Using complex numbers to solve for a current in this circuit

In summary: Precise.In summary, the OP is trying to solve an equation for current in a circuit, but he can't seem to get it right. He started with a complex number and simplified it down to 5. Then he plugged it back in to the original equation and got the correct answer.
  • #1
GJ1
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4
Homework Statement
I am having trouble solving this equation, it is for an physics/electronics course
Relevant Equations
[50/(4+j3)(50)+100]x150
First I solved 4+j3, which I squared 4 and 3 to equal 16 and 9 then I added them to get 25 and then I got the square root of 25 = 5.

Then I plugged it back in to the equation.
[50/(5)(50)+100] x 150 to get 50/350x 150= 1/7(150)= 21.42. I've attached the correct answer.
 

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  • #2
GJ1 said:
Homework Statement:: I am having trouble solving this equation, it is for an physics/electronics course
Relevant Equations:: [50/(4+j3)(50)+100]x150
That doesn't look like an unambiguous mathematical expression.
GJ1 said:
First I solved 4+j3, which I squared 4 and 3 to equal 16 and 9 then I added them to get 25 and then I got the square root of 25 = 5.
Well, okay. That's the magnitude of that complex number.
GJ1 said:
Then I plugged it back in to the equation.
[50/(5)(50)+100] x 150 to get 50/350x 150= 1/7(150)= 21.42. I've attached the correct answer.
I can't make any sense of what you've attached. What are you actually trying to do here?
 
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  • #3
The equation is for current in a circuit, it's in the picture that I've attached. It includes a complex number for the reactive properties in the current "4+3i" or 4+3j. When, I sive the equation for I1, I get 21.42, the actual number is 20 - 10j or 20-10i.
 
  • #4
GJ1 said:
The equation is for current in a circuit, it's in the picture that I've attached. It includes a complex number for the reactive properties in the current "4+3i" or 4+3j. When, I sive the equation for I1, I get 21.42, the actual number is 20 - 10j or 20-10i.
Okay, so the answer is a complex number, but you somehow got rid of the imaginary part and replaced the complex number ##4 + 3j## with ##5##?
 
  • #5
But, how is the answer 20-10j
 
  • #6
GJ1 said:
But, how is the answer 20-10j
GJ1 said:
First I solved 4+j3, which I squared 4 and 3 to equal 16 and 9 then I added them to get 25 and then I got the square root of 25 = 5.
Does that mean that you got ##4 + j3 = 5##?

Note that ##|4 + j3| = 5##.
 
  • #7
Yes
 
  • #8
GJ1 said:
Yes
You're obviously misunderstanding something fundamental about complex numbers. ##4 +j3 \ne 5##
 
  • #9
How would I add this complex number in this equation
 
  • #10
GJ1 said:
How would I add this complex number in this equation
It's difficult to help you, because the expression you wrote in the original post is unclear. Although, I think I can see from your calculations what you mean.

Basically, the answer is a complex number. It's a simplification of the complex number you started with. The key is to express ##\dfrac 1 {4 +j3}## as a complex number in standard (Cartesian) form.

The standard method for doing this is to use the complex conjugate:
$$\dfrac 1 {4 +j3} = \dfrac {4 - j3}{(4+j3)(4-j3)} = \dots$$Does any of that look familiar?
 
  • #11
1) Make sure that you make your formulas unambiguous. Use parentheses; they are cheap.
Is that 50/( (4+j3)(50) )or ( 50/(4+j3) )(50)?
2) Often the first step in manipulating a fraction with a complex denominator ##1/(a+bj)## is to multiply both the numerator and denominator by the complex conjugate of the denominator to get ##(a-bj)/(a^2+b^2)##.
 
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  • #12
I have limited experience with complex numbers, this is the first that I've had to use them. If I see how this equation is solved, it will go along way in my understanding of them.
 
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  • #13
GJ1 said:
I have limited experience with complex numbers, this is the first that I've had to use them. If I see how this equation is solved, it will go along way in my understanding of them.
I doubt that!

Technically, we can't do your homework for you, and we expect you to have the required skills for the courses you are taking. You need to revise (or learn) Complex Numbers asap.
 
  • #14
FactChecker said:
1) Make sure that you make your formulas unambiguous. Use parentheses; they are cheap.
Is that 50/( (4+j3)(50) )or ( 50/(4+j3) )(50)?
2) Often the first step in manipulating a fraction with a complex denominator ##1/(a+bj)## is to multiply both the numerator and denominator by the complex conjugate of the denominator to get ##(a-bj)/(a^2+b^2)##.
It's (50/(4+j3)(50)
 
  • #15
GJ1 said:
It's (50/(4+j3)(50)
Which is simply ##\frac 1 {4 +j3}##.
 
  • #16
GJ1 said:
It's (50/(4+j3)(50)
Although, what I think it should be is:
$$\frac{50}{(4 + j3)(50) + 100}$$
 
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  • #17
Complex numbers have all the properties of real numbers and more.
## (50/(4+3j))*50 = 1/(4+3j) = \frac {4-3j}{(4+3j)(4-3j)} = \frac {4-3j}{4^2+3^2} = \frac {4}{25} - \frac {3}{25}j##
 
  • #18
PeroK said:
Although, what I think it should be is:
$$\frac{50}{(4 + j3)(50) + 100}$$
The OP must make clear how much is in the numerator and in the denominator using carefully placed parenthesis.
 
  • #19
GJ1 said:
It's (50/(4+j3)(50)
That clarifies nothing. Is that final (50) in the numerator or in the denominator? Also, placing '()' around that final 50 doesn't help anything.
 
  • #20
PeroK said:
I doubt that!

Technically, we can't do your homework for you, and we expect you to have the required skills for the courses you are taking. You need to revise (or learn) Complex Numbers as see

FactChecker said:
This is invalid. Please be more careful with your parenthesis. Where is the ')' that matches the first '('?
[50/(4+j3)(50)+100]x150 This is the equation
 
  • #21
GJ1 said:
[50/(4+j3)(50)+100]x150 This is the equation
I would normally interpret that as ##[\frac {50(50)}{(4+j3)} + 100 ]150##. Is that right?
 
  • #22
50 divided by (4+j3) x 50 +100
 
  • #23
The denominator is(4+j3)x50+100
 
  • #24
The answer that was given is 20-j10
 
  • #25
$$\bigg [\frac{50}{(4 + j3)(50) + 100} \bigg ] \times 150 = 20 -j10$$
 
  • #26
Yes, That is how the equation is given.
 
  • #27
GJ1 said:
It's (50/(4+j3)(50)
As already noted, this is ambiguous. Also, the parentheses aren't matched -- there are three left parens and two right parens.
PeroK said:
Which is simply 14+j3.
Assuming that the ambiguous expression above is ##\frac {50}{(4 + j3)50}##. As written, it could be interpreted as ##\frac{50}{4 + j3}\cdot 50## or ##\frac{2500}{4 + j3}##.
GJ1 said:
[50/(4+j3)(50)+100]x150 This is the equation
That is not an equation. An equation always has an = symbol somewhere in the middle.
 
  • #28
50/[(4+3j)x50+100]x150=20-10j
 
  • #29
Babadag said:
50/[(4+3j)x50+100]x150=20-10j
How did the get 20-10j as the answer
 
  • #30
For homework, we should only give hints and guidance. Use the fact that ##\frac {1}{a+bj} = \frac {a-bj}{(a+bj)(a-bj)} = \frac {a-bj}{a^2+b^2} = \frac {a}{a^2+b^2} - \frac {b}{a^2-b^2} j## and see what you get.
 
  • #31
This was not a homework assignment, it's not necessary to understand the equation for this course, I was just interested in how they derived their answer.
 
  • #32
GJ1 said:
This was not a homework assignment, it's not necessary to understand the equation for this course, I was just interested in how they derived their answer.
I should have said that we are only supposed to give hints and directions in a homework type of question. I think you should give it another try using the hint in my post #30 above.
 
  • #33
Babadag said:
50/[(4+3j)x50+100]x150=20-10j
Although you have parentheses and brackets in your work, you don't have enough of them.

50/[(4+3j)x50+100]x150 means this:
$$\frac{50}{(4 + 3j) \cdot 50 + 100} \cdot 150$$
The above equals $$\frac {7500}{(4 + 3j) \cdot 50 + 100}$$
and does not equal 20 - 10j.
 
  • #34
Mark44 said:
50/[(4+3j)x50+100]x150 means this:
$$\frac{50}{(4 + 3j) \cdot 50 + 100} \cdot 150$$
The above equals $$\frac {7500}{(4 + 3j) \cdot 50 + 100}$$
and does not equal 20 - 10j.
And yet:
$$\bigg [\frac{50}{(4 + j3)(50) + 100} \bigg ] \times 150 = \frac{150}{(4 + j3) + 2}= \frac{50}{2+ j} = \frac{(50)(2 - j)}{5} = 20 - j10 $$
 
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  • #35
Can you just explain how you get 50/2+j from 150/(4+j)+2
 
<h2>1. What are complex numbers?</h2><p>Complex numbers are numbers that consist of a real part and an imaginary part. The imaginary part is expressed using the imaginary unit, denoted by "i". Complex numbers are written in the form a + bi, where a is the real part and bi is the imaginary part.</p><h2>2. How are complex numbers used in solving circuits?</h2><p>Complex numbers are used in solving circuits because they allow us to represent both the magnitude and phase of a current or voltage. This is particularly useful in alternating current (AC) circuits, where the current and voltage are constantly changing.</p><h2>3. Can complex numbers be used in all types of circuits?</h2><p>Yes, complex numbers can be used in all types of circuits. They are particularly useful in AC circuits, but can also be used in DC circuits for calculations such as impedance and power calculations.</p><h2>4. How do you use complex numbers to solve for a current in a circuit?</h2><p>To solve for a current in a circuit using complex numbers, you first need to represent all the circuit elements (resistors, capacitors, inductors) as impedances, which are complex numbers. Then, you can use Ohm's Law and Kirchhoff's Laws to set up and solve a system of equations to find the current.</p><h2>5. Are there any limitations to using complex numbers in circuit analysis?</h2><p>While complex numbers are a powerful tool in circuit analysis, they do have some limitations. For example, they cannot account for non-linear circuit elements, and they may not accurately represent real-world conditions such as temperature and noise. In these cases, more advanced techniques may be needed.</p>

1. What are complex numbers?

Complex numbers are numbers that consist of a real part and an imaginary part. The imaginary part is expressed using the imaginary unit, denoted by "i". Complex numbers are written in the form a + bi, where a is the real part and bi is the imaginary part.

2. How are complex numbers used in solving circuits?

Complex numbers are used in solving circuits because they allow us to represent both the magnitude and phase of a current or voltage. This is particularly useful in alternating current (AC) circuits, where the current and voltage are constantly changing.

3. Can complex numbers be used in all types of circuits?

Yes, complex numbers can be used in all types of circuits. They are particularly useful in AC circuits, but can also be used in DC circuits for calculations such as impedance and power calculations.

4. How do you use complex numbers to solve for a current in a circuit?

To solve for a current in a circuit using complex numbers, you first need to represent all the circuit elements (resistors, capacitors, inductors) as impedances, which are complex numbers. Then, you can use Ohm's Law and Kirchhoff's Laws to set up and solve a system of equations to find the current.

5. Are there any limitations to using complex numbers in circuit analysis?

While complex numbers are a powerful tool in circuit analysis, they do have some limitations. For example, they cannot account for non-linear circuit elements, and they may not accurately represent real-world conditions such as temperature and noise. In these cases, more advanced techniques may be needed.

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