Using Kirchhoff's rules to calculate current

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Homework Help Overview

The discussion revolves around using Kirchhoff's laws to calculate the currents in a circuit diagram. The original poster seeks assistance with the third loop of the circuit, indicating some confusion regarding the application of these laws.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Kirchhoff's laws, with one member asking for clarification on the original poster's confusion regarding the third loop. There is a request for the original poster to share their attempts at a solution to facilitate better assistance.

Discussion Status

The discussion includes attempts to solve for the currents I1, I2, and I3, with specific equations provided for each loop. Guidance has been offered regarding the application of Kirchhoff's Current Law (KCL) at the nodes where currents join or split, suggesting a need for careful labeling of net currents.

Contextual Notes

Participants note the importance of applying KCL in conjunction with Kirchhoff's Voltage Law (KVL) to accurately analyze the circuit, particularly at the junctions where currents converge.

trevorwisc
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1. Calculate the three currents indicated in the circuit diagram below.

I have attached the diagram.

How would I use Kirchhoff's laws to find the currents I am getting thrown off by the third loop.
 

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trevorwisc said:
1. Calculate the three currents indicated in the circuit diagram below.

I have attached the diagram.

How would I use Kirchhoff's laws to find the currents I am getting thrown off by the third loop.


How are you "getting thrown off by the third loop"? You'll have to show your attempt at a solution so we can see how to help.
 
gneill said:
How are you "getting thrown off by the third loop"? You'll have to show your attempt at a solution so we can see how to help.

Solution attempt:

I1 loop 9.00V-9.00(I1)=0 so I1=1A
I2 loop 12.00v-6.00(I2)=0 so I2=2A
For I3 loop 12.00V-10(I3)-8(I1)-1(I2)=0 by plugging in I1 and I2 I get I3 to be 5A.
 
trevorwisc said:
Solution attempt:

I1 loop 9.00V-9.00(I1)=0 so I1=1A
I2 loop 12.00v-6.00(I2)=0 so I2=2A
For I3 loop 12.00V-10(I3)-8(I1)-1(I2)=0 by plugging in I1 and I2 I get I3 to be 5A.

Ah. Don't forget to apply KCL at the nodes where currents join/split. For the "I1 loop", for example, I3 joins the party at the rightmost node, so the current through the 8.00 Ohm resistor is actually I1 + I3. Label the net currents on wires around the nodes.
 

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