Compound circuits often seem difficult to solve at first, but once you start solving small parts of the total circuit, everything starts to fall into place. In this compound circuit, you have an overall parallel arrangement. In one branch of the parallel circuit, you have a single resistor, in the other branch of the parallel circuit, you have two resistors in series.
The first step I always suggest doing is finding the total/equivalent resistance of the entire circuit. Once you have this piece to the puzzle, you can solve for the total current in the circuit. This information is often useful to check you solution. In this example, the current in each of the branches of the parallel circuit must add up to the total current in the circuit.
After finding the total resistance and current, I would then solve for the current in the branch containing the one resistor. Because this is a parallel arrangement and the voltage across this resistor is the same as the voltage of the battery, this is a simple calculation using ohms law. From there, you can solve for the current through the branch containing the two resistors in series. You know their equivalent resistance (simply the sum because they are arranged in series), and you know the voltage across the two resistors. Using the current through this branch, you can then solve for the voltage drop in each of the resistors.
As mentioned above, you last step should be to verify you results. The voltage drop in each of the two resistors in series must equal the total voltage. Also, the current in each of the branches must add up to the total current.
It seems that students often have difficulty with these circuit problems where you have to solve for the voltage and current in each of the branches of a compound circuit. Solving these types of problems is almost like completing a sudoku puzzle. The first thing you need to do is solve for anything possible, even if it doesn’t initially seem to be going in the direction of your answer. Often times, this is simply solving for the equivalent resistance of the circuit. Once you have the equivalent resistance, you then use that information to solve for the total current in the circuit.
From there, you have to take a look at the schematic and problem to find out where to go next. If there is a branch that is in a parallel arrangement with the batter, you can use ohms law to find the current in that branch. If you have a resistor that is in series with the battery, you know that the total current is the current flowing through this resistor. If you know the current and the value of the resistor, you can easily solve for the voltage drop across the resistor.
As you move around the circuit counterclockwise, you encounter two resistors that are in parallel. Because you solved for the voltage drop for the one resistor in series, and you know the total voltage of the circuit, you can find the voltage drop in the parallel part by simply taking the difference. Once you have the voltage across each of these resistors (same for both because they are parallel), you can find the current in each of the resistors using ohms law.
Again, the best advice is to look at what is given, and solve for some part of the circuit using this information. Each time you use ohms law, you uncover another piece of information that will help you solve another part of the whole circuit. I find that keeping all of the information in a table makes it much easier to keep track of the things you have solved for as well as the things you are still trying to figure out.
Once you move past static electricity, and start thinking about how charges move, you move into the realm of ohm’s law. Before you get there though, there are a couple important things to think about. First is why charges move. When an object becomes charged, either positively or negatively, we say that it has some electrical potential. If you have two objects that have charges, one of two things can happen. Case one: both objects have exactly the same charge. In this situation, we say that they have exactly the same potential. Obviously, if we were to try to find the difference in potentials, you would get zero. Hypothetically, if both objects have a charge of (-1 charge unit), the difference between -1 and -1 is zero. Case two: the two objects have different charge, say -1 and -2. Here, if we were to find the difference, we would fine that there is a difference of 1 charge unit.
Now just having two objects with different charges in not enough to make the charge move. There needs to be one more thing. That thing is some path between the two charged objects. This path usually takes the form of a conductor, or wire, between the two charged objects. Once this conductor is in place, charge begins to flow. The charge will continue to flow as long as there is some difference in potential, and there is a path for the charge to move.
We’re not done yet. A potential difference is measured as voltage (V) with the unit of volts. The rate at which the current moves from one place to another is defined as the current (I), and that is measured in amperes. The final piece to this puzzle is the fact that even materials that do a good job at conducting electricity don’t do it with 100% efficiency. There is always some resistance to the flow of electricity. This is defined as resistance (R), and it is measured in ohms. The amount of resistance present in a material is a function of a bunch of things, like, the type of material itself, the temperature, the length of the conductor, and the thickness of the conductor.
These three pieces come together and are related by Ohm’s Law. Ohm’s Law simply states that V = I * R, or, voltage = current * resistance.
Things that are electrically charged behave very similarly to magnets. Using this analogy, we can look at how objects with charges are either attracted or repelled. Most everyone has played with magnets as a kid, and you may have noticed that magnet usually have ends which are labeled (N)orth (S)outh. With very little effort, you can figure out that north attracts south, north repels north and south repels south. Electrical charges are exactly the same. Positive charges attract negative charges, positive charges repel positive charges, and negative charges repel negative charges.
It can be useful (as in trying to complete a homework assignment, but I am sure there are other uses which I just can’t think of at the moment) to be able to calculate the forces present between two charged objects. In order to do this, all you need to know is the charge on each of the objects, and the distance between the objects. In the above solution video, I use Coulomb’s Law to calculate the repulsive force of two balloons which have become negatively charged.
In this problem, we look at how to solve a problem dealing with the bending of a light wave as it changes from one medium to another. Here light moves from air t lucite. The angle of incidence is 30 degrees, and we need to use the indices of refraction and snells law to find the angle of refraction. Remember, all angles are measured from the normal (a line which is perpendicular to the surface of incidence.)
In this problem, we look at a simple machines and figure out the mechanical advantage present. All of these types of problems revolve around the idea that the work you put into a system, is equal to the work you get out of a system, and that work is equal to force times distance. The first solution deals with a lever, and the second problem deals with a simple pulley.
In this problem solution, we look at an amusement park ride and calculate various things regarding the rides circular motion. We look at period, frequency, tangential/linear velocity, centripetal acceleration, and centripetal force. This problem is consistent with a high school conceptual physics course.
If you are taking a science class and are seeking tutoring, please drop me an email and I would be very happy to arrange tutoring sessions with you. There are a variety of ways that we can set up the tutoring sessions in order to make them as convenient as possible for you. I offer tutoring sessions via real time one way or two way video/audio chat. If you choose, I will also record solutions to specific problems and then send them to you so that you may see the steps that need to be take in order to solve the problem. Flexibility is key, and this method of tutoring allows for efficient student-teacher contact without the limitations of in person meetings. If you live in the Westchester, NY area, in person meetings can be arranged.
In this video, I explain how to calculate frequency and then the number of waves in a give time period give the wavelength of a beam of light. Also, there is a conversion from nanometers to meters in the problem. This problem fits in with a high school conceptual physics class.
Hello and welcome to the new mrlovescience.com. I wanted to test out using the wordpress format for my web site to see if it would facilitate me updating it. In the past, I was using iWeb, and I was finding that the content was stagnating because it was taking too much effort to redesign, and then republish the site each and every time that I wanted to make a change. Hopefully, I will now be able to keep the content fresh and add new things with ease.
This site will feature a variety of topics, but it will revolve around science and surroundings. It is my intention to make this site useful to students and instructors alike, and I look forward to reading comments about my postings. If there is something that you would like to see here, don’t hesitate to drop me a line, and I will do my best to put it up.