An AC Supply's RMS Voltage

solved ResistorOnAc

Q. The peak voltage of an ac supply is 300 V $v_{m} = 300$ V. What is the rms voltage, in volts?

Inputs

peak

v_{m}

300 V

Outputs

rms

V

How it's solved

Solved with ResistorOnAc

Given

$v_{m} = 300$ V

Find $V$ .

Working

peak is root two times rms(supply) $v_{m} = \sqrt{2} \cdot V$
$300 = \sqrt{2} \cdot V$
$V = 212.1$ V
$V \approx 212.1$ Vto 4 significant figures

Conclusion

A sine wave's rms value is its peak shrunk by the square root of two.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :an_ac_supply_s_rms_voltage do4  kind :numerical5  describes "The peak voltage of an ac supply is %{peak}. %{rms}, in volts?"6  given peak: "300 V"7  asks rms: "What is the rms voltage"8end

solution.rb

1require_relative "../questions/an_ac_supply_s_rms_voltage"2require_relative "../science/alternating_current/resistor_on_ac"34solve :an_ac_supply_s_rms_voltage, using: ResistorOnAc do5  objects :supply67  take peak: supply.peak_voltage89  rms = supply.rms_voltage10  answer rms:, in: "V", figures: 41112  conclusion "A sine wave's rms value is its peak shrunk by the square root of two."13end

playground.rb

1require_relative "../solutions/an_ac_supply_s_rms_voltage"23play_with :an_ac_supply_s_rms_voltage do4  title "An AC Supply's RMS Voltage"5end

The laws it drew on

quantity	written	units	within	from
rms voltage	$V$	V	0 … ∞	PeakAndRms
peak voltage	$v_{m}$	V	—	PeakAndRms
rms current	$I$	A	0 … ∞	PeakAndRms
peak current	$i_{m}$	A	—	PeakAndRms
resistance	$R$	Ω	0 … ∞	ResistorOnAc
power drawn	$P$	W	—	ResistorOnAc
peak voltage of ground	$v_{m,g}$	V	—
rms voltage of ground	$V_{g}$	V	0 … ∞
peak voltage of supply	$v_{m}$	V	—
rms voltage of supply	$V$	V	0 … ∞
peak current of ground	$i_{m,g}$	A	—
rms current of ground	$I_{g}$	A	0 … ∞
peak current of supply	$i_{m}$	A	—
rms current of supply	$I$	A	0 … ∞
power drawn of ground	$P_{g}$	W	—
resistance of ground	$R_{g}$	Ω	0 … ∞
power drawn of supply	$P$	W	—
resistance of supply	$R$	Ω	0 … ∞

v_{m,c} = \sqrt{2} \cdot V_{c}

peak is root two times rms

i_{m,c} = \sqrt{2} \cdot I_{c}

peak current is root two times rms

P_{l} = \frac{{V_{l}}^{2}}{R_{l}}

power of a resistor

V_{l} = I_{l} \cdot R_{l}

current through a resistor

v_{m,g} = \sqrt{2} \cdot V_{g}

peak is root two times rms(ground)

v_{m} = \sqrt{2} \cdot V

peak is root two times rms(supply)

i_{m,g} = \sqrt{2} \cdot I_{g}

peak current is root two times rms(ground)

i_{m} = \sqrt{2} \cdot I

peak current is root two times rms(supply)

P_{g} = \frac{{V_{g}}^{2}}{R_{g}}

power of a resistor(ground)

P = \frac{V^{2}}{R}

power of a resistor(supply)

V_{g} = I_{g} \cdot R_{g}

current through a resistor(ground)

V = I \cdot R

current through a resistor(supply)