A Resistor's Power over a Cycle

solved ResistorOnAc

Q. A 100 Ω $R = 100$ Ω resistor is connected to a 220 V $V = 220$ V, 50 Hz ac supply. What is the net power consumed over a full cycle, in watts?

Inputs

resistance

R

100 Ω

supply

V

220 V

Outputs

power

P

How it's solved

Solved with ResistorOnAc

Given

$R = 100$ Ω
$V = 220$ V

Find $P$ .

Working

power of a resistor(resistor) $P = \frac{V^{2}}{R}$
$P = \frac{220^{2}}{100}$
$P = 484$ W
$P \approx 484$ Wto 3 significant figures

Conclusion

The rms voltage is the steady voltage that would heat the resistor as much, so the average power over a cycle is its square over the resistance.

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_resistor_s_power_over_a_cycle do4  kind :numerical5  describes "A %{resistance} resistor is connected to a %{supply}, 50 Hz ac supply. %{power}, in watts?"6  given resistance: "100 Ω"7  given supply: "220 V"8  asks power: "What is the net power consumed over a full cycle"9end

solution.rb

1require_relative "../questions/a_resistor_s_power_over_a_cycle"2require_relative "../science/alternating_current/resistor_on_ac"34solve :a_resistor_s_power_over_a_cycle, using: ResistorOnAc do5  objects :resistor67  take resistance: resistor.resistance8  take supply: resistor.rms_voltage910  power = resistor.power_drawn11  answer power:, in: "W", figures: 31213  conclusion "The rms voltage is the steady voltage that would heat the resistor as much, so the average " \14             "power over a cycle is its square over the resistance."15end

playground.rb

1require_relative "../solutions/a_resistor_s_power_over_a_cycle"23play_with :a_resistor_s_power_over_a_cycle do4  title "A Resistor's Power over a Cycle"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 resistor	$v_{m}$	V	—
rms voltage of resistor	$V$	V	0 … ∞
peak current of ground	$i_{m,g}$	A	—
rms current of ground	$I_{g}$	A	0 … ∞
peak current of resistor	$i_{m}$	A	—
rms current of resistor	$I$	A	0 … ∞
power drawn of ground	$P_{g}$	W	—
resistance of ground	$R_{g}$	Ω	0 … ∞
power drawn of resistor	$P$	W	—
resistance of resistor	$R$	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(resistor)

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(resistor)

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

power of a resistor(ground)

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

power of a resistor(resistor)

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

current through a resistor(ground)

V = I \cdot R

current through a resistor(resistor)