Hi, I again. I'm studying, but I really stop in the basics (and important) things.
First problem:
Using the nodal analysis, what is the power in the resistor of 4 ohm?
Answer: 64 W
///
1 (leftest) - \(i_1 = i_2 + i_3\)
2 - 4 V
3 - \( i_4 + i_6 + i_5= i_7\)
And the great problem for me (I think):
What is \(v_1\)?
I put \(v_1 = -4 V\)
\(i_1 = 2A
i_2 = \frac {v_1 - 4}{2}
i_3 = \frac {v_1 - v_3}{4}
i_4 = 4 - v_3
i_5 = i_3 = \frac {v_1 - v_3}{4}
i_6 =3v_1 = -12A
i7 = \frac {v_3}{2}
\)
So...
1 - \(2 = \frac {v_1 - 4}{2} + \frac {v_1 - v_3}{4}\)
3 - \(4 - v_3 - 12 + \frac {v_1 - v_3}{4} = \frac {v_3}{2} \)
I have this system of linear equations:
\(16 = 3v_1 - v_3
-32 = -v_1 + 7v_3
v_1 = 4 V , v_2 = -4 V
\)
what is the power in the resistor of 4 ohm?
\(
i_3 = \frac {v_1 - v_3}{4} = \frac {4 + 4}{4} = 2A
P = i i R\) ->\(P = 2*2* 4 = 16W\)
////
And a little help in this:
So, how I can write \(v_1\)?
Final answer (if help): \(v = -2 V, v_1 = -4 V\)
I really appreciate your support!
Thanks!
First problem:
Using the nodal analysis, what is the power in the resistor of 4 ohm?
Answer: 64 W
///
1 (leftest) - \(i_1 = i_2 + i_3\)
2 - 4 V
3 - \( i_4 + i_6 + i_5= i_7\)
And the great problem for me (I think):
What is \(v_1\)?
I put \(v_1 = -4 V\)
\(i_1 = 2A
i_2 = \frac {v_1 - 4}{2}
i_3 = \frac {v_1 - v_3}{4}
i_4 = 4 - v_3
i_5 = i_3 = \frac {v_1 - v_3}{4}
i_6 =3v_1 = -12A
i7 = \frac {v_3}{2}
\)
So...
1 - \(2 = \frac {v_1 - 4}{2} + \frac {v_1 - v_3}{4}\)
3 - \(4 - v_3 - 12 + \frac {v_1 - v_3}{4} = \frac {v_3}{2} \)
I have this system of linear equations:
\(16 = 3v_1 - v_3
-32 = -v_1 + 7v_3
v_1 = 4 V , v_2 = -4 V
\)
what is the power in the resistor of 4 ohm?
\(
i_3 = \frac {v_1 - v_3}{4} = \frac {4 + 4}{4} = 2A
P = i i R\) ->\(P = 2*2* 4 = 16W\)
////
And a little help in this:
So, how I can write \(v_1\)?
Final answer (if help): \(v = -2 V, v_1 = -4 V\)
I really appreciate your support!
Thanks!