A Car Battery Shorted

solved OhmsLaw

Q. The storage battery of a car has an emf of 12 V $V = 12$ V. If the internal resistance of the battery is 0.4 Ω $R = 0.4$ Ω, what is the maximum current that can be drawn from the battery, in amperes?

Inputs

emf

V

12 V

internal

R

0.4 Ω

Outputs

current

I

How it's solved

Solved with OhmsLaw

Given

$V = 12$ V
$R = 0.4$ Ω

Find $I$ .

Working

ohms law(battery) $V = I \cdot R$
$12 = I \cdot 0.4$
$I = 30$ A
$I \approx 30$ Ato 2 significant figures

Conclusion

The most current flows when nothing outside the battery resists it, so the whole emf drives the current through the battery's own internal 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_car_battery_shorted do4  kind :numerical5  describes "The storage battery of a car has an emf of %{emf}. If the internal resistance of the battery is " \6            "%{internal}, %{current}, in amperes?"7  given emf: "12 V"8  given internal: "0.4 Ω"9  asks current: "what is the maximum current that can be drawn from the battery"10end

solution.rb

1require_relative "../questions/a_car_battery_shorted"2require_relative "../science/current_electricity/ohms_law"34solve :a_car_battery_shorted, using: OhmsLaw do5  objects :battery67  take emf: battery.voltage8  take internal: battery.resistance910  current = battery.current11  answer current:, in: "A", figures: 21213  conclusion "The most current flows when nothing outside the battery resists it, so the whole emf drives the " \14             "current through the battery's own internal resistance."15end

playground.rb

1require_relative "../solutions/a_car_battery_shorted"23play_with :a_car_battery_shorted do4  title "A Car Battery Shorted"5end

The laws it drew on

quantity	written	units	within	from
voltage	$V$	V	—	OhmsLaw
current	$I$	A	—	OhmsLaw
resistance	$R$	Ω	0 … ∞	OhmsLaw
length of the conductor	$l$	m	—	OhmsLaw
cross section	$A$	m²	—	OhmsLaw
resistivity	$ρ$	Ω m	—	OhmsLaw
voltage of ground	$V_{g}$	V	—
current of ground	$I_{g}$	A	—
resistance of ground	$R_{g}$	Ω	0 … ∞
voltage of battery	$V$	V	—
current of battery	$I$	A	—
resistance of battery	$R$	Ω	0 … ∞
resistivity of ground	$ρ_{g}$	Ω m	—
length of the conductor of ground	$l_{g}$	m	—
cross section of ground	$A_{g}$	m²	—
resistivity of battery	$ρ$	Ω m	—
length of the conductor of battery	$l$	m	—
cross section of battery	$A$	m²	—

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

ohms law

R_{c} = \frac{ρ_{c} \cdot l_{c}}{A_{c}}

resistance from resistivity

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

ohms law(ground)

V = I \cdot R

ohms law(battery)

R_{g} = \frac{ρ_{g} \cdot l_{g}}{A_{g}}

resistance from resistivity(ground)

R = \frac{ρ \cdot l}{A}

resistance from resistivity(battery)