Electric Circuits Legacy Problem #21 Guided Solution

As electric circuit problems become more complicated, it becomes even more important that the beginning student of physics begin to visualize the situation and even diagram the situation. Schematic circuit diagrams are really useful when it comes to solving problems like these. I've developed a picture of a power source or a voltage source, 12 volts, and it's got two resistors connected in series or in parallel. It doesn't take me much time to actually sketch out the situation with a little circuit symbol of a battery, a little squiggly line circuit symbol for a resistor, and some wires which are just lines. I sketch it out and begin to get a picture of what's going on here. I have a 12 volt source and I have two resistors. In one case they're connected in series and in one case they're connected in parallel. What I'm asked to do is find the overall current in the circuit. Another way to translate that is I want to know the current through that battery. So to find the current through the battery, I need to take the battery's voltage and divide it by the overall or equivalent resistance. In the case of series circuits, the equivalent resistance is simply the sum of the individual resistances. So for the first part of this question, part A, where they're connected in series, the overall resistance is just the 6 ohms plus the 8 ohms or the 14 ohms. In finding the current, it's a matter of using the equation delta V equals I times R, where the delta V is 12 volts and the R I use in there is the 14 ohms. Plugging and chugging and solving for I gets me an answer of about 0.8571 amperes or amps. I can round that to two significant digits and that's the explanation for the 0.86A. Now in part B, I have the resistors connected in parallel. Solving a problem like this demands I first find the overall or equivalent resistance of this collection of two resistors. The equation for parallel circuits for finding the overall resistance goes like this. 1 over R eq is equal to 1 over R1 plus 1 over R2. So 1 over the total resistance is simply equal to 1 over 6 plus 1 over 8. I need to solve this for total resistance first. Then I can solve for current. I need to use my calculator or find a common denominator. Then I can add these two fractions on the right side of the equation. When I do, I simply have 1 over R equivalent. I then need to take the reciprocal of both sides of the equation so that I have R equivalent. It comes out to be about 3.4286 ohms. Now I can say delta V equals IR for this parallel circuit where delta V is 12 and R is equal to 3.4286. Plugging and chugging gives me 3.5 amps. That's the answer to part B.