A Bulb on the Mains

Q. A light bulb is rated at 100 W for a 220 V supply. What is the peak voltage of the source, in volts? expects 311 V

What the draft establishes

Fills in once a draft is checked.

The web of concepts

given read from the words equation asked

The equations it used

None yet.

Check an answer

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?

The playground

Inputs

power

P

100 W

supply

V

220 V

Outputs

peak

v_{m}

The solution

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 Ruby

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

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

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

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)