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olga2289 [7]
3 years ago
11

Charging method .Constant current method​

Engineering
1 answer:
mina [271]3 years ago
4 0

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.

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Answer:

Answer explained below

Explanation:

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4 years ago
Consider a fully-clamped circular diaphragm poly-Si with a radius of 250 μm and a thickness of 4 μm. Assume that Young’s modulus
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Answer:

Explanation:

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4 0
3 years ago
All of the dimensions on an aircraft drawing are_________<br> to the bottom of the drawing.
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All of the dimensions on an aircraft drawing are _________ to the bottom of the drawing


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7 0
2 years ago
1. A glass window of width W = 1 m and height H = 2 m is 5 mm thick and has a thermal conductivity of kg = 1.4 W/m*K. If the inn
emmasim [6.3K]

Answer:

1. \dot Q=19600\ W

2. \dot Q=120\ W

Explanation:

1.

Given:

  • height of the window pane, h=2\ m
  • width of the window pane, w=1\ m
  • thickness of the pane, t=5\ mm= 0.005\ m
  • thermal conductivity of the glass pane, k_g=1.4\ W.m^{-1}.K^{-1}
  • temperature of the inner surface, T_i=15^{\circ}C
  • temperature of the outer surface, T_o=-20^{\circ}C

<u>According to the Fourier's law the rate of heat transfer is given as:</u>

\dot Q=k_g.A.\frac{dT}{dx}

here:

A = area through which the heat transfer occurs = 2\times 1=2\ m^2

dT = temperature difference across the thickness of the surface = 35^{\circ}C

dx = t = thickness normal to the surface = 0.005\ m

\dot Q=1.4\times 2\times \frac{35}{0.005}

\dot Q=19600\ W

2.

  • air spacing between two glass panes, dx=0.01\ m
  • area of each glass pane, A=2\times 1=2\ m^2
  • thermal conductivity of air, k_a=0.024\ W.m^{-1}.K^{-1}
  • temperature difference between the surfaces, dT=25^{\circ}C

<u>Assuming layered transfer of heat through the air and the air between the glasses is always still:</u>

\dot Q=k_a.A.\frac{dT}{dx}

\dot Q=0.024\times 2\times \frac{25}{0.01}

\dot Q=120\ W

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Answer:

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