A Resistor's Power over a Cycle

Q. A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply. What is the net power consumed over a full cycle, in watts? expects 484 W

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The web of concepts

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The equations it used

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Check an answer

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?

The playground

Inputs

resistance

R

100 Ω

supply

V

220 V

Outputs

power

P

The solution

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 Ruby

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

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

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

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)