: Finding the Current and Power across each Resistor in a Complicated Network

In summary, the conversation discusses a complicated electrical circuit with a 12V battery connected across two points. The individual is asked to determine the equivalent resistance, total current, and total power of the circuit, as well as the current and power for each individual resistor. The conversation includes a discussion on how to approach the problem and the importance of understanding basic concepts in solving it.
  • #1
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URGENT: Finding the Current and Power across each Resistor in a Complicated Network

Homework Statement


For the circuit below, a 12 V batter is connected across the points a and b. Determine the following:

a) the equivalent resistance between points a and b
b) the total current flowing through the resistor network
c) the total power dissipated by the resistor network
d) the current through each individual resistor and the power dissipated by each resistor. Sum up the power dissipation from each and compare that sum to your result from c. Is this what you expect?
pp2.png


Homework Equations


series: Req=R1+R2+...
parallel: 1/Req=1/R1+1/R2+...
I=V/R
P=I^2R


The Attempt at a Solution


a) I know how to do this part. I get 6 ohms.
b) using I=V/R=12/6=2A
c) P=I^2R=2^2*6=24 Watts
d) I have no idea how to do this part. How does the Voltage split up? Does 6V go into the top part then the 6 splits into 3 for each part within the part? Then combine to form 6 again once it comes out the square in the top part?
 
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  • #2


There's 12 across the top part and 12 V across the bottom part as well.
Find the equivalent resistance of the top and the bottom part and then
calculate the currents through both parts.
 
  • #3


for top Req=9 ohms
bottom Req=18 ohms

top (using V=IR) I=12/9=4/3 Amps
bottom I=18/9=2 Amps

now how do i find the currents in each resistor?
 
  • #4


the best way to do a problem like this is to add up all the resistors and redraw the circuit each time you combined two resistors. This way, once you have the final resistance, you can find the total current, total voltage, total power, etc. then you can work your way backwards by using the circuits you drew.

Remember that current is the same for resistors in series and voltage is the same for resistors in parallel

know your concepts, it will make problems like these 100% easier.
 
  • #5


johnnyies said:
the best way to do a problem like this is to add up all the resistors and redraw the circuit each time you combined two resistors. This way, once you have the final resistance, you can find the total current, total voltage, total power, etc. then you can work your way backwards by using the circuits you drew.

Remember that current is the same for resistors in series and voltage is the same for resistors in parallel

know your concepts, it will make problems like these 100% easier.

I did as you suggested. The only part that's confusing me is the box on the top part of the circuit. with the 6, 2, and 4 ohm resistors. Since they are in parallel, the top and bottom both have 12 Volts going through them. So (using V=IR) does that mean there are 2 Amps going through the 6 ohm resistor. And 6 and 3 amps going through the 2 and 4 ohm resistors respectively?
 

1. How do I find the current across each resistor in a complicated network?

To find the current across each resistor in a complicated network, you can use Kirchhoff's Current Law (KCL) which states that the sum of all currents entering and exiting a node must equal zero. By setting up a system of equations using KCL, you can solve for the current in each branch of the network.

2. What is Kirchhoff's Current Law and how is it used to find current?

Kirchhoff's Current Law (KCL) is a fundamental law in circuit analysis that states that the algebraic sum of all currents entering and exiting a node in a circuit must equal zero. This law is used to find the current in each branch of a network by setting up a system of equations based on the currents entering and exiting a given node.

3. How do I find the power across each resistor in a complicated network?

To find the power across each resistor in a complicated network, you can use the formula P = I^2*R, where P is power in watts, I is current in amps, and R is resistance in ohms. By calculating the current through each resistor using KCL and then plugging it into the formula, you can determine the power across each resistor.

4. Is it necessary to use Kirchhoff's Current Law to find the current and power in a complicated network?

Yes, it is necessary to use Kirchhoff's Current Law to accurately determine the current and power in a complicated network. This law is based on the principle of conservation of charge and is a fundamental tool in circuit analysis.

5. Are there any other methods or laws that can be used to find the current and power in a complicated network?

Yes, there are other methods and laws that can be used to find the current and power in a complicated network, such as Kirchhoff's Voltage Law (KVL) and Ohm's Law. However, KCL is often the most efficient and accurate method for solving complicated networks. It is also important to note that these laws are all interconnected and can be used together to solve for different variables in a circuit.

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