An AC Circuit's Peak Current

solved ResistorOnAc

Q. The rms value of current in an ac circuit is 10 A $I = 10$ A. What is the peak current, in amperes?

Inputs

rms

I

10 A

Outputs

peak

i_{m}

How it's solved

Solved with ResistorOnAc

Given

$I = 10$ A

Find $i_{m}$ .

Working

peak current is root two times rms(circuit) $i_{m} = \sqrt{2} \cdot I$
$i_{m} = \sqrt{2} \cdot 10$
$i_{m} = 14.14$ A
$i_{m} \approx 14.1$ Ato 3 significant figures

Conclusion

A sinusoidal current swings higher than its rms value 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 :an_ac_circuit_s_peak_current do4  kind :numerical5  describes "The rms value of current in an ac circuit is %{rms}. %{peak}, in amperes?"6  given rms: "10 A"7  asks peak: "What is the peak current"8end

solution.rb

1require_relative "../questions/an_ac_circuit_s_peak_current"2require_relative "../science/alternating_current/resistor_on_ac"34solve :an_ac_circuit_s_peak_current, using: ResistorOnAc do5  objects :circuit67  take rms: circuit.rms_current89  peak = circuit.peak_current10  answer peak:, in: "A", figures: 31112  conclusion "A sinusoidal current swings higher than its rms value by the square root of two."13end

playground.rb

1require_relative "../solutions/an_ac_circuit_s_peak_current"23play_with :an_ac_circuit_s_peak_current do4  title "An AC Circuit's Peak Current"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 circuit	$v_{m}$	V	—
rms voltage of circuit	$V$	V	0 … ∞
peak current of ground	$i_{m,g}$	A	—
rms current of ground	$I_{g}$	A	0 … ∞
peak current of circuit	$i_{m}$	A	—
rms current of circuit	$I$	A	0 … ∞
power drawn of ground	$P_{g}$	W	—
resistance of ground	$R_{g}$	Ω	0 … ∞
power drawn of circuit	$P$	W	—
resistance of circuit	$R$	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(circuit)

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(circuit)

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

power of a resistor(ground)

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

power of a resistor(circuit)

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

current through a resistor(ground)

V = I \cdot R

current through a resistor(circuit)