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Ostrovityanka [42]
3 years ago
10

Determine the voltages at all nodes and the currents through all branches. Assume that the transistor B is 100,

Engineering
1 answer:
iren [92.7K]3 years ago
4 0

Answer:

The voltages of all nodes are, IE = 4.65 mA, IB =46.039μA,  IC=4.6039 mA, VB = 10v, VE =10.7, Vc =4.6039 v

Explanation:

Solution

Given that:

V+ = 20v

Re = 2kΩ

Rc = 1kΩ

Now we will amke use of the method KVL in the loop.

= - Ve + IE . Re + VEB + VB = 0

Thus

IE = V+ -VEB -VB/Re

Which gives us the following:

IE = 20-0.7 - 10/2k

= 9.3/2k

so, IE = 4.65 mA

IB = IE/β +1 = 4.65 m /101

Thus,

IB = 0.046039 mA

IB = 46.039μA

IC =βIB

Now,

IC = 100 * 0.046039

IC is 4.6039 mA

Now,

VB = 10v

VE = VB + VEB

= 10 +0.7 = 10.7 v

So,

Vc =Ic . Rc = 4.6039 * 1k

=4.6039 v

Finally, this is the table summary from calculations carried out.

Summary Table

Parameters          IE       IC           IB            VE       VB         Vc

Unit                     mA     mA          μA            V           V          V

Value                  4.65    4.6039   46.039    10.7      10     4.6039

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Two steel plates of 15 mm thickness each are clamped together with an M14 x 2 hexagonal head bolt, a nut, and one 14R metric pla
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Answer:

a) 50 mm

b) 808.24 MN/m

Explanation:

Given:

Thickness of each steel plate = 15mm

a) To find suitable length of bolt, we'll use:

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To find the grip length since there is a washer, we'll use:

Grip length = plate thickness + washer thickness

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b) Bolt stiffness:

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LT = 2*d + 6

Where d = 14mm, from table(8-7)

= 2*14 + 6

= 34 mm

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Ad= \frac{\pi}{4} d^2 = \frac{pi}{4} * 14^2 = 153.94 mm^2

Length of unthreaded portion in grip:

Ld = 50 - 34 = 16mm

Length of threaded portion in grip:

Lt = 33.5 - 16 = 17.5mm

Bolt stiffness = \frac{A_d A_t E}{A_d l_t + A_t l_d}

= \frac{153.94 * 115 * 207}{(153.94*17.5)+(115*16)} = 808.24

Bolt stiffness = 808.24 MN/m

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