Kirchhoff's law: Circuit problem with multiple batteries and resistors

In summary: Your answer is correct.In summary, Bolter redid the simultaneous equations to take into account the corrections and got the current as 47/44 A through 6 ohms, 16/11 A through 1 ohms, and –17/44 A through 4 ohms.
  • #1
Bolter
262
31
Homework Statement
See image attached
Relevant Equations
Kirchhoff's 1st and 2nd law
Struggling to see if I am doing this right or not from a section of a question I have been given
Screenshot 2020-03-20 at 23.38.14.png

I have drawn out a sketch of the circuit again, and labelled as to what my 1st and 2nd closed loop to be to then apply Kirchhoff's 2nd law

IMG_4244.JPG

IMG_4245.JPG

IMG_4246.JPG


I get a negative I_2 current which just indicates that my original assumption of the currents direction was wrong and it is in fact going the opposite way

So I believe that the current going through:

6 ohm resistor = 0.703 A
1 ohm resistor = 1.162 A
4 ohm resistor = 0.459 A

Is this right or have I miserably gone wrong somewhere? :oldconfused:

Any help would be really appreciated! Thanks
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
I followed through your workings and arrived at the same conclusion as you.
I found your error. It is in the step where you expand the parenthesis:
8(I1 + I2) + I1 = 10

Check how you expand the parenthesis and take care of the 8 coefficient.
 
Last edited:
  • #3
Hi @Bolter I originally made the same error, and didn't catch your mistake. It was when I took those currents, plugged them back into the original circuit - I realized that the loop voltage drops did not sum to zero, so then started looking for an error.

Lesson: always go back and plug your answers into your original problem, as a check.
 
  • Like
Likes Leo Liu
  • #4
scottdave said:
Hi @Bolter I originally made the same error, and didn't catch your mistake. It was when I took those currents, plugged them back into the original circuit - I realized that the loop voltage drops did not sum to zero, so then started looking for an error.

Lesson: always go back and plug your answers into your original problem, as a check.
Hi, can I know what value you got for I? I got ##47\over44##.
 
  • #5
scottdave said:
I followed through your workings and arrived at the same conclusion as you.
I found your error. It is in the step where you expand the parenthesis:
8(I1 + I2) + I1 = 10

Check how you expand the parenthesis and take care of the 8 coefficient.

Thank you for catching in on my mistake :)

I have redone solving the simultaneous equations again using the corrections and I got these as my currents

IMG_4247.JPG


So I can say that the current through:

6 ohm resistor = 47/44 A
1 ohm resistor = 16/11 A
4 ohm resistor = –17/44 A

Is it also possible if you could help me out with part c) of this question?

Screenshot 2020-03-21 at 09.51.56.png

I first worked out total charge that passes in 10 seconds, then divided that by the charge of a single electron to get total number of electrons in 10s?

IMG_4248.JPG
 
  • #6
All those calculations (current in resistors & number of electrons) look correct.
 
  • Like
Likes Bolter

FAQ: Kirchhoff's law: Circuit problem with multiple batteries and resistors

1. What is Kirchhoff's law?

Kirchhoff's law, also known as Kirchhoff's circuit laws, are two fundamental principles in electrical circuit analysis. They are used to determine the current and voltage in a circuit with multiple batteries and resistors.

2. What are Kirchhoff's two laws?

Kirchhoff's first law, also known as the "junction rule," states that the total current flowing into a junction must be equal to the total current flowing out of the junction. Kirchhoff's second law, also known as the "loop rule," states that the sum of the voltage drops in a closed loop circuit must be equal to the sum of the voltage sources in that same loop.

3. How do Kirchhoff's laws apply to circuits with multiple batteries and resistors?

Kirchhoff's laws are essential in analyzing circuits with multiple batteries and resistors. They allow us to determine the current and voltage at any point in the circuit by applying the junction and loop rules. By using these laws, we can calculate the total resistance, current, and voltage in the circuit.

4. Can Kirchhoff's laws be applied to circuits with only one battery and resistor?

Yes, Kirchhoff's laws can be applied to circuits with only one battery and resistor. In this case, the junction rule states that the current flowing into the resistor must be equal to the current flowing out of the resistor, while the loop rule states that the voltage drop across the resistor must be equal to the voltage of the battery.

5. What are some practical applications of Kirchhoff's laws?

Kirchhoff's laws are used in various electrical engineering applications, such as circuit design, power systems, and electronic devices. They are also used in troubleshooting and analyzing complex circuits to determine the cause of malfunctions. Additionally, Kirchhoff's laws are crucial in understanding and predicting the behavior of electrical circuits in everyday objects, such as household appliances and electronic devices.

Back
Top