16:

You will construct a series circuit with the resistor you are given and another resistor, let’s call it X. Draw a picture. We know X has a value less than your given resistor (as per the problem) so we know the voltage across it is less than the that of the given resistor. I’m going to be general since I am actually answer in this question for several people. Let’s say you have a 6V battery and you need a 5V power supply. What this means in practical purposes is that you need 5V across the given (larger) resistor. The remaining voltage (1V in this case) will be across resistor X. Use your values to find the current in your given resistor. Since you have constructed a series circuit (a voltage divider) you can use this same current in resistor X. You already know the voltage across resistor X is the difference between your power supply and what you needed (in this case 1V), so using your current and this voltage you can find the value of resistor X.

This is actually a common problem in the construction of electronics. You need a particular voltage somewhere, but you only have a certain power supply. This is the process you have to follow.

17c&d:

C: Since the hairdryer and the TV are plugged into the same outlet, this implies that they are in parallel (does your TV shut off if you turn off the hairdryer? Hopefully not, although I do wonder why you are plugging your TV and hair dryer into the same outlet to begin with…a little odd). So you simply need to find the equivalent resistance of the two items in parallel.

D: This is basically part B all over again, but using your equivalent resistance of the two appliances from part C. You need count the contribution of resistance in the wires (for most people, 2-3 Ohm’s) once.

22:

The problem mentions two batteries in series. This is essentially what you did in the Ohm’s law lab by adding batteries to a circuit. It is what you do when you put two batteries in a flashlight, or four in your calculator. By placing batteries in series (facing the same direction) you are providing additional increases in electric potential to the circuit than a single battery. In this case, you have two 1.5V cells, giving you have 3 total volts, and therefore TWO times the internal resistance stated, plus the resistance of the bulb given in the problem. A and B are easy. Use ohm’s law to find the current (part A) and P=IV to find the power (part B). Part C is the part I mentioned in class that is acting strange in webassign. All you should have to do is get rid of the internal resistance mentioned in the problem and re-do parts A and B and find the difference between your original and new value, but webassign is spitting out 90% of the correct answers. Treat the “internal resistance” of the batteries as a small resistor (each) in series with the rest of the circuit.