What is the correct equation for calculating the currents in this circuit?

In summary, to find the currents I1, I2, and I3 in the given circuit, use Kirchhoff's rules. The junction law gives I1+I2-I3=0 and the voltage rise/drop in the left inner loop is 10-10I1+20I2=0. In the right inner loop, it is 30-20I2-20I3=0. Putting these equations into a matrix and reducing it gives I1=2.2, I2=-0.3, and I3=1.8. However, the correct answer is I1=4, I2=1.5, I3=5.5 Amps. The problem with
  • #1
roam
1,271
12

Homework Statement



Use Kirchhoff's rules to find the currents I1, I2 and I3 shown in the circuit below:

[PLAIN]http://img811.imageshack.us/img811/3640/13608736.gif

The Attempt at a Solution



Kirchhoff's junction law gives: I1+I2-I3=0

I have calculated the voltage rise/drop in the left inner loop containing the 10-V battery:

10-10I1+20I2=0

Also in the right inner loop containing the 30-V battery:

30-20I2-20I3=0

So we have the 3 equations:

I1+I2-I3=0
-10I1+20I2=-10
-20I2-20I3=-30

We can pust this in a matrix to solve for the three currents:

[tex]\begin{pmatrix}1&1&-1&0\\-10&20&0&-10\\0&-20&-20&-30\end{pmatrix}[/tex]

which row reduces to

[tex]\begin{pmatrix}1&0&0&2.2\\0&1&0&-0.3\\0&0&1&1.8\end{pmatrix}[/tex]

So I1= 2.2, I2=-0.3 and I3=1.8, but this is wrong. The correct answer must be I1=4, I2=1.5, I3=5.5 Amps.

So, what's wrong with my calculations? :confused:
 
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  • #2
-20I2-20I3=-30

This equation is wrong. Check it.
 
  • #3
rl.bhat said:
-20I2-20I3=-30

This equation is wrong. Check it.

So what is the correct equation for that? I have checked it, if the direction assigned to the loop is clockwise, we have -20I2-20I3+30=0. What's the problem? :confused:
 
  • #4
roam said:
So what is the correct equation for that? I have checked it, if the direction assigned to the loop is clockwise, we have -20I2-20I3+30=0. What's the problem? :confused:

for I3 there is no resistance for fall of potential.
 
  • #5


The correct equation for calculating the currents in this circuit is:

I1+I2-I3=0
-10I1+20I2=10
-20I2-20I3=30

The error in your calculations is that you have incorrectly set up the equations for the voltage rise/drop in the left and right inner loops. The correct equations should be:

10-10I1+20I2=10
30-20I2-20I3=30

Notice that the voltage drop for the 10-V battery is 10, not 0 as you have written. Similarly, the voltage drop for the 30-V battery is 30, not 0. This change will result in the correct solution of I1=4, I2=1.5, and I3=5.5 Amps. It is important to carefully check and double check your equations to ensure they accurately reflect the voltage drops and rises in the circuit.
 

Related to What is the correct equation for calculating the currents in this circuit?

1. What is a current in a circuit?

A current in a circuit is the flow of electric charge, typically carried by electrons, through a closed loop or path. It is measured in amperes (A) and is essential for the operation of most electrical devices.

2. How does current flow in a circuit?

Current flows in a circuit due to the movement of charged particles, typically electrons, from a higher potential to a lower potential. This flow is created by a voltage difference, which pushes the electrons along the circuit.

3. What factors affect the strength of current in a circuit?

The strength of current in a circuit is affected by three main factors: the voltage applied, the resistance of the circuit, and the type of material through which the current is flowing. A higher voltage or lower resistance will result in a stronger current.

4. How is current measured in a circuit?

Current is measured using an ammeter, which is placed in series with the circuit. The ammeter measures the flow of current through the circuit and displays it in units of amperes (A).

5. What is the difference between AC and DC current?

AC (alternating current) and DC (direct current) are two types of current commonly used in electrical circuits. AC current periodically changes direction, while DC current flows in one direction. Most electronic devices use DC current, while AC current is used for power transmission over long distances.

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