An AC Circuit's Peak Current

Q. The rms value of current in an ac circuit is 10 A. What is the peak current, in amperes? expects 14.1 A

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The web of concepts

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

The playground

Inputs

rms

I

10 A

Outputs

peak

i_{m}

The solution

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 Ruby

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

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

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

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)