A Ten Volt Battery's Terminal Voltage

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 $I_{r} = I_{c}$ is 0.5 A $I_{c} = 0.5$ A, what is the terminal voltage of the battery when the circuit is closed, in volts?

Inputs

emf

V_{c}

10 V

internal

R_{b}

3 Ω

current

I_{c}

0.5 A

Outputs

terminal

V_{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”
$I_{r} = I_{c}$ from “the current in the circuit”

Find $V_{r}$ .

Working

the current in the circuit $I_{r} = I_{c}$
$I_{r} = 0.5$
$I_{r} = 0.5$ A
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
ohms law(resistor) $V_{r} = I_{r} \cdot R_{r}$
$V_{r} = 0.5 \cdot 17$
$V_{r} = 8.5$ V
$V_{r} \approx 8.5$ Vto 2 significant figures

Conclusion

Part of the emf is spent driving the current through the battery itself; what is left across its terminals is the voltage across the resistor.

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_ten_volt_battery_s_terminal_voltage 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}, %{terminal}, in volts?"7  given emf: "10 V"8  given internal: "3 Ω"9  given current: "0.5 A"10  asks terminal: "what is the terminal voltage of the battery when the circuit is closed"11end

solution.rb

1require_relative "../questions/a_ten_volt_battery_s_terminal_voltage"2require_relative "../science/current_electricity/ohms_law"34solve :a_ten_volt_battery_s_terminal_voltage, 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.resistance12  deduce_from "the current in the circuit", resistor.current => circuit.current1314  terminal = resistor.voltage15  answer terminal:, in: "V", figures: 21617  conclusion "Part of the emf is spent driving the current through the battery itself; what is left across its " \18             "terminals is the voltage across the resistor."19end

playground.rb

1require_relative "../solutions/a_ten_volt_battery_s_terminal_voltage"23play_with :a_ten_volt_battery_s_terminal_voltage do4  title "A Ten Volt Battery's Terminal Voltage"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

I_{r} = I_{c}

the current in the circuit