Please help! If the current in a circuit is 3 + j2 amps and the resistance is 4 - j3 ohms, what is the voltage?
1 answer:
The value of the voltage is 18 - j volts if the current in a circuit is 3 + j2 amps and the resistance is 4 - j3 ohms option (c) is correct.
<h3>What is a complex number?</h3>
It is defined as the number which can be written as x+iy where x is the real number or real part of the complex number and y is the imaginary part of the complex number and i is the iota which is nothing but a square root of -1.
We have:
The current in a circuit is 3 + j2 amps and the resistance is 4 - j3 ohms.
As we know,
V = IR
V = (3 + j2)(4 - j3)
j² = -1 (complex part)
After simplification:
V = 18 - j volts
Thus, the value of the voltage is 18 - j volts if the current in a circuit is 3 + j2 amps and the resistance is 4 - j3 ohms option (c) is correct.
Learn more about the complex number here:
#SPJ1
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Answer:
a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.
c) This last case the histogram is not symmetrical and the data seems to be skewed.
Step-by-step explanation:
For this case we have the following data:
(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
We can use the following R code to get the histogram
> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
> hist(x1,main="Histogram a)")
The result is on the first figure attached.
For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> hist(x2,main="Histogram b)")
The result is on the first figure attached.
For this case the data is skewed to the left and we can't assume that we have the normality assumption.
(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> hist(x3,main="Histogram c)")
The result is on the first figure attached.
This last case the histogram is not symmetrical and the data seems to be skewed.
