Circuit Question - Driving me crazy

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In summary, the problem is trying to find the current through resistor g when switch B has been closed and switch A has been open for a long time. The homework statement is trying to find the potential difference between the resistors in series.
  • #1
Allura
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Homework Statement



The problem is "What is the current through resistor g when switch B has been closed and switch A has been open for a long time?" (see diagram)

Homework Equations



V = IR
I/Reff = 1/R1 + 1/R2
Reff = R1 + R2


The Attempt at a Solution



Because f+g are in series, the Reff = 6 ohms. Since f/g is in parallel with c, the new Reff is 4 ohms. Then because f/g/c are in series with a/b, the new total resistance is 10 ohms. As a result, we can find through V = IR that the current is 9A (90 V = I x 10 ohms).

So then, if V = IR, to find the current through each respective resistor (in parallel), I did 90 V = I x 12 ohms, and 90 = I 6 ohms, however, this gives me currents that are incorrect. Instead, the answers are 6 A, and 3 A through the 6 ohm resistor and 12 ohm resistor. I understand this conceptually, i.e. yes, the resistor 12 ohm is 2X the resistor 6 ohm, therefore current would be half as large - however, I can't figure this out mathematically. Literally, I have spent an hour on this problem, I've read various tutorials, watched khanacademy, read the solution in the back of the book (which gives a conceptual answer) but I can't figure this out.

Please help, because I really want to figure this out and I can't move on until I do.
 

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  • #2
Allura said:
Because f+g are in series, the Reff = 6 ohms. Since f/g is in parallel with c, the new Reff is 4 ohms. Then because f/g/c are in series with a/b, the new total resistance is 10 ohms. As a result, we can find through V = IR that the current is 9A (90 V = I x 10 ohms).

Correct till here .

Allura said:
So then, if V = IR, to find the current through each respective resistor (in parallel), I did 90 V = I x 12 ohms, and 90 = I 6 ohms,

The P.D across 12 and 6 Ω is not 90 V . Look carefully in the picture .
 
  • #3
Tanya Sharma said:
Correct till here .
The P.D across 12 and 6 Ω is not 90 V . Look carefully in the picture .

How do you find the potential difference? I thought it was the same between any set of resistors in series (i.e. only the current changes with differing resistances). Please help, at this point I am totally lost.

Thanks
 
  • #4
Allura said:
How do you find the potential difference? I thought it was the same between any set of resistors.

Thanks

You have correctly calculated the current 9A flowing through the battery.The current 9A from positive terminal of battery passes through the series combination of 'a' and 'b' and then divides in the parallel combination of 12 Ω and 6Ω (f+g) .

How does current divide in the parallel resistors ?
 
  • #5
Tanya Sharma said:
You have correctly calculated the current 9A flowing through the battery.The current 9A from positive terminal of battery passes through the series combination of 'a' and 'b' and then divides in the parallel combination of 12 Ω and 6Ω (f+g) .

How does current divide in the parallel resistors ?

I = V / R ? Then the currents add up to the total current, but I still can't get the right answers.

I know total current in is equal to the total current out..?
 
  • #6
Tanya Sharma said:
You have correctly calculated the current 9A flowing through the battery.The current 9A from positive terminal of battery passes through the series combination of 'a' and 'b' and then divides in the parallel combination of 12 Ω and 6Ω (f+g) .

How does current divide in the parallel resistors ?

I thought maybe, okay find a new voltage? But V= IR --> (9A) (4 ohms) where R = Total effective resistance. Then it's 36 V, so V = IR --> 36/12 = 3 A for c? Right?

But why do you find a new V and do the problem this way? But 4 ohms is for f, and not for C. I'm so confused :( :(
 
  • #7
The P.D across parallel resistors is same.

Let the potential difference across the parallel resistors 12 Ω and 6Ω be V .

Then V=I1(12) and V=I2(6) , where I1+I2 =9A .

Now solve for I2 .

Does this make sense ?
 
Last edited:
  • #8
Tanya Sharma said:
The P.D across parallel resistors is same.

Let the potential difference across the parallel resistors 12 Ω and 6Ω be V .

Then V=I1(12) and V=I2(6) , where I1+I2 =9A

Does this make sense ?

Yes, got that. Then what?
 
  • #9
Solve and obtain value of I2 .This is what you want.
 
  • #10
Tanya Sharma said:
Solve and obtain value of I2 .This is what you want.

Why are we solving for I2 though?
 
  • #11
Allura said:
Why are we solving for I2 though?

Because this is what the question is asking .I2 is flowing through the resistor g .
 

1. What is a circuit and how does it work?

A circuit is a closed loop of conductive material that allows electricity to flow through it. It typically includes a power source, such as a battery, and various components, such as resistors and capacitors. Electricity flows from the power source through the components and back to the source, completing the circuit. This flow of electricity is controlled by the components, which can either allow or block the flow based on their properties.

2. What is the difference between a series and parallel circuit?

In a series circuit, the components are connected end-to-end, creating a single pathway for electricity to flow through. This means that if one component fails, the entire circuit will stop working. In a parallel circuit, the components are connected in multiple branches, allowing for multiple pathways for electricity to flow through. This means that if one component fails, the others will still continue to function.

3. How do I calculate the total resistance of a circuit?

The total resistance of a circuit can be calculated by adding up the individual resistances of all the components in the circuit. In a series circuit, the total resistance is equal to the sum of all the individual resistances. In a parallel circuit, the total resistance is calculated using the formula 1/Rtotal = 1/R1 + 1/R2 + 1/R3... where R1, R2, R3... are the individual resistances.

4. What is the purpose of a capacitor in a circuit?

A capacitor is a component that stores electrical energy in the form of an electric field. It is often used in circuits to smooth out voltage fluctuations, filter out unwanted frequencies, and store energy for later use. It can also be used to block direct current while allowing alternating current to pass through.

5. How can I troubleshoot a circuit that is not working?

First, check all connections to make sure they are secure. Then, use a multimeter to check for continuity and proper voltage at each component. If a component is not functioning properly, replace it. If all components are functioning correctly, the issue may be with the power source or the wiring. It may also be helpful to refer to the circuit diagram to ensure all components are connected correctly.

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