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Bad White [126]
2 years ago
6

the increase of current when 15 V is applied to 10000ohm rheostat which is adjusted to 1000ohm value​

Engineering
1 answer:
Anastasy [175]2 years ago
4 0
Given data:
•) applied voltage = 15 V
•). Resistance = 1000 ohm

Required:
•). The magnitude of current= ?

•••••••••••••SOLUTION•••••••••••••

We can find the relation ship between current, voltage and resistance with the help of Ohms law.

According to ohms law;

V= IR.

Rearranging the above equation;

I= V/ R

Putt the values in the above equation; we get

I= 15V/ 1000ohm

I = 0.015 A( ampere)

••••••••••••••• CONCLUSION•••••••

The value of the current would be 0.15 ampere when Resistance is equal to 1000 and that of Voltage is equal to 15 V.
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A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.
denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is 8.621 \times 10^{-4} in/in

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

\frac{M}{I}=\frac{F}{y}

Here

  • M is the moment which is to be calculated
  • I is the moment of inertia given as

                         I=\frac{bd^3}{12}

Here

  • b is the breath given as 0.75"
  • d is the depth which is given as 8"

                     I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4

  • y is given as

                     y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\

  • Force is 50 ksi

\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in

The yield moment is 400 k.in.

Part b:

The strain is given as

Strain=\frac{Stress}{Elastic Modulus}

The stress at the station 2" down from the top is estimated by ratio of triangles as

                        F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi

Now the steel has the elastic modulus of E=29000 ksi

Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in

So the strain is 8.621 \times 10^{-4} in/in

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi

The plastic moment is 600 ksi.

3 0
3 years ago
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