A Ten Volt Battery's Terminal Voltage

Q. A battery of emf 10 V and internal resistance 3 Ω is connected to a resistor. If the current in the circuit is 0.5 A, what is the terminal voltage of the battery when the circuit is closed, in volts? expects 8.5 V

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The web of concepts

given read from the words equation asked

The equations it used

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Check an answer

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?

The playground

Inputs

emf

V_{c}

10 V

internal

R_{b}

3 Ω

current

I_{c}

0.5 A

Outputs

terminal

V_{r}

The solution

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 Ruby

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

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

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

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