step 3: Write down the equation for average power

Average power = I_rms × V_rms

Step 4: Rearrange the equation for I_rms

Step 5: Substitute in the values and calculate I_rms

= 10 A

Step 6: Write down the equation for I₀

Step 7: Rearrange for I₀ and substitute in the values

= 14.1 A