If you were to plot the voltage versus the current for a given circuit, what would you expect the slope of the line to be? If no
w the resistance of the circuit were a function of temperature would you still expect to see a straight line when plotting voltage versus current if the temperature was allowed tovary?
1 answer:
Answer:
Part 1: It would be a straight line, current will be directly proportional to the voltage.
Part 2: The current would taper off and will have negligible increase after the voltage reaches a certain value. Graph attached.
Explanation:
For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.
V=I*R
where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.
In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.
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Answer:
- <u>0.59</u>
Explanation:
The <em>batch</em> is <em>rejected</em> if any of the <em>random samples are found defective</em>, or, what is the same, it will be accepted only if all 15 samples are good.
The probability that none be defective is the same probability that all the samples are good. Thus, start to calculate the probability that the batch is accepted.
The probability that the first sample is good is 214 /225, because there are 225 - 11 = 214 good samples in 225 doses.
The probability that the second samples is good too is 213/224, because there is 1 less good sample, in the 224 remaining samples.
By the same process, you conclude that the consecutive probabilities of selecting a good sample are: 212/223, 211/222, 210/221, . . . up to 199/211.
The joint probability of all the samples are good is the product of each probability:
The result is: 0.41278 ≈ 0.41
The conclusion is that the probability that all the samples are good and the batch is accepted is 0.41.
Therefore, <em>the probability that the batch is rejected</em> is 1 - 0.41 = 0.59.
Answer: hello some parts of your question is missing attached below is the missing information
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube
answer : Total surface area = 3/2 * area of old radiator
Explanation:
we will use this relation
K =
change in T = ΔT
therefore New Area ( A ) = 3/2 * area of old radiator
Given that the thermal conductivity is the same in the new and old radiators
Answer:
0.5m^2/Vs and 0.14m^2/Vs
Explanation:
To calculate the mobility of electron and mobility of hole for gallium antimonide we have,
(S)
Where
e= charge of electron
n= number of electrons
p= number of holes
mobility of electron
mobility of holes
electrical conductivity
Making the substitution in (S)
Mobility of electron
Mobility of hole in (S)
Then, solving the equation:
(1)
(2)
We have,
Mobility of electron
Mobility of hole is
Answer:
Explanation:
The python code to generate this is quite simple to run.
i hope you understand everything written here, you can as well try out other problems to understand better.
First to begin, we import the package;
Code:
import pandas as pd
import matplotlib.pyplot as plt
name = input('Enter name of the file: ')
op = input('Enter name of output file: ')
df = pd.read_csv(name)
df['Date'] = pd.to_datetime(df["Date"].apply(str))
plt.plot(df['Date'],df['Absent']/(df['Present']+df['Absent']+df['Released']),label="% Absent")
plt.legend(loc="upper right")
plt.xticks(rotation=20)
plt.savefig(op)
plt.show()
This should generate the data(plot) as seen in the uploaded screenshot.
thanks i hope this helps!!!
