Calculating r.m.s. Values | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Calculating r.m.s. Values

The shortcut depends on wave shape. Use √2 only for a sinusoidal current or voltage.

For a sine wave, use the √2 relations

I_{\mathrm{rms}}=\dfrac{I_0}{\sqrt{2}}

Symbols

$I_{\mathrm{rms}}$ = root-mean-square current (A)
$I_0$ = peak current (A)

V_{\mathrm{rms}}=\dfrac{V_0}{\sqrt{2}}

Peak to r.m.s.: divide by √2.
r.m.s. to peak: multiply by √2.
A peak current of 3.5 A has an r.m.s. value of 2.5 A (9702/42/O/N/24 Q8(b)(iii)).

Common mistake

A half-wave output is not a complete sine wave. Its r.m.s. value does not follow the ordinary sine-wave √2 shortcut.

For any waveform: square, mean, square root

x_{\mathrm{rms}}=\sqrt{\operatorname{mean}(x^2)}

Square each value. The sign no longer matters.
Find the mean of the squared values over one cycle.
Take the square root of that mean.

Worked example

A non-sinusoidal waveform

A current has magnitude $2I_0$ for half a cycle and magnitude $I_0$ for the other half.

\operatorname{mean}(I^2)=\dfrac{4I_0^2+I_0^2}{2}=\dfrac{5}{2}I_0^2

Answer

I_{\mathrm{rms}}^2=\dfrac{5}{2}I_0^2

I_{\mathrm{rms}}=1.58I_0

This first-principles method was required in 2023 (9702/41/O/N/23 Q7).

Your turnfirst-principles check

During one cycle, a current has magnitude 6.0 A for one quarter of the time and magnitude 2.0 A for the remaining three quarters. Calculate its r.m.s. value.

Show worked answer

I_{\mathrm{rms}}^2=\dfrac{36+12}{4}

I_{\mathrm{rms}}^2=12\ \mathrm{A^2}

Answer

I_{\mathrm{rms}}=\sqrt{12}=3.5\ \mathrm{A}

Calculating r.m.s. Values

The shortcut depends on wave shape. Use √2 only for a sinusoidal current or voltage.

For a sine wave, use the √2 relations

I_{\mathrm{rms}}=\dfrac{I_0}{\sqrt{2}}

Symbols

$I_{\mathrm{rms}}$ = root-mean-square current (A)
$I_0$ = peak current (A)

V_{\mathrm{rms}}=\dfrac{V_0}{\sqrt{2}}

Peak to r.m.s.: divide by √2.
r.m.s. to peak: multiply by √2.
A peak current of 3.5 A has an r.m.s. value of 2.5 A (9702/42/O/N/24 Q8(b)(iii)).

Common mistake

A half-wave output is not a complete sine wave. Its r.m.s. value does not follow the ordinary sine-wave √2 shortcut.

For any waveform: square, mean, square root

x_{\mathrm{rms}}=\sqrt{\operatorname{mean}(x^2)}

Square each value. The sign no longer matters.
Find the mean of the squared values over one cycle.
Take the square root of that mean.

Worked example

A non-sinusoidal waveform

A current has magnitude $2I_0$ for half a cycle and magnitude $I_0$ for the other half.

\operatorname{mean}(I^2)=\dfrac{4I_0^2+I_0^2}{2}=\dfrac{5}{2}I_0^2

Answer

I_{\mathrm{rms}}^2=\dfrac{5}{2}I_0^2

I_{\mathrm{rms}}=1.58I_0

This first-principles method was required in 2023 (9702/41/O/N/23 Q7).

Your turnfirst-principles check

During one cycle, a current has magnitude 6.0 A for one quarter of the time and magnitude 2.0 A for the remaining three quarters. Calculate its r.m.s. value.

Show worked answer

I_{\mathrm{rms}}^2=\dfrac{36+12}{4}

I_{\mathrm{rms}}^2=12\ \mathrm{A^2}

Answer

I_{\mathrm{rms}}=\sqrt{12}=3.5\ \mathrm{A}