Circuit with 2 batteries and 6 resistors

In summary: I see the mistake I made in my calculation. In summary, the circuit consists of six resistors and two batteries with voltages V1 = 18 V and V2 = 12 V. The values for the resistors are R1 = R5 = 73 Ω, R2 = R6 = 121 Ω, R3 = 62 Ω, and R4 = 92 Ω. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows. The relationships between the currents are I2=I6 and I4=I5. By using a combination of KVL and KCL equations, we can solve for the unknown currents I1, I
  • #1
riane
6
0

Homework Statement



A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 73 Ω, R2 = R6 = 121 Ω R3 = 62 Ω, and R4 = 92 Ω. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows. What is V4, I1, I2, and I3?

Homework Equations


V=IR


The Attempt at a Solution


-I3R3+I1R1-ΔV1=0
+I6R6-ΔV2+I2R2+I1R1=0
-I1R1-I2R2-I5R5-I4R4-I6R6=0
-ΔV1-I3R3-I2R2+ΔV2-I6R6=0
-ΔV2-I5R5-I4R4=0
-I4R4-I6R6-ΔV1-I3R3-I2R2-I5R5=0

Link to picture of circuit:
https://www.physicsforums.com/attachment.php?attachmentid=43882&d=1329194903
 
Physics news on Phys.org
  • #2
riane said:

Homework Statement



A circuit is constructed with six resistors and two batteries as shown. The battery voltages are V1 = 18 V and V2 = 12 V. The positive terminals are indicated with a + sign, The values for the resistors are: R1 = R5 = 73 Ω, R2 = R6 = 121 Ω R3 = 62 Ω, and R4 = 92 Ω. The positive directions for the currents I1, I2 and I3 are indicated by the directions of the arrows. What is V4, I1, I2, and I3?

Homework Equations


V=IR


The Attempt at a Solution


-I3R3+I1R1-ΔV1=0
+I6R6-ΔV2+I2R2+I1R1=0
-I1R1-I2R2-I5R5-I4R4-I6R6=0
-ΔV1-I3R3-I2R2+ΔV2-I6R6=0
-ΔV2-I5R5-I4R4=0
-I4R4-I6R6-ΔV1-I3R3-I2R2-I5R5=0

Link to picture of circuit:
https://www.physicsforums.com/attachment.php?attachmentid=43882&d=1329194903

You've written a number of KVL equations which appear to be correct assuming that the directions of the currents not specified in the diagram are chosen appropriately.

You should now consider writing some KCL equations at junctions so that you can eliminate superfluous currents (for example, note that I4 and I5 are really the same current since the branch they're in is series-connected. How are I2 and I6 related?

Once you've established the relationships between the currents you should be in a position to choose a subset of your equations and solve for the unknowns (simultaneous equations). Choose a set of equations such that each component appears at least once in the set.

Note: The placement of voltage source V2 effectively isolates the rightmost loop from the rest of the circuit. Can you see why?
 
  • #3
I already established the relationships between the currents: I2=I6 and I4=I5; I2=I1+I3; R2=R6

I already have worked out the problem using:
-I3R3+I1R1-ΔV1=0,
+I6R6-ΔV2+I2R2+I1R1=0, and I2=I1+I3, and I keep getting I1=.147 or .10 or .11, and all three answers for I1 are wrong. I see that V2 isolates the rightmost loop from the rest of the circuit, but I don't know what That is supposed to do for me. Is it supposed to give me I4 and I5 with the equation -ΔV2-I5R5-I4R4=0? If I have that, then how is it going to help me find I1 I2 and I3 any different than solving for individual variables like I already have above? I feel like I'm just going in circles.
 
  • #4
riane said:
I already established the relationships between the currents: I2=I6 and I4=I5; I2=I1+I3; R2=R6

I already have worked out the problem using:
-I3R3+I1R1-ΔV1=0,
+I6R6-ΔV2+I2R2+I1R1=0, and I2=I1+I3, and I keep getting I1=.147 or .10 or .11, and all three answers for I1 are wrong. I see that V2 isolates the rightmost loop from the rest of the circuit, but I don't know what That is supposed to do for me. Is it supposed to give me I4 and I5 with the equation -ΔV2-I5R5-I4R4=0? If I have that, then how is it going to help me find I1 I2 and I3 any different than solving for individual variables like I already have above? I feel like I'm just going in circles.

The effect of V2 means that you can completely ignore R4 and R5 and their current when you are looking to find I1, I2, I3.

Can you show the steps you used to solve your equations to find I1? Maybe we can locate where things are going astray.
 
  • #5
Here is an attatched image of the work done in altering the equations and plugging in the unknowns.
 

Attachments

  • homework image.jpg
    homework image.jpg
    15.3 KB · Views: 1,772
  • #6
riane said:
Here is an attatched image of the work done in altering the equations and plugging in the unknowns.

Voltage supply mishap:
attachment.php?attachmentid=55854&stc=1&d=1361149417.gif


Otherwise, looks okay.
 

Attachments

  • Fig2.gif
    Fig2.gif
    70.2 KB · Views: 4,539
  • #7
Thank you!
 

FAQ: Circuit with 2 batteries and 6 resistors

1. What is a circuit with 2 batteries and 6 resistors?

A circuit with 2 batteries and 6 resistors is a type of electrical circuit that consists of two power sources (batteries) and six components that resist the flow of electricity (resistors).

2. How does a circuit with 2 batteries and 6 resistors work?

In this type of circuit, the batteries supply the electrical energy, while the resistors control the flow of electricity by reducing the voltage. The resistors are connected in series, which means that the electrical current must pass through each one in order to complete the circuit.

3. What is the purpose of using 2 batteries in this circuit?

The use of two batteries in this circuit allows for a higher voltage to be supplied, which can be useful for powering more complex electronic devices or for longer periods of time.

4. How do the resistors affect the current in this circuit?

The resistors act as obstacles to the flow of electrical current, causing a decrease in the current as it passes through each one. This is known as resistance and is measured in ohms. The more resistors in a circuit, the greater the overall resistance and the lower the current.

5. What happens if one of the resistors in this circuit fails?

If one of the resistors in this circuit fails, it will create an open circuit and the flow of electricity will be interrupted. This can cause the circuit to stop working or for the remaining resistors to experience a higher voltage, potentially damaging them. It is important to regularly check and replace any faulty components in a circuit to ensure proper functioning.

Back
Top