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3 years ago

Can i join three 12 volts batteriesto give me 24 volts output

Engineering

1 answer:

bulgar [2K]3 years ago

8 0

Answer:

YES

Explanation:

If we connect batteries in series then the output voltage is the sum of the individual voltage of each battery i.e if you connect three 12 volts batteries in series then their output voltage will be 12+12+12=36 volts, but the current rating of the batteries in series will be same of the individual battery rating in 'mah'.

But when we connect the batteries in parallel their voltage is not added but their current rating in mah is addition of their individual rating.

So, If you want 24 volts from three 12 volts battery then you can connect two of them in series and the other one in parallel with them this will give 24 volts and the current will be addition of the two series batteries and the third which is in parallel with them. You can use this configuration if current value is not a big factor.

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3 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe

ankoles [38]

Answer:

F = 8849 N

Explanation:

Given:

Load at a given point = F = 4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ $_{fs} = F_{f} L / \pi R^{3}$

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³

= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³

= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

= 187 / 3.141593 (0.0056)³

= 338943767.745358

= 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

$F_{f}$ = 2 σ $_{fs}$ d³/3 L

F = 2σd³/3L

= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)

= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)

= 146016 / 125 / 3 ( 44 x 1 / 1000 )

= ( 146016 / 125 ) / (3 ( 11 / 250 ))

= 97344 / 11

F = 8849 N

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3 years ago

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3 years ago

A cylindrical bar of metal having a diameter of 20.2 mm and a length of 209 mm is deformed elastically in tension with a force o

Rus_ich [418]

Answer:

A) ΔL = 0.503 mm

B) Δd = -0.016 mm

Explanation:

A) From Hooke's law; σ = Eε

Where,

σ is stress

ε is strain

E is elastic modulus

Now, σ is simply Force/Area

So, with the initial area; σ = F/A_o

A_o = (π(d_o)²)/4

σ = 4F/(π(d_o)²)

Strain is simply; change in length/original length

So for initial length, ε = ΔL/L_o

So, combining the formulas for stress and strain into Hooke's law, we now have;

4F/(π(d_o)²) = E(ΔL/L_o)

Making ΔL the subject, we now have;

ΔL = (4F•L_o)/(E•π(d_o)²)

We are given;

F = 50500 N

L_o = 209mm = 0.209m

E = 65.5 GPa = 65.5 × 10^(9) N/m²

d_o = 20.2 mm = 0.0202 m

Plugging in these values, we have;

ΔL = (4 × 50500 × 0.209)/(65.5 × 10^(9) × π × (0.0202)²)

ΔL = 0.503 × 10^(-3) m = 0.503 mm

B) The formula for Poisson's ratio is;

v = -(ε_x/ε_z)

Where; ε_x is transverse strain and ε_z is longitudinal strain.

So,

ε_x = Δd/d_o

ε_z = ΔL/L_o

Thus;

v = - [(Δd/d_o)/(ΔL/L_o)]

v = - [(Δd•L_o)/(ΔL•d_o)]

Making Δd the subject, we have;

Δd = -[(v•ΔL•d_o)/L_o]

We are given v = 0.33; d_o = 20.2mm

So,

Δd = -[(0.33 × 0.503 × 20.2)/209]

Δd = -0.016 mm

8 0

3 years ago

Can i join three 12 volts batteriesto give me 24 volts output​

Can i join three 12 volts batteriesto give me 24 volts output