Solve Current Through 3 Ohm Resistor: Kirchoff's Rule & I=V/R

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In summary, the conversation discusses a problem involving determining the current through a 3 Ohm resistor in a circuit with two voltage sources. The problem is approached using the Branch-Current method and Kirchoff's Voltage and Current laws. The Superposition theorem is also mentioned as a possible method for solving the problem.
  • #1
TitaniumX
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Homework Statement



What is the current through the 3 Ohms resistor in the diagram below?

30cqhkx.jpg



Homework Equations


I=V/R and Kirchoff's Rule



The Attempt at a Solution


I tried to simplify the circuit by adding the 5ohm and 1ohm resistors because they are in series, an the same thing with the 3ohm and 1ohm resistor. This problem has two voltage source so it's kind of confusing. I think I may have to use Kirchoff's Rules but I'm not sure how to approach this to get the answer. Any help would be great.
 
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  • #2
So the best way to answer this without just giving you the answer it to solve a similar question so with that in mind, I will use this circuit.

cFhwlQY.png


There are other methods out there but I'm going to use the Branch-Current method (look up loop current method).

Step 1 is to define the branch currents and the direction that they are acting in.
Steb 2 Define the voltages in terms of the currents
[tex]V_{ab} = 2I_1[/tex]
[tex]V_{bd} = 8I_3[/tex]
[tex]V_{cb} = 4I_2[/tex]

Step 3 using Kirchoffs Voltage law we get

[tex]ƩV_{abda} = 0 = V_{ab} + V_{bd} + V_{da}[/tex]
[tex]ƩV_{bcdb} = 0 = V_{cb} + V_{bd} + V_{dc}[/tex]

These can be re written as

[tex]ƩV_{abda} = V_{ab} + V_{bd} + V_{da} = 2I_1 + 8I_3 - 32 [/tex]
[tex]ƩV_{bcdb} = V_{cb} + V_{bd} + V_{dc} = 4I_2 + 8I_3 - 20 [/tex]

then finally You use Kichchoffs current laws at node b to get your final equation.

[tex]ƩI_b = 0 = I_1 + I_2 - I_3[/tex]

this gives the 3 simultaneous equations

[tex]I_1 + I_2 - I_3= 0[/tex]
[tex]2I_1 + 8I_3 = 32 [/tex]
[tex]4I_2 + 8I_3 = 20 [/tex]

Solving gives the currents [itex]I_1 = 4[/itex], [itex]I_2 = -1[/itex] and [itex]I_3= 3[/itex].

Hope that helps you.
 
  • #3
TitaniumX said:
1This problem has two voltage source so it's kind of confusing.


If two voltage sources confuse you, this is an excellent time to invoke the Superposition theorem.

So make the first source zero volts (short) and calculate the current thru the 3 ohm = i1.

Then repeat, shorthing out the second source & working with the first. Calculate i2.

Then when both sources are active the answer is i1 + i2.
 

FAQ: Solve Current Through 3 Ohm Resistor: Kirchoff's Rule & I=V/R

What is Kirchoff's Rule?

Kirchoff's Rule, also known as Kirchoff's Current Law, states that the sum of currents entering a junction in a circuit must equal the sum of currents leaving the junction. In other words, charge is conserved within a closed circuit.

How do I use Kirchoff's Rule to solve for current through a 3 Ohm resistor?

To solve for current through a 3 Ohm resistor using Kirchoff's Rule, you will need to set up a system of equations based on the currents entering and leaving the junction where the resistor is located. Use Ohm's Law (I=V/R) to determine the voltage drop across the resistor, and then use Kirchoff's Rule to solve for the current.

What is the formula for Ohm's Law?

The formula for Ohm's Law is I=V/R, where I represents current in amps, V represents voltage in volts, and R represents resistance in ohms.

Can Kirchoff's Rule be used for both series and parallel circuits?

Yes, Kirchoff's Rule can be used for both series and parallel circuits. In series circuits, the total current entering a junction will equal the total current leaving the junction. In parallel circuits, the sum of the currents entering the junction will equal the sum of the currents leaving the junction.

What are some common mistakes when using Kirchoff's Rule to solve for current?

Some common mistakes when using Kirchoff's Rule include forgetting to account for the direction of the current, not setting up the equations correctly, and not using the correct values for voltage and resistance. It is important to carefully consider the direction of the current and double check all calculations to avoid errors.

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