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1 year ago

How many hours should I charge a 4.8v 600mah battery (I need it by today please)

Engineering

1 answer:

prohojiy [21]1 year ago

5 0

The number of hours that will be needed to charge a 600mah battery will be 1.5 hours.

<h3>What is a battery?</h3>

It should be noted that an electric battery simply means a source of electric power that consist of one or more electrochemical cells that are with external connections that are important for powering electrical devices.

It should be noted that when a battery is supplying power, then the positive terminal is the cathode while the negative terminal is the anode.

In conclusion, the number of hours that will be needed to charge a 600mah battery will be 1.5 hours.

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Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

Elodia [21]

Answer:

1.25 cm/day

Explanation:

An air thickness , (l) = 0.15 cm

Air Temperature =

$(T_a)=20^0C = (20+273)K\\(T_a)=293K$

Mass Diffusion coefficient (D) = $0.25cm^2/sec$

If the air pressure $(P_a) = 0.5 P_{sat}$

We are to determine how fast will the water $(H_2O)$ level drop in a day.

From the property of air at T = 20° C

$P_{sat} = 2.34$ from saturated water properties.

The mass flow of $(H_2O)$ can be calculated as:

$H_2O = \frac{D}{\phi} \delta C$

where:

$\delta C = \frac{P_{sat}*P_a}{RT }$

R(constant) = 8.314 kJ/mol.K

$\delta C = \frac{2.34*0.5}{8.314*293 }$

$\delta C = 4.803*10^{-4}\\ \delta C =0.48*10^{-3} mol/m^3\\ \delta C = 0.48*10^{-6} mol/cm^3$

Since 1 mole = 18 cm ³ of water

$0.48*10^{-6}mol/cm^3$ will be: $(0.48*10^{-6}mol/cm^3 *18)cm^3/cm^3$

$\delta C = 8.64 * 10^{-6}$

Again:

$H_2O = \frac{D}{\phi} \delta C$

$= \frac{0.25}{0.15}*8.69*10^{-6} \frac{cm^2/sec}{cm}$

$=1.4481*10^{-5} \frac{cm^2/sec}{cm}$

Converting the above value to cm/day: we have:

$1.448*10^{-5}*3600*24\frac{cm}{s}*\frac{s}{yr}*\frac{yr}{day}$

= 1.25 cm/day

∴ the rate at which the water level drop in a day = 1.25 cm/day

7 0

2 years ago

The current at resonance in a series L-C-R circuit is 0.2mA. If the applied voltage is 250mV at a frequency of 100 kHz and the c

iVinArrow [24]

Answer:

The resistance of the circuit is 1250 ohms

The inductance of the circuit is 0.063 mH.

Explanation:

Given;

current at resonance, I = 0.2 mA

applied voltage, V = 250 mV

resonance frequency, f₀ = 100 kHz

capacitance of the circuit, C = 0.04 μF

At resonance, capacitive reactance ( $X_c$ ) is equal to inductive reactance ( $X_l$ ),

$Z = \sqrt{R^2 + (X_ l - X_c)^2} \\\\But \ X_l= X_c\\\\Z = R$

Where;

R is the resistance of the circuit, calculated as;

$R = \frac{V}{I} \\\\R = \frac{250 \ \times \ 10^{-3}}{0.2 \ \times \ 10^{-3}} \\\\R = 1250 \ ohms$

The inductive reactance is calculated as;

$X_l = X_c = \frac{1}{\omega C} = \frac{1}{2\pi f_o C} = \frac{1}{2\pi (100\times 10^3)(0.04\times 10^{-6} ) } = 39.789 \ ohms\\$

The inductance is calculated as;

$X_l = \omega L = 2\pi f_o L\\\\L = \frac{X_l}{2\pi f_o}\\\\L = \frac{39.789}{2\pi (100 \times 10^3)} \\\\L= 6.3 \ \times \ 10^{-5} \ H\\\\L = 0.063 \times \ 10^{-3} \ H\\\\L = 0.063 \ mH$

5 0

2 years ago

The ruler game, HELPPPP PLS

Rasek [7]

Answer:

Explanation:

8 0

2 years ago

Read 2 more answers

Explain how you would solve for total resistance in a parallel circuit versus a series circuit. How would you apply and solve fo

Mekhanik [1.2K]

Answer:

No, series parallel, as it implies has components of the circuit configured in both series and parallel. This is typically done to achieve a desired resistance in the circuit. A parallel circuit is a circuit that only has the components hooked in parallel, which would result in a lower total resistance in the circuit than if the components were hooked up in a series parallel configuration.

Explanation:

8 0

2 years ago

?Why the efficiency of Class A amplifier is very poor

horsena [70]

Sorry not sure about that

6 0

2 years ago