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2 years ago

You test a diode with an ohmmeter. The two readings indicate a very low value of resistance. The diode is probably __________.

Engineering

1 answer:

Angelina_Jolie [31]2 years ago

3 0

Answer:

shorted

Explanation:

A diode is a device that passes current much better in one direction than in the reverse direction.

<h3>Ohmmeter test</h3>

Conceptually, an ohmmeter applies a voltage to the device under test and reports the current through the device on a scale calibrated in ohms. It typically has a low enough open-circuit voltage, and a high enough internal resistance so as to avoid damage to parts under test.

A "continuity" tester may have an open-circuit voltage of a few tens of millivolts and/or a short-circuit current of a few microamps, so as to properly detect continuity and avoid contact damage in "dry" circuits. Such a tester is virtually useless for diode testing.

An ohmmeter suitable for diode testing will generally have an open-circuit voltage of a few volts, and a short-circuit current of a few milliamps. When such a meter is used to test a diode, it will indicate a few kilohms (or less) in the "forward" direction, and several 10s or 100s of megohms in the reverse direction.

<h3>Low-resistance readings</h3>

If the "forward" resistance reading is unusually low (a few ohms), the diode may be <em>damaged</em>. If both forward and reverse readings are unusually low, the diode my be considered to be <em>shorted</em>.

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We need to design a logic circuit for interchanging two logic signals. The system has three inputs I1I1, I2I2, and SS as well as

Salsk061 [2.6K]

Explanation:

Inputs and Outputs:

There are 3 inputs = I₁, I₂, and S

There are 2 outputs = O₁ and O₂

The given problem is solved in three major steps:

Step 1: Construct the Truth Table

Step 2: Obtain the logic equations using Karnaugh map

Step 3: Draw the logic circuit

Step 1: Construct the Truth Table

The given logic is

When S = 0 then O₁ = I₁ and O₂ = I₂

When S = 1 then O₁ = I₂ and O₂ = I₁

I₁ | I₂ | S | O₁ | O₂

0 | 0 | 0 | 0 | 0

0 | 0 | 1 | 0 | 0

0 | 1 | 0 | 0 | 1

0 | 1 | 1 | 1 | 0

1 | 0 | 0 | 1 | 0

1 | 0 | 1 | 0 | 1

1 | 1 | 0 | 1 | 1

1 | 1 | 1 | 1 | 1

Step 2: Obtain the logic equations using Karnaugh map

Please refer to the attached diagram where Karnaugh map is set up.

The minimal SOP representation for output O₁

$O_1 = I_1 \bar{S} + I_2 S$

The minimal SOP representation for output O₂

$O_2 = I_2 \bar{S} + I_1 S$

Step 3: Draw the logic circuit

Please refer to the attached diagram where the circuit has been drawn.

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3 years ago

A tensile test was made on a tensile specimen, with a cylindrical gage section which had a diameter of 10 mm, and a length of 40

tamaranim1 [39]

Answer:

The answers are as follow:

a) 10 mm

b) 12.730 N/ $mm^{2}$

c) 127.307 N/ $mm^{2}$

d) 0.25

Explanation:

d1 = 10mm , L1 = 40 mm, L2 = 50 mm, reduction in area = 90% = 0.9

Force = F =1000 N

let us find initial area first, A1 = pi* $r^{2}$ = 78.55 $mm^{2}$

using reduction in area formula : 0.9 = (A1 - A2 ) / A1

solving it will give, A2 = 0.1 A1 = 7.855 $mm^{2}$

a) The specimen elongation is final length - initial length

50 - 40 = 10 mm

b) Engineering stress uses the original area for all stress calculations,

Engineering stress = force / original area = F / A1 = 1000 / 78.55

Engineering stress = 12.730 $N / mm^{2}$

c) True stress uses instantaneous area during stress calculations,

True fracture stress = force / final area = F / A2 = 1000 / 7.855

True Fracture stress = 127.30 $N / mm^{2}$

e) strain = change in length / original length

strain = 10 / 40 = 0.25

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3 years ago

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m_a_m_a [10]

Answer:

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4 years ago

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avanturin [10]

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For more information on soft ware development process kindly visit

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3 years ago

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saw5 [17]

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