The Resistor on a Ten Volt Battery

solved OhmsLaw

Q. A battery of emf 10 V $V_{c} = 10$ V and internal resistance 3 Ω $R_{b} = 3$ Ω is connected to a resistor $R_{c} = R_{b} + R_{r}$ . If the current in the circuit is 0.5 A $I_{c} = 0.5$ A, what is the resistance of the resistor, in ohms?

Inputs

emf

V_{c}

10 V

internal

R_{b}

3 Ω

current

I_{c}

0.5 A

Outputs

resistance

R_{r}

How it's solved

Solved with OhmsLaw

Given

$V_{c} = 10$ V
$R_{b} = 3$ Ω
$I_{c} = 0.5$ A
$R_{c} = R_{b} + R_{r}$ from “connected to a resistor”

Find $R_{r}$ .

Working

ohms law(circuit) $V_{c} = I_{c} \cdot R_{c}$
$10 = 0.5 \cdot R_{c}$
$R_{c} = 20$ Ω
connected to a resistor $R_{c} = R_{b} + R_{r}$
$20 = 3 + R_{r}$
$R_{r} = 17$ Ω
$R_{r} \approx 17$ Ωto 2 significant figures

Conclusion

The emf drives the current through the resistor and the battery's own resistance in a row, so the two resistances together are the emf over the current.

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 :the_resistor_on_a_ten_volt_battery do4  kind :numerical5  describes "A battery of emf %{emf} and internal resistance %{internal} is connected to a resistor. If the current " \6            "in the circuit is %{current}, %{resistance}, in ohms?"7  given emf: "10 V"8  given internal: "3 Ω"9  given current: "0.5 A"10  asks resistance: "what is the resistance of the resistor"11end

solution.rb

1require_relative "../questions/the_resistor_on_a_ten_volt_battery"2require_relative "../science/current_electricity/ohms_law"34solve :the_resistor_on_a_ten_volt_battery, using: OhmsLaw do5  objects :battery, :resistor, :circuit67  take emf: circuit.voltage8  take internal: battery.resistance9  take current: circuit.current1011  deduce_from "connected to a resistor", circuit.resistance => battery.resistance + resistor.resistance1213  resistance = resistor.resistance14  answer resistance:, in: "Ω", figures: 21516  conclusion "The emf drives the current through the resistor and the battery's own resistance in a row, so " \17             "the two resistances together are the emf over the current."18end

playground.rb

1require_relative "../solutions/the_resistor_on_a_ten_volt_battery"23play_with :the_resistor_on_a_ten_volt_battery do4  title "The Resistor on a Ten Volt Battery"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	$V_{b}$	V	—
current of battery	$I_{b}$	A	—
resistance of battery	$R_{b}$	Ω	0 … ∞
voltage of resistor	$V_{r}$	V	—
current of resistor	$I_{r}$	A	—
resistance of resistor	$R_{r}$	Ω	0 … ∞
voltage of circuit	$V_{c}$	V	—
current of circuit	$I_{c}$	A	—
resistance of circuit	$R_{c}$	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	$ρ_{b}$	Ω m	—
length of the conductor of battery	$l_{b}$	m	—
cross section of battery	$A_{b}$	m²	—
resistivity of resistor	$ρ_{r}$	Ω m	—
length of the conductor of resistor	$l_{r}$	m	—
cross section of resistor	$A_{r}$	m²	—
resistivity of circuit	$ρ_{c}$	Ω m	—
length of the conductor of circuit	$l_{c}$	m	—
cross section of circuit	$A_{c}$	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_{b} = I_{b} \cdot R_{b}

ohms law(battery)

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

ohms law(resistor)

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

ohms law(circuit)

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

resistance from resistivity(ground)

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

resistance from resistivity(battery)

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

resistance from resistivity(resistor)

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

resistance from resistivity(circuit)

R_{c} = R_{b} + R_{r}

connected to a resistor