A Bulb on the Mains

solved ResistorOnAc

Q. A light bulb is rated at 100 W $P = 100$ W for a 220 V $V = 220$ V supply. What is the peak voltage of the source, in volts?

Inputs

power

P

100 W

supply

V

220 V

Outputs

peak

v_{m}

How it's solved

Solved with ResistorOnAc

Given

$P = 100$ W
$V = 220$ V

Find $v_{m}$ .

Working

peak is root two times rms(bulb) $v_{m} = \sqrt{2} \cdot V$
$v_{m} = \sqrt{2} \cdot 220$
$v_{m} = 311.1$ V
$v_{m} \approx 311$ Vto 3 significant figures

Conclusion

A supply is rated by its rms voltage, the steady voltage that heats a resistor as much; the sine wave peaks higher 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 :a_bulb_on_the_mains do4  kind :numerical5  describes "A light bulb is rated at %{power} for a %{supply} supply. %{peak}, in volts?"6  given power: "100 W"7  given supply: "220 V"8  asks peak: "What is the peak voltage of the source"9end

solution.rb

1require_relative "../questions/a_bulb_on_the_mains"2require_relative "../science/alternating_current/resistor_on_ac"34solve :a_bulb_on_the_mains, using: ResistorOnAc do5  objects :bulb67  take power: bulb.power_drawn8  take supply: bulb.rms_voltage910  peak = bulb.peak_voltage11  answer peak:, in: "V", figures: 31213  conclusion "A supply is rated by its rms voltage, the steady voltage that heats a resistor as much; the " \14             "sine wave peaks higher by the square root of two."15end

playground.rb

1require_relative "../solutions/a_bulb_on_the_mains"23play_with :a_bulb_on_the_mains do4  title "A Bulb on the Mains"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 bulb	$v_{m}$	V	—
rms voltage of bulb	$V$	V	0 … ∞
peak current of ground	$i_{m,g}$	A	—
rms current of ground	$I_{g}$	A	0 … ∞
peak current of bulb	$i_{m}$	A	—
rms current of bulb	$I$	A	0 … ∞
power drawn of ground	$P_{g}$	W	—
resistance of ground	$R_{g}$	Ω	0 … ∞
power drawn of bulb	$P$	W	—
resistance of bulb	$R$	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(bulb)

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(bulb)

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

power of a resistor(ground)

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

power of a resistor(bulb)

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

current through a resistor(ground)

V = I \cdot R

current through a resistor(bulb)