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3 years ago

What happens to battery when it produces current to the system

Engineering

1 answer:

olasank [31]3 years ago

3 0

Answer:

Electricity, as you probably already know, is the flow of electrons through a conductive path like a wire. This path is called a circuit. ... The chemical reactions in the battery causes a build up of electrons at the anode. This results in an electrical difference between the anode and the cathode

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Foam weather stripping is often placed in the frames of doors and

Firdavs [7]

Answer:

prevents weathering

Explanation:

6 0

3 years ago

Determine whether or not each of the following four transaction execution histories is serializable. If a history is serializabl

ludmilkaskok [199]

Answer:

Option D. w1[x] w2[u] w2[y] w1[y] w3[x] w3[u] w1[z]

Explanation:

The execution in the option D is correct. This is because there is more than one reasonable criterion.

8 0

2 years ago

Find the time-domain sinusoid for the following phasors:_________

sattari [20]

Answer:

a. r(t) = 6.40 cos (ωt + 38.66°) units

b. r(t) = 6.40 cos (ωt - 38.66°) units

c. r(t) = 6.40 cos (ωt - 38.66°) units

d. r(t) = 6.40 cos (ωt + 38.66°) units

Explanation:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = $\sqrt{a^2 + b^2}$

Ф = direction = tan⁻¹ ( $\frac{b}{a}$ )

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

(i) convert to polar form

r = $\sqrt{5^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

(i) convert to polar form

r = $\sqrt{5^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + 4^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{4}{-5}$ )

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

(i) convert to polar form

r = $\sqrt{(-5)^2 + (-4)^2}$

r = $\sqrt{25 + 16}$

r = $\sqrt{41}$

r = 6.40

Φ = tan⁻¹ ( $\frac{-4}{-5}$ )

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid

r(t) = 6.40 cos (ωt + 38.66°)

3 0

3 years ago

Charging method .Constant current method

mina [271]

Answer:

There are three common methods of charging a battery; constant voltage, constant current and a combination of constant voltage/constant current with or without a smart charging circuit.

Constant voltage allows the full current of the charger to flow into the battery until the power supply reaches its pre-set voltage. The current will then taper down to a minimum value once that voltage level is reached. The battery can be left connected to the charger until ready for use and will remain at that “float voltage”, trickle charging to compensate for normal battery self-discharge.

Constant current is a simple form of charging batteries, with the current level set at approximately 10% of the maximum battery rating. Charge times are relatively long with the disadvantage that the battery may overheat if it is over-charged, leading to premature battery replacement. This method is suitable for Ni-MH type of batteries. The battery must be disconnected, or a timer function used once charged.

Constant voltage / constant current (CVCC) is a combination of the above two methods. The charger limits the amount of current to a pre-set level until the battery reaches a pre-set voltage level. The current then reduces as the battery becomes fully charged. The lead acid battery uses the constant current constant voltage (CC/CV) charge method. A regulated current raises the terminal voltage until the upper charge voltage limit is reached, at which point the current drops due to saturation.

4 0

2 years ago

The ice on the rear window of an automobile is defrosted by attaching a thin, transparent, film type heating element to its inne

pshichka [43]

Answer:

A)Q = 1208.33 W/m²

B)K = 0.138 W/m.K

Explanation:

We are given;

inside air temperature;T_∞,i =25 °C = 25 + 273 = 298K

outside air temperature;T_∞,o = -10°C = - 10 + 273 = 263K

Inner surface temperature;T_s,i = 15 °C = 15 + 273 = 288K

Thickness, L = 4mm = 0.004m

convection heat transfer coefficient ; hi = 25 W/(m².K)

A) From an energy balance at the inner surface and the thermal circuit, the electric power required per unit window area is given as;

Q = [(T_s,i - T_∞,o)/((L/k) + (1/hi))] - [(T_∞,o - T_s,i)/(1/hi)]

Plugging in the relevant values with k for glass as 1.4 W/m.k, we have;

Q = [(288 - 263)/((0.004/1.4) + (1/25))] - [(263 - 288)/(1/25)]

Q = 583.33 + 625

Q = 1208.33 W/m²

B) The formula for thermal conductivity is;

K = (QL)/(AΔT)

Where;

K is the thermal conductivity in W/m.K

Q is the amount of heat transferred through the material

L is the distance between the two isothermal planes

A is the area of the surface in square meters

ΔT is the difference in temperature in Kelvin

ΔT = 298K - 263K = 35K

Now, since we have value of heat per unit area to be Q = 1208.33 W/m², let's rearrange the equation to reflect that; Thus ;

k = (Q/A) x (L/ΔT)

K = 1208.33 x (0.004/35)

K = 0.138 W/m.K

5 0

3 years ago