Any complaints email to janovi@wm.edu.
AnswerThe light will be dimmer. The energy needed to power the lamp is carried in the current in the wire. If you split the wire with a metal object, creating a short circuit, a good portion of the current will flow through the metal object since it is a conductor and is a path of less resistance. So the lamp will recieve a smaller amount of energy and the light will dim.
6.)The two prongs of a power cord are meant to carry current to and from a lamp. If you were to plug only one of the prongs into an outlet, the lamp wouldn't light at all. Why wouldn't it at least glow at half its normal brightness?AnswerSimply put, you wouldn't have a complete circuit. Charge would accumulate at one of the prongs and not reach the outlet to pick up more energy. Also, you can say that the conduction path is cut so electrons cannot flow anymore.
10.)Suppose that you have two 1.5 V batteries, some wires, and a light bulb. One of the batteries, however, is unlabeled and there is nothing on it to indicate which is its poitive terminal and which is its negative terminal. The other battery is properly labeled. How can you determine which terminal of the unlabeled battery is positive and which is negative?
AnswerWe know that when we put two batteries together such that the negative terminal of one touches the positive terminal of the other, charge flowing through both batteries will experience a voltage increase of 3 V. And if we line up the batteries so either the positive or negative terminals are touching, the charge will experience an increase and drop in voltage (or no work is done on the current). To figure out which terminal is which we first see how bright the light bulb is with JUST ONE battery, then place the second battery between the bulb and the first battery. Try two circuits, where you swtich the terminals of the second battery and you should be able to label the terminals by the brightness of the light bulb.
12.)Electric power is often sold on the basis of kilowatt-hours. A house that uses an average of 1000 W for an hour has consumed 1kWh. Explain why a kilowatt-hour is a unit of energy.
AnswerA watt is a joule per second. By changing hours to seconds and multiplying a watt by seconds, you get a unit of energy.
13.)If you overloaded an extension cord so that it's carrying too much electric current, why does it become so hot?
AnswerIncreasing a current means increasing the amount of charge in the current. As you increase the amount of charge without increasing giving the current more room in which to move, the electrons will transmit their kinetic energy to each other and eventually to their surroundings. As kinetic energy is transferred to the cord, the cord's temperature will increase.
15.)Your friends are installing a loft in their room and are using thin speaker wires to provide power to an extra outlet. If they only draw a small amount of current from the outlet, the voltage drop in each of the wires will remain small. Why?
Answer
This is simply Ohm's law again. The current in a circuit element is directly proportional to the voltage drop across the element.18.)The electric company makes sure that the voltage drop acros an incandescent light bulb is always 120 V. But as the bulb's filament warms up, its electric resistance increases. What happens to the current passing through the filament?
AnswerThe amount of work being done on the current does not increase but the resistance is increasing. So less current is flowing through the wire. Look at ohm's law, V/R = I. If V remains constant and R increases, then I must decrease.
20.)Why will the power surge in exercise 19 cause the bulb to consume approximately 20 % more power than normal?
AnswerThere are two major steps to solving this problem. First, we start with Ohm's Law. Before the surge, Ohm's law gives Vo=RIo, where Vo and Io are the initial voltage and current, respectively. After the surge, there is a 10% increase in the voltage drop. We can express this as V = IR, where V and I are the voltage and current after the surge. Now, we're given that V = `1.1Vo.
1.)An automobile has a 12-W reading lamp in the ceiling. This lamp operates with a voltage drop of 12 V across it. How much current flows through the lamp?
AnswerUse equation 9.2.2. The power consumed is equal to the voltage drop times the current, or P = VI. We want to calculate the current flowing through the reading lamp, so we divide both sides by V, obtaining I = P/V. Pluggin in the given values for P and V, we find the current is 1 A.
2.)The rear defroster of your car operates on a current of 5 A. If the voltage drop across it is 10 V, how much electric power is it consuming as it melts the frost?
AnswerWe'll use the same equation as we did in the first problem, only this time we don't have to do any algebra. We are given the values for V and I, plugging these into the equation yields 50 W.