2i-6i1+4i2=0 3i+4i1-9i2=0 5i-3i1-5i2=0 check i1

• ritzmax72
In summary, according to the author, if you assume the current direction as shown in the diagram, you get three equations. One of the equations (with I subscript 1) is i=0, meaning that i1=0 and i2=0. The author tried solving this equation using matrix methods, but didn't get a solution. The author also tried solving the equation using row echelon form, but didn't get any results. The author thinks that there might be something wrong with him, but he doesn't know what it is.
ritzmax72
In this diagram(made at last) if i assume the curren direction as shown, I get 3 equations:
2i-6i1+4i2=0
3i+4i1-9i2=0
5i-3i1-5i2=0
check i1 is i subscript 1.(i don't know how to write subscript).
if i put in matrix form( i want to solve via matrix method only).
i am getting i=0 i1=0 i2=0, used cramer,s rule.
also tried row echeleon form. but didn't get any solution.
am i right in giving curret direction? is there any wrong convection? am i forgetting any rule.
in book the author has taken x y z(x and y in leftmost two branch AND Z IN MIDDLE branch).
he gets a good matrix with 0,0,2 in right hand side after "=" sign.(i m not writing those equations)
tell me what is the wrong with me?please neede within 1 day!

I'm sorry but that diagram is hard to read. I can tell you though that the system of equations represent voltages in a loop that sum to zero. Your diagram shows a 2v battery and your system of equations do not show this voltage.

That maybe one of the problems...

That means i should write
equation=2 volts?
But kvl says "the algebric sum of the products of currents and resistances in each of the conductor in any closed path(mesh) in a network plus the algebric sum of the emfs in that path is zero"
taking this def. Along the rhombhus path. I don't get any emf. There's only one emf i.e outermost.

The voltages across either two of the outer series resistors equals the supply voltage.

So, you would get I1 * 1 + (I1 - I2) * 2 = 2 ...Where the final 2 is the supply voltage

Yes u have done right thing.
I1*1-(I1-I2)=2. I UNDERSTAND.
BUT WHEN WE USE KVL, THEN WE CHOOSE A LOOP.THEN WE TRAVEL AROUND THAT LOOP ADD THE VOLTAGE DROPS. WE TAKE PROPER SIGN CONVENTION. THAT IS IF WE TRAVEL IN DIRECTION OF ASSUMED CURRENT THEN WE TAKE VOLTAGE DROP AS NEGATIVE AND VICE VERSA. TAKING THESE THINGS AND THREE LOOPS I HAVE WRITTEN 3 EQUATIONS. EACH EQUALS TO 0. SO I DON'T GET ANY RESULT. WHAT WRONG HAVE I DONE?

You haven't done anything wrong, but the currents in the circuit must depend on the supplied voltage, so you have to include equations that have this voltage in them.

There are two resistors at the top and two at the bottom that are across 2 volts. So you can make equations using that information.

The total current from the 2 V supply in your circuit is 1.068 amps.

Fine. Thanks all.

1. What does "i" represent in the equations?

"i" represents the imaginary unit, which is defined as the square root of -1 in mathematics.

2. What do the subscripts "1" and "2" indicate in the equations?

The subscripts "1" and "2" represent different variables or components in the equations. For example, in the equation 2i-6i1+4i2=0, "i1" and "i2" represent the coefficients of the imaginary unit "i".

3. What does it mean to "check i1" in the last equation?

To "check i1" in the last equation means to substitute the value of i1 into the equation and solve for the remaining variables. This is done to verify the solution and ensure that it satisfies all three equations.

4. How many solutions are there for this system of equations?

There are infinitely many solutions for this system of equations. This is because there are three equations but only two unknown variables, which means that one variable can be expressed in terms of the other two.

5. Can this system of equations be solved using traditional algebraic methods?

No, this system of equations cannot be solved using traditional algebraic methods. It involves complex numbers and the equations are not linear, so special methods such as substitution or elimination of variables must be used to solve it.

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