An AC Supply's RMS Voltage

Q. The peak voltage of an ac supply is 300 V. What is the rms voltage, in volts? expects 212.1 V

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

The playground

Inputs

peak

v_{m}

300 V

Outputs

rms

V

The solution

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 Ruby

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

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

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

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)