RC circuits - Current and electric potential

[msemsey]

Homework Statement

Given the drawing, calculate the current going through the 2Ω Resistor and then calculate the difference in electric potential from point a to b.

Homework Equations

V = IR
juction rule:
ƩI = 0 = I_in - I_out
ƩV = 0

The Attempt at a Solution

Well, I attempted to draw all the currents. The Green is from the 3Volts on the bottom, the orange is from the 12V on the top. I would think that I could use a sum of the different V=IR equations (12/4 + 3/2 + 3/2), but 6 is not the answer. I'm pretty lost clearly.

 

[Basic_Physics]

For starters use just one current in each "branch", that is the current will change to another value if you continue through a junction. Label each of these i1, i2...
Lets try it with a new drawing then.

 

[msemsey]

Here is my updated drawing. I'm confident that I found all the currents and that the junction equations are right, but the loop rule equations confuse me... I'm not sure where the currents that don't pass through resistors go in the loop equations.

 

[aralbrec]

Here is my updated drawing. I'm confident that I found all the currents and that the junction equations are right, but the loop rule equations confuse me... I'm not sure where the currents that don't pass through resistors go in the loop equations.

You have too many unknowns in your KCL equations. You should recognize, eg, that i4=i5. Note with the two equations you have, you can only solve for two unknowns so the first step you would need to take with your equations is to eliminate all the duplicate currents. Make current a property of the branch.

For your loop equations, there are two loops and two currents. You have them identified as i1 and i2. i1 flows around the bottom loop and i2 flows around the top loop. Notice, eg, that for the i1 loop, the same current i1 must flow through the battery e, the resistance R and the 2 ohm resistor. The 2 ohm resistor is special because the current i2 must also flow through it, but it is the net current that causes the voltage drop. What justifies this net current calculation is KCL.

 

[Nickie]

I would approach this problem by first analyzing the circuit and identifying the components and their values. In this case, we have a series circuit with a 2Ω resistor and two voltage sources, 3V and 12V.

To calculate the current through the 2Ω resistor, we can use Ohm's Law (V=IR) and the junction rule (ƩI=0). We know the total current in a series circuit is the same at all points, so we can set up the equation: 12V = 3V + I2Ω(2Ω). Solving for I2Ω, we get a current of 4.5A.

To calculate the difference in electric potential from point a to b, we can use the sum of voltage drops around the circuit (ƩV=0). Starting at point a, we have a drop of 3V across the 3Ω resistor, then a drop of 4.5V across the 2Ω resistor, and finally a drop of 12V across the 12V source. This gives us a total drop of 19.5V from point a to b.

In summary, the current through the 2Ω resistor is 4.5A and the difference in electric potential from point a to b is 19.5V. It's important to carefully analyze the circuit and use the appropriate equations to solve for the desired quantities.