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A LevelPhysicsCIERevision Notes21. Alternating Currents21.1 Characteristics of Alternating CurrentsRoot-Mean-Square Current & Voltage

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Root-Mean-Square Current & Voltage (CIE A Level Physics)

Revision Note

Katie M

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Katie M

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Root-mean-square current & voltage

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

The value of a constant current that produces the same power in a resistor as the alternating current

  • The r.m.s current Ir.m.s is defined by the equation:

I subscript r. m. s end subscript space equals space fraction numerator space I subscript 0 over denominator square root of 2 end fraction

  • The r.m.s value of an alternating voltage is defined as:

The value of a constant voltage that produces the same power in a resistor as the alternating voltage

  • The r.m.s voltage Vr.m.s is defined by the equation:

V subscript r. m. s end subscript space equals space fraction numerator space V subscript 0 over denominator square root of 2 end fraction

  • Where:
    • I0 = peak current (A)
    • V0 = peak voltage (V)
  • So, r.m.s current is equal to 0.707 × I0, which is about 70% of the peak current I0
  • The r.m.s 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 an electrical component with resistance e.g. a lamp

Peak voltage and RMS voltage

RMS v Peak grap, downloadable AS & A Level Physics revision notes

Vr.m.s and peak voltage. The r.m.s voltage is about 70% of the peak voltage

Worked example

An alternating current is I is represented by the equation

I space equals space 410 space sin open parentheses 100 straight pi t close parentheses

where I is measured in amps and t is in seconds.

For this alternating current, determine

(a)
the r.m.s current
(b)
the frequency.
 

Answer:

(a)

Step 1: Write out the equation for r.m.s current

I subscript r. m. s end subscript space equals space fraction numerator I subscript 0 space over denominator square root of 2 end fraction

Step 2: Determine the peak voltage I0

  • The alternating current equation is in the form: I space equals space I subscript 0 sin open parentheses omega t close parentheses
  • Comparing this to I space equals space bold 410 sin open parentheses 100 straight pi t close parenthesesmeans the peak current is I0 = 410 A

Step 3: Substitute into the Ir.m.s equation

I subscript r. m. s end subscript space equals space fraction numerator 410 over denominator square root of 2 end fraction space equals space 290 space straight A

(b)

Step 1: Write out the equation for angular frequency

omega space equals space 2 straight pi f

Step 2: Determine the angular frequency

  • The alternating current equation is in the form: I space equals space I subscript 0 sin open parentheses omega t close parentheses
  • Comparing this to I space equals space 410 sin open parentheses bold 100 bold pi t close parenthesesmeans the angular frequency is ω = 100π rad s−1

Step 3: Rearrange and substitute into the equation to calculate frequency

f space equals space fraction numerator omega over denominator 2 straight pi end fraction space equals space fraction numerator 100 straight pi over denominator 2 straight pi end fraction space equals space 50 space Hz

Examiner Tip

These equations are not given on your data sheet, so make sure you learn them!

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    Katie M

    Author: Katie M

    Expertise: Physics

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.