A Battery on Charge

solved OhmsLaw

Q. A storage battery of emf 8.0 V $V_{b} = 8.0$ V and internal resistance 0.5 Ω $R_{i} = 0.5$ Ω is being charged by $V_{c} = V_{s} - V_{b}$ a 120 V $V_{s} = 120$ V dc supply using a series resistor $R_{c} = R_{i} + R_{r}$ of 15.5 Ω $R_{r} = 15.5$ Ω. What is the terminal voltage of the battery during charging $V_{t} = V_{b} + V_{i}$ , in volts?

Inputs

emf

V_{b}

8 V

internal

R_{i}

0.5 Ω

supply

V_{s}

120 V

series

R_{r}

15.5 Ω

Outputs

terminal

V_{t}

How it's solved

Solved with OhmsLaw

Given

$V_{b} = 8.0$ V
$R_{i} = 0.5$ Ω
$V_{s} = 120$ V
$R_{r} = 15.5$ Ω
$V_{c} = V_{s} - V_{b}$ from “is being charged by”
$R_{c} = R_{i} + R_{r}$ from “using a series resistor”
$I_{i} = I_{c}$ from “a series resistor of”
$V_{t} = V_{b} + V_{i}$ from “terminal voltage of the battery during charging”

Find $V_{t}$ .

Working

is being charged by $V_{c} = V_{s} - V_{b}$
$V_{c} = 120 - 8$
$V_{c} = 112$ V
using a series resistor $R_{c} = R_{i} + R_{r}$
$R_{c} = 0.5 + 15.5$
$R_{c} = 16$ Ω
ohms law(circuit) $V_{c} = I_{c} \cdot R_{c}$
$112 = I_{c} \cdot 16$
$I_{c} = 7$ A
a series resistor of $I_{i} = I_{c}$
$I_{i} = 7$
$I_{i} = 7$ A
ohms law(inside) $V_{i} = I_{i} \cdot R_{i}$
$V_{i} = 7 \cdot 0.5$
$V_{i} = 3.5$ V
terminal voltage of the battery during charging $V_{t} = V_{b} + V_{i}$
$V_{t} = 8 + 3.5$
$V_{t} = 11.5$ V
$V_{t} \approx 11.5$ Vto 3 significant figures

Conclusion

The supply pushes current backward through the battery against its emf, so the voltage across its terminals is the emf plus what the current takes to cross its 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_battery_on_charge do4  kind :numerical5  describes "A storage battery of emf %{emf} and internal resistance %{internal} is being charged by a %{supply} dc " \6            "supply using a series resistor of %{series}. %{terminal}, in volts?"7  given emf: "8.0 V"8  given internal: "0.5 Ω"9  given supply: "120 V"10  given series: "15.5 Ω"11  asks terminal: "What is the terminal voltage of the battery during charging"12end

solution.rb

1require_relative "../questions/a_battery_on_charge"2require_relative "../science/current_electricity/ohms_law"34solve :a_battery_on_charge, using: OhmsLaw do5  objects :battery, :inside, :supply, :resistor, :circuit, :terminals67  take emf: battery.voltage8  take internal: inside.resistance9  take supply: supply.voltage10  take series: resistor.resistance1112  deduce_from "is being charged by", circuit.voltage => supply.voltage - battery.voltage13  deduce_from "using a series resistor", circuit.resistance => inside.resistance + resistor.resistance14  deduce_from "a series resistor of", inside.current => circuit.current15  deduce_from "terminal voltage of the battery during charging", terminals.voltage => battery.voltage + inside.voltage1617  terminal = terminals.voltage18  answer terminal:, in: "V", figures: 31920  conclusion "The supply pushes current backward through the battery against its emf, so the voltage across its " \21             "terminals is the emf plus what the current takes to cross its internal resistance."22end

playground.rb

1require_relative "../solutions/a_battery_on_charge"23play_with :a_battery_on_charge do4  title "A Battery on Charge"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 inside	$V_{i}$	V	—
current of inside	$I_{i}$	A	—
resistance of inside	$R_{i}$	Ω	0 … ∞
voltage of supply	$V_{s}$	V	—
current of supply	$I_{s}$	A	—
resistance of supply	$R_{s}$	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 … ∞
voltage of terminals	$V_{t}$	V	—
current of terminals	$I_{t}$	A	—
resistance of terminals	$R_{t}$	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 inside	$ρ_{i}$	Ω m	—
length of the conductor of inside	$l_{i}$	m	—
cross section of inside	$A_{i}$	m²	—
resistivity of supply	$ρ_{s}$	Ω m	—
length of the conductor of supply	$l_{s}$	m	—
cross section of supply	$A_{s}$	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	—
resistivity of terminals	$ρ_{t}$	Ω m	—
length of the conductor of terminals	$l_{t}$	m	—
cross section of terminals	$A_{t}$	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_{i} = I_{i} \cdot R_{i}

ohms law(inside)

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

ohms law(supply)

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

ohms law(resistor)

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

ohms law(circuit)

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

ohms law(terminals)

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_{i} = \frac{ρ_{i} \cdot l_{i}}{A_{i}}

resistance from resistivity(inside)

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

resistance from resistivity(supply)

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_{t} = \frac{ρ_{t} \cdot l_{t}}{A_{t}}

resistance from resistivity(terminals)

V_{c} = V_{s} - V_{b}

is being charged by

R_{c} = R_{i} + R_{r}

using a series resistor

I_{i} = I_{c}

a series resistor of

V_{t} = V_{b} + V_{i}

terminal voltage of the battery during charging