Solve Circuit Problems: Find Current & Potential | RC Circuits Explained

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In summary, the student found that if two 3 ohm resistors are connected in series, the voltage across them is 8 V. The current through each resistor is determined by the resistance and the voltage drop across the resistor.
  • #1
Firestrider
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Ok I'm not sure if I did these right, and my teacher isn't exactly sure how it's done either. Here are the problems:

1. http://img153.imageshack.us/img153/7456/circuit10al.jpg
Find the current that flows through the [tex]3\Omega[/tex] resistor.
2. http://img153.imageshack.us/img153/9011/circuit22wq.jpg
What is the potential of point D relative to point C?

This is what I did for #1:

I solved the parallel branch into a series equivilent by finding the resistance equivilent of the 3, 4, and 5 [tex]\Omega[/tex] resistors and got [tex](\frac{1}{3}+\frac{1}{4}+\frac{1}{5})^{-1}=1.27\Omega[/tex]. Since all the resistors were in a series now I took the sum and got [tex]R_{eq}=1.27+2+ 1=4.27\Omega[/tex]. To find the current throughout I did [tex]\frac{V}{R_{eq}}=\frac{12}{4.27}=2.8A[/tex].
Now I had to find the current in the [tex]3\Omega[/tex] resistor so I think I had to find the voltage at the point before it went into the parallel circuit which came out to be [tex]V_{3\Omega}=\xi-(V_{d2\Omega}+V_{d1\Omega})=12-((2.8*2)+(2.8*1))=3.6V[/tex]
and then divide that by 3 to get 1.2A

For #2:

I had the same problem on this one as the last one. Basically on how to find the current in a parallel circuit. I know Kirchhoff's junction rule in that the sum of all the currents going in, must be the sum of all the currents coming out, and I know that the voltage stays the same and current changes in parallel circuits. I've heard that the current is split evenly between each branch, is proportional to each branches resistance, and/or it is calculated by the voltage divided by the resistance at that point. I'm not sure which, if any, to use. My answer, which stays consistant with #1 of 8V is probably not correct since that would mean more voltage than the
terminal. But if the split between the two branches it stays consistant with Kirchhoff's rules and not #1

My teacher didn't go over anything about RC circuits, just that the resistance is calculated in the direct opposite manner of resistors. I think RC circuits might be on the AP exam, and my book doesn't have any RC circuit problems. Anyone know where I can find some?

Also does anyone know what AP Physics B is in college?
 
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  • #2
Any help, please?
 
  • #3
Your work on the first part is correct. If you repeat the methodology on the 2nd you should get the correct result. As you are aware your answer for the second part is not correct. In the second problem since the resistance of each branch is the same, 6 ohms, the current through each branch will be the same.

Remember the voltage drop across a single resistor is determined by the resistance and the current.
 
  • #4
Firestrider said:
Any help, please?
Your solution for the first problrem is correct. Just apply the same method for the second one. Both 4 ohm resistors are connected in series, the voltage across them is known, so you get the current. Multiply it by 4 ohm, you get the voltage across the resistor CD (3 V)
The voltage is obtained as potential difference. VDC = UD-UC. You need the potential of D (UD) with respect to UC. The magnitude is 3 V, you have to decide the sign. A is connected to the positive terminal of the battery, so A is more positive than C. The current flows in the direction A->D->C. The current across a resistor flows along decreasing potential, so D is more positive than D, that is the potential of D with respect to C is 3 V.

ehild

Sorry, I misread the 2 ohm resistor... so both the voltage and potential are 4 V.
 
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  • #5
http://photo-origin.tickle.com/image/69/7/5/O/69751595O802495255.jpg
 
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1. How do I determine the total resistance of a circuit with multiple resistors?

To determine the total resistance of a circuit with multiple resistors, you can use the formula: R(total) = R1 + R2 + R3 + ..., where R1, R2, R3, etc. are the resistances of each individual resistor. This only works for resistors connected in series. For resistors connected in parallel, you can use the formula: 1/R(total) = 1/R1 + 1/R2 + 1/R3 + ...

2. How do I calculate the current in a circuit?

To calculate the current in a circuit, you can use Ohm's Law, which states that current (I) is equal to voltage (V) divided by resistance (R): I = V/R. You can also use Kirchhoff's Current Law, which states that the sum of the current entering a node is equal to the sum of the current leaving the node.

3. How do I determine the voltage drop across a specific resistor in a circuit?

To determine the voltage drop across a specific resistor in a circuit, you can use Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R): V = I*R. You can also use Kirchhoff's Voltage Law, which states that the sum of the voltage drops in a closed loop is equal to the sum of the voltage sources.

4. What is the difference between series and parallel circuits?

Series circuits have all components connected in a single loop, so the current is the same through each component. In parallel circuits, the components are connected in multiple branches, so the current is split between each branch. Additionally, the total resistance in a series circuit is equal to the sum of individual resistors, while the total resistance in a parallel circuit is less than the smallest individual resistor.

5. How do I calculate the power dissipated by a resistor in a circuit?

To calculate the power dissipated by a resistor in a circuit, 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. You can also use the formula: P = V^2/R, where V is voltage in volts. This is known as the power law.

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