Equation

[shermaine]

May i know how to get this answer for these two equation sloving simultaneously?

B2 = j0.99a1 - 0.1b2 + 0.02b3

B3 = 0.1a1 - j0.99b2 + 0.02b3

And sub them into

B1 = 0.1a1 - 0.1b2 + 0.02b3

Please advise. Thanks.

 

[normad]

what is it that you want to do? do you want to write B1 interms of B2 & B3?

 

[Triode]

So Im guessing that the "j" is a typo, otherwise you have 3 equations with 4 unknowns and you cannot solve. I'm guessing this is part of a homework assignment, so I'll just give you a tip. You need to solve all three simultaneously, you cannot solve two and then plug them into the third one. The two main methods of doing this are substitiution, and putting it into a matrix and converting it to row reduced eschelon form.

 

[MrAl]

Hello there,


Here is one way to solve those equations...


Start with the two equations:
B2=j*0.99*a1-0.1*b2+0.02*b3
B3=0.1*a1-j*0.99*b2+0.02*b3

Subtract the b3 term from both sides, and we get:
B2-0.02*b3=j*0.99*a1-0.1*b2
B3-0.02*b3=0.1*a1-j*0.99*b2

Note that we now have two equations in two unknowns a1 and b2, treating b3 as a constant.

To work in whole numbers, multiply both sides of both equations by 100, and we get:
100*B2-2*b3=99*j*a1-10*b2
100*B3-2*b3=10*a1-99*j*b2

Next we want to solve for a1 so to get rid of b2 we multiply the top equation by (9.9*j), and we get:
990*j*B2-19.8*j*b3=-980.1*a1-99*j*b2
100*B3-2*b3=10*a1-99*j*b2

Now subtract the bottom equation from the top and that eliminates b2, and we get:
-100.0*B3+990*j*B2-19.8*j*b3+2*b3=-990.1*a1

To work again in whole numbers multiply both sides by 10, and we get:
-1000*B3+9900*j*B2+(20-198*j)*b3=-9901*a1

To solve for a1 divide both sides by -9901, and we get:
a1=(-1000*B3+9900*j*B2+(20-198*j)*b3)/-9901

At this point we now have a1 solved for, in terms of the B constants and the variable b3.
Next we would solve for b2 in the same manner, and that would give us both a1 and b2 in
terms of the B constants and the variable b3, which we could then substitute into the last
equation B1=0.1*a1-0.1*b2+0.02*b3, and after doing that we would have one equation in
only one variable b3. We could then solve for that b3 and then go back and insert it into the
previous solutions for a1 and b2 and that would give us the solutions for all three variables
a1, b2, and b3.

 

[shermaine]

So Im guessing that the "j" is a typo, otherwise you have 3 equations with 4 unknowns and you cannot solve. I'm guessing this is part of a homework assignment, so I'll just give you a tip. You need to solve all three simultaneously, you cannot solve two and then plug them into the third one. The two main methods of doing this are substitiution, and putting it into a matrix and converting it to row reduced eschelon form.

j is an imaginary

 

[shermaine]

Hello there,


Here is one way to solve those equations...


Start with the two equations:
B2=j*0.99*a1-0.1*b2+0.02*b3
B3=0.1*a1-j*0.99*b2+0.02*b3

Subtract the b3 term from both sides, and we get:
B2-0.02*b3=j*0.99*a1-0.1*b2
B3-0.02*b3=0.1*a1-j*0.99*b2

Note that we now have two equations in two unknowns a1 and b2, treating b3 as a constant.

To work in whole numbers, multiply both sides of both equations by 100, and we get:
100*B2-2*b3=99*j*a1-10*b2
100*B3-2*b3=10*a1-99*j*b2

Next we want to solve for a1 so to get rid of b2 we multiply the top equation by (9.9*j), and we get:
990*j*B2-19.8*j*b3=-980.1*a1-99*j*b2
100*B3-2*b3=10*a1-99*j*b2

Now subtract the bottom equation from the top and that eliminates b2, and we get:
-100.0*B3+990*j*B2-19.8*j*b3+2*b3=-990.1*a1

To work again in whole numbers multiply both sides by 10, and we get:
-1000*B3+9900*j*B2+(20-198*j)*b3=-9901*a1

To solve for a1 divide both sides by -9901, and we get:
a1=(-1000*B3+9900*j*B2+(20-198*j)*b3)/-9901

At this point we now have a1 solved for, in terms of the B constants and the variable b3.
Next we would solve for b2 in the same manner, and that would give us both a1 and b2 in
terms of the B constants and the variable b3, which we could then substitute into the last
equation B1=0.1*a1-0.1*b2+0.02*b3, and after doing that we would have one equation in
only one variable b3. We could then solve for that b3 and then go back and insert it into the
previous solutions for a1 and b2 and that would give us the solutions for all three variables
a1, b2, and b3.

hi mral,

thanks so much! i will look through it

 

[Triode]

j is an imaginary

oh, haha, I should have caught that, I'm an engineer. I just still expect i when theres an imaginary cause I'm not an electrical engineer.