EECS: Find i_0 in a circuit with 4 resistors and 1 indep. current source

In summary: It's easy to make a mistake like this, so always double check your signs and equations. Keep up the good work!In summary, in order to determine i_0 in a circuit using nodal analysis, the first step is to use KCL at each node and set the sum of currents entering the node equal to zero. Then, using the values from the resulting equations, a coefficient matrix can be created and solved to find the values of v_1 and v_2. Finally, the value of i_0 can be calculated by using linearity and plugging in the values of v_1 and v_2. It is important to pay attention to signs and equations in order to avoid mistakes and ensure accurate results.
  • #1
VinnyCee
489
0
(a) Using nodal analysis, determine [itex]i_0[/itex] in the circuit.

(b) Now use linearity to find [itex]i_0[/itex].


ch4prob4.jpg



Work so far:

[tex]i_0\,=\,\frac{v_1\,-\,0}{6\,\ohm}\,=\,\frac{v_1}{6\,\ohm}[/tex]

KCL@[itex]v_1[/itex]: [tex]\frac{v_1\,-\,0}{3\,\ohm}\,+\,\frac{v_1\,-\,0}{6\,\ohm}\,+\,\frac{v_1\,-\,v_2}{2\,\ohm}\,=\,0[/tex]

KCL@[itex]v_2[/itex]: [tex]\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,+\,9A\,=\,0[/tex]

[tex]v_1\,-\,\frac{1}{2}\,v_2\,=\,0[/tex]

[tex]-\,\frac{1}{2}\,v_1\,+\,\frac{3}{4}\,v_2\,=\,-9[/tex]

Whe I put these two equaitons into a 3 X 2 coefficient matrix, I get

[tex]v_1\,=\,-\,9V[/tex]
[tex]v_2\,=\,-\,18V[/tex]

Then I plug into the first equation:

[tex]i_0\,=\,\frac{v_1}{6\,\ohm}\,=\,\frac{(-\,9V)}{6\,\ohm}\,=\,-\,\frac{3}{2}A[/tex]

Is this the correct value for [itex]i_0[/itex] or have I got the sign wrong?
 
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  • #2
VinnyCee said:
KCL@[itex]v_2[/itex]: [tex]\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,+\,9A\,=\,0[/tex]
You are repeating the same mistake here as you did in the previous exercise.
 
  • #3
What is that, can you explain?
 
  • #4
VinnyCee said:
What is that, can you explain?
The sign on the 9A is wrong. The other two terms in that KCL are for currents *out* of the node. So since the 9A is into the node, it has to be -9A in that equation. Makes sense?
 
  • #5
OIC! There has to be a minus in there somewhere if all of the terms are on one side and it's equal to zero. Whoops!

Using your correction:

KCL @ [itex]v_2[/itex]: [tex]\frac{v_2\,-\,v_1}{2\,\ohm}\,+\,\frac{v_2\,-\,0}{4\,\ohm}\,-\,9A\,=\,0[/tex]

[tex]-\,\frac{1}{2}\,v_1\,+\,\frac{3}{4}\,v_2\,=\,9[/tex]

Which switches the signs on the [itex]v_1[/itex] and [itex]v_2[/itex]!

[tex]i_0\,=\,\frac{v_1}{6\,\ohm}\,=\,\frac{(9V)}{6\,\ohm}\,=\,\frac{3}{2}A[/tex]
 
  • #6
Yes, that's much better. And be careful when you formulate the KCLs next time.
 

1. How do I calculate i_0 in a circuit with 4 resistors and 1 independent current source?

To calculate i_0 in this type of circuit, you will need to use Ohm's Law, which states that current (i) is equal to voltage (V) divided by resistance (R). In this case, you will need to calculate the total resistance of the circuit using the resistors and then use the value of the independent current source to solve for i_0.

2. Can I use Kirchoff's Laws to find i_0 in this circuit?

Yes, you can use Kirchoff's Laws to solve for i_0 in this circuit. Kirchoff's Current Law states that the sum of all currents entering a node must equal the sum of all currents leaving the node. Kirchoff's Voltage Law states that the sum of all voltage drops in a closed loop must equal the sum of all voltage sources in the loop.

3. How do I determine the direction of i_0 in this circuit?

The direction of i_0 will depend on the polarity of the independent current source. If the current source is supplying current, then the direction of i_0 will be in the same direction as the current source. If the current source is sinking current, then the direction of i_0 will be opposite to the direction of the current source.

4. Can I simplify this circuit to make it easier to calculate i_0?

Yes, you can simplify this circuit by using series and parallel resistance rules. For example, if there are resistors in parallel, you can use the equation 1/R_total = 1/R_1 + 1/R_2 + ... to calculate the total resistance. Similarly, if there are resistors in series, you can simply add their values to find the total resistance.

5. What units will i_0 be measured in?

i_0 will be measured in amps (A), as current is typically measured in this unit. If you are using different units, such as milliamps (mA) or microamps (uA), make sure to convert the units accordingly in your calculations.

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