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Nady [450]
3 years ago
10

Two wires of the same length and cross sectional area have finite but different E-field magnitudes inside them when the same cur

rent is flowing in each. The E-field magnitude in wire 1 is less than the field magnitude in wire 2. Select all of the following statements that are true based on this situation:
The voltage drop in wire 1 is the same as the voltage drop in wire 2
The conductivity of wire 1 is less than the conductivity in wire 2
There is not enough information given in the problem to definitely determine the answer to the other 9 statements.
The power dissipation in wire 1 is more than the power dissipation in wire 2
The conductivity of wire 1 is more than the conductivity in wire 2
The conductivity of wire 1 is the same as the conductivity in wire 2
The power dissipation in wire 1 is less than the power dissipation in wire 2
The power dissipation in wire 1 is the same as the power dissipation in wire 2
The voltage drop in wire 1 is greater than the voltage drop in wire 2
The voltage drop in wire 1 is less than the voltage drop in wire 2
Engineering
1 answer:
morpeh [17]3 years ago
3 0

Answer:

The voltage drop in wire 1 is more than the voltage drop in wire 2.

The conductivity of wire 1 is less than the conductivity of wire 2.

The power dissipation in wire 1 is less than the power dissipation in wire 2.

Explanation:

The electric field magnitude inside the wire is dependent on the current flow. Voltage drop can cause load regulation errors in wiring. The conductor is then used to maintain the regulated flow of current in the wires. In the given scenario the wire  magnitude is greater than wire 1 magnitude. If the magnitude is smaller there will be less conductivity therefore more voltage drops will be observed.

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It looks like... A machine that reads electric pulse and surge... Not sure though.

Explanation:

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3 years ago
99 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
qwelly [4]

Answer:

1. Can you tell me something about yourself?

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7 0
3 years ago
Read 2 more answers
The following program includes fictional sets of the top 10 male and female baby names for the current year. Write a program tha
otez555 [7]

Answer:

Define Variables and Use List methods to do the following

Explanation:

#<em>Conjoins two lists together</em>

all_names = male_names.union(female_names)

#<em>Finds the names that appear in both lists, just returns those</em>

neutral_names = male_names.intersection(female_names)

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3 0
3 years ago
Read 2 more answers
Find the number of Btu conducted through a wall in 8 hours. The wall is 8 feet high by 24 feet long and has a total R-value of 1
dedylja [7]

Answer:

ΔQ = 4930.37 BTu

Explanation:

given data

height h = 8ft

Δt = 8  hours

length L = 24 feet

R value = 16.2 hr⋅°F⋅ft² /Btu

inside temperature t1 = 68°F

outside temperature t2 = 16°F

to find out

number of Btu conducted

solution

we get here number of Btu conducted by this expression that s

\frac{\Delta Q}{\Delta t} =\frac{-A}{R} (t2 -t1)     ......................1

here A is area that is = h × L = 8 × 24 = 1492 ft²

put here value we get

\frac{\Delta Q}{8} =\frac{-192}{16.2} (16-68)

solve it we get

ΔQ = 4930.37 BTu

7 0
3 years ago
P10.12. A certain amplifier has an open-circuit voltage gain of unity, an input resistance of and an output resistance of The si
klio [65]

complete question

A certain amplifier has an open-circuit voltage gain of unity, an input resistance of 1 \mathrm{M} \Omega1MΩ and an output resistance of 100 \Omega100Ω The signal source has an internal voltage of 5 V rms and an internal resistance of 100 \mathrm{k} \Omega.100kΩ. The load resistance is 50 \Omega.50Ω. If the signal source is connected to the amplifier input terminals and the load is connected to the output terminals, find the voltage across the load and the power delivered to the load. Next, consider connecting the load directly across the signal source without the amplifier, and again find the load voltage and power. Compare the results. What do you conclude about the usefulness of a unity-gain amplifier in delivering signal power to a load?

Answer:

3.03 V  0.184 W

2.499 mV  125*10^-9 W

Explanation:

First, apply voltage-divider principle to the input circuit: 1

V_{i}= (R_i/R_i+R_s) *V_s = 10^6/10^6+(0.1*10^6)\\*5

    = 4.545 V

The voltage produced by the voltage-controlled source is:

A_voc*V_i = 4.545 V

We can find voltage across the load, again by using voltage-divider principle:  

V_o = A_voc*V_i*(R_o/R_l+R_o)

      = 4.545*(100/100+50)

      = 3.03 V  

Now we can determine delivered power:  

P_L = V_o^2/R_L

      = 0.184 W

Apply voltage-divider principle to the circuit:  

V_o = (R_o/R_o+R_s)*V_s

       = 50/50+100*10^3*5

       = 2.499 mV

Now we can determine delivered power:  

P_l = V_o^2/R_l

     = 125*10^-9 W

Delivered power to the load is significantly higher in case when we used amplifier, so a unity gain amplifier can be useful in situation when we want to deliver more power to the load. It is the same case with the voltage, no matter that we used amplifier with voltage open-circuit gain of unity.  

4 0
3 years ago
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