Root-Mean-Square Current & Voltage (Cambridge (CIE) A Level Physics): Revision Note

Exam code: 9702

Katie M

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on

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:

Ir.m.s =  I02

  • 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:

Vr.m.s =  V02

  • 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 = 410 sin(100πt)

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

Ir.m.s = I0 2

Step 2: Determine the peak voltage I0

  • The alternating current equation is in the form: I = I0sin(ωt)

  • Comparing this to I = 410sin(100πt)means the peak current is I0 = 410 A

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

Ir.m.s = 4102 = 290 A

(b)

Step 1: Write out the equation for angular frequency

ω = 2πf

Step 2: Determine the angular frequency

  • The alternating current equation is in the form: I = I0sin(ωt)

  • Comparing this to I = 410sin(100πt)means the angular frequency is ω = 100π rad s−1

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

f = ω2π = 100π2π = 50 Hz

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Katie M

Author: Katie M

Expertise: Curriculum Expert

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.

Caroline Carroll

Reviewer: Caroline Carroll

Expertise: Head of Content Delivery

Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about delivering high-quality resources to help students achieve their full potential.