Using Kirchhoff's Rules to Solve a Circuit Problem with Resistors and Batteries

In summary, three resistors and two 10.0 volt batteries are arranged as shown in the circuit diagram. The current to the node supplied by battery 1 is equal to 10V/40 ohms which gives ".25 amps". The current to the node supplied by battery 2 is equal to 10V/10 ohms which gives "1 amp". Adding the two results together would be 1.25 amps at the node. If I use ohms law: I= V/R, the current supplied by battery 1 isn't 10/40 because the 40 ohms isn't connected to ground. The current supplied by battery 2 is equal to 10V/10 ohms which gives "1 amp". If I use the node voltage method: I
  • #1
William Bush
29
0
1. Three resistors and two 10.0 volt batteries are arranged as shown in the circuit diagram. Which of the following entries in the table is correct?



2. P= I^2 X R, P= V^2/R, P= I X V



3. I've been working on this problem for 45 minutes and can't find a way forward. My instructor has not lectured on this material yet and my textbook hasn't been much help. I think this problem involves Kirchhoff's junction rule or his loop rule. I've attached a pic of the circuit and the related table. The correct answer is highlighted but I haven't been able to solve the problem. Any assistance will be greatly appreciated. Thanks in advance!
 

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  • #2
The power delivered by a voltage source... is the voltage * current coming out of the positive terminal of the voltage source...

Hint: find the voltage at the node joining the 3 resistors... then you can get all the currents..
 
  • #3
I'm not quite sure how to go about finding the voltage at the node joining the 3 resistors.

If I use ohms law: I= V/R

The current to the node supplied by battery 1 is equal to 10V/40 ohms which gives ".25 amps". The current to the node supplied by battery 2 is equal to 10V/10 ohms which gives "1 amp". Adding the two results together would be 1.25 amps at the node. Am I on the right track? If so, how does the 12 ohm resistor in the middle come into play?
 
  • #4
William Bush said:
I'm not quite sure how to go about finding the voltage at the node joining the 3 resistors.

If I use ohms law: I= V/R

The current to the node supplied by battery 1 is equal to 10V/40 ohms which gives ".25 amps". The current to the node supplied by battery 2 is equal to 10V/10 ohms which gives "1 amp". Adding the two results together would be 1.25 amps at the node. Am I on the right track? If so, how does the 12 ohm resistor in the middle come into play?

No, the current supplied by battery 1 isn't 10/40 because the 40 ohms isn't connected to ground... have you studied the node voltage method... let the voltage at the node by x... Now using KCL (sum of currents coming out = 0) at that node... can you get an expression for x?
 
  • #5
I have not studied the node voltage method yet. However, I am looking at my textbook and I see that there is a rule that states that the voltage into the node is equal to the voltage coming out of the node. Is this what you are talking about?
 
  • #6
William Bush said:
I have not studied the node voltage method yet. However, I am looking at my textbook and I see that there is a rule that states that the voltage into the node is equal to the voltage coming out of the node. Is this what you are talking about?

must be current in and current out... yeah... kirchhoffs current law...

There's another way you can do this... use the principle of superposition... short out one of the voltage sources... then find the currents through the 40 ohm and 10 ohm resistors... then turn that one back on, and short out the other voltage source... again find the currents through the 40 ohm and 10 ohm...

The sum of the two values for current through the 40ohm resistor (careful of direction) is the actually current through the 40 ohm resistor... likewise with the 10 ohm resistor...
 
  • #7
How and when will the 12 ohm resistor come into play?
 
  • #8
Also, why is direction important?...I thought that current is constant everywhere in a closed system?
 
  • #9
William Bush said:
Also, why is direction important?...I thought that current is constant everywhere in a closed system?

No, that's not true... when there is a single loop current is the same everywhere within that one loop...
 
  • #10
Thank-you for helping me out...it's past midnight my time and my brain isn't working very well anymore. You have given me a direction to pursue in the morning so hopefully this will make more sense then. If I sleep with my book under my pillow, do you think I will wake up with the solution?!
 
  • #11
William Bush said:
Thank-you for helping me out...it's past midnight my time and my brain isn't working very well anymore. You have given me a direction to pursue in the morning so hopefully this will make more sense then. If I sleep with my book under my pillow, do you think I will wake up with the solution?!

lol! maybe... I recommend checking out any examples on "node voltage" method in your text... it gives a real quick method to solve this problem. good luck!
 
  • #12
Big thanks goes to "learningphysics"! I was able to solve this problem after another two hours of work. Way to long for me to list the steps but basically I had to use Kirchhoff's junction and loop rules to develope equations that allowed me to solve for current from batteries 1 & 2. Once I had the currents for both batteries, I plugged their values into the Power = V X I formula to solve the problem.
 

1. What is power in a circuit?

Power in a circuit refers to the rate at which electrical energy is transferred from a power source to a load. It is measured in watts (W) and is the product of voltage (V) and current (I).

2. How do you calculate power in a circuit?

The formula for calculating power in a circuit is P = VI, where P is power in watts, V is voltage in volts, and I is current in amps.

3. What is the relationship between power and voltage in a circuit?

The relationship between power and voltage in a circuit is directly proportional. This means that as voltage increases, power also increases, and vice versa.

4. How does resistance affect power in a circuit?

Resistance in a circuit affects power by reducing the amount of current that can flow at a given voltage. This leads to a decrease in power, as power is the product of voltage and current.

5. How can power be increased in a circuit?

Power can be increased in a circuit by either increasing the voltage or the current. This can be achieved by using a higher voltage power source or by decreasing the resistance in the circuit, respectively.

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