Solve Network of Resistors: R1, R2, R3, R4, V=6V

In summary, The given circuit consists of four resistors (R1 = 3W, R2 = 6W, R3 = 11W, R4 = 8W) connected in a series-parallel configuration and a battery with a voltage of 6V. The current through R1 is -2A. To find an equivalent resistor for the entire circuit, R1 and R4 are combined in series and then added to R2 and R3 using the equation for parallel resistors. The total resistance is 2.87W. The current supplied by the battery is unknown. To find the current through R2, the voltage across R2 must be determined by calculating the voltage at the node between R2
  • #1
sarah895
3
0

Homework Statement



This is the diagram:
https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1202/summer/homework/Ch-21-DC-Circuits/resistor_network/ex1s95p3.gif [Broken]

R1 = 3W, R2 = 6W, R3 = 11W, R4 =8W, V=6V

A) What is the current through the resistor, R1 in the above circuit? -2A

B) What single, equivalent resistor could replace all of the resistors in this circuit?

C) What is the current supplied by the battery?

D) What is the current through the resistor, R2 ?


Homework Equations



1/R (parallel) = 1/R1 + 1/R2 + 1/R3 + 1/R4
R (series) = R1 + R2 + R3 + R4
E = IR

The Attempt at a Solution



I solved part a. For part be, it seemed to me that R1 and R4 were connected in series, so I combined those two resistors by adding the resistances. Then I added that resistance to R2 and R3 using the equation for parallel resistors.

So, I had: 1/R = (1/11W) + (1/6W) + (1/11W) = 2.87W. This answer is wrong, and I am not sure what to do. Any help would be greatly appreciated :)
 
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  • #2
Okay, so I solved parts b and c also. Now I am working on part d.

I thought that I2 would = V/R2 = 6/6 = 1, but this is wrong. Any ideas?

Thanks,
Sarah
 
  • #3
You need the correct potential drop across the resistor. I'll redraw the circuit for you, convince yourself that the one I have drawn is equivalent. I hope it will then be clearer to you.

Redrawing the circuit into this form is often a good idea when solving such problems. It's now very clear which parts are in parallel, and which are in series.
 

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  • #4
sarah895 said:
Okay, so I solved parts b and c also. Now I am working on part d.

I thought that I2 would = V/R2 = 6/6 = 1, but this is wrong. Any ideas?

Thanks,
Sarah

The voltage across R2 is necessarily different from the 6V. Because the voltage at that node between R2||R4 and R3 is determined by 6v * R3 / (R3 + (R2||R4)). Subtract that from 6v to give the drop across R2. With that voltage drop divided by R2 you should have I through R2.
 

What is a network of resistors?

A network of resistors is a circuit that consists of multiple resistors connected in a specific pattern. The resistors work together to regulate the flow of electrical current in the circuit.

How do I solve for the total resistance in a network of resistors?

To solve for the total resistance in a network of resistors, you can use the formula R = R1 + R2 + R3 + ... where R represents the total resistance and R1, R2, R3, etc. represent the individual resistances of each resistor. Alternatively, you can use Kirchhoff's laws and Ohm's law to solve for the total resistance.

What is the role of voltage in a network of resistors?

Voltage is the driving force that causes current to flow through a network of resistors. It is the potential difference between the two ends of the resistors that allows electrons to flow through the circuit.

How do I calculate the current flowing through each resistor in a network?

To calculate the current flowing through each resistor in a network, you can use Ohm's law, which states that current (I) is equal to voltage (V) divided by resistance (R). So, I = V/R. You can also use Kirchhoff's current law, which states that the total current entering a junction in a circuit is equal to the total current leaving the junction.

What happens to the total resistance in a network when resistors are connected in series or parallel?

In a series circuit, the total resistance is equal to the sum of the individual resistances. This means that the total resistance will increase as more resistors are added. In a parallel circuit, the total resistance is less than the individual resistances since the resistors are connected across the same voltage source. This means that the total resistance will decrease as more resistors are added.

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