Lecture 37 Power Circuits Parallel and Series Resistances Cutnell+Johnson: 0.4-0.8 Circuits Now that we know about voltages, currents and resistance, we can talk about circuits. A circuit is simply a closed loop of current. There is a battery to supply the voltage, a wire to carry the current, and something attached to the wire (like a washing machine, a light bulb, etc.) A material with resistance is called a resistor. Insulators have a large resistance, conductors a small one. You put resistors in a circuit to lower the current in that part of the device. The resistance of copper is very small relative to most things, so it is common to just neglect the resistance in the wire, and just worry about the other things (like the light bulb or washing machine) connected to the circuit. The conventional way to draw a circuit is to draw the + and terminals of the battery, a line for the wires, and a squiggly line for any resistors on the circuit. Circuits may also have switches, which allow you to stop the current by breaking the wire. You may notice that sometimes light switches spark. That happens when the switch is not quite on, but close enough so that the current can jump across the air. Another common part of a real circuit is a fuse. The point of a fuse is that it can carry only so much current. It blows up if you put too much current in it, and breaks the circuit. Back in the old days, people used fuses in their houses, but circuit breakers are more convenient, because they can be easily reset. Still, fuses are used in most electronic devices it s better to blow an fuse than to blow up the whole thing. Problem A wall outlet has a fixed voltage, which is 0 V in the USA. The current varies, depending on the resistance of the objects attached. At what resistance will a A fuse blow? Answer R = V I = 0 V A = 60Ω For any resistances lower than 60Ω, the fuse will blow. By the way, the wall voltage is significantly higher in Europe, which is why people more frequently get electrocuted. You have a fixed
resistance, more or less, so if you stick your finger in a wall socket at twice the voltage, twice the current flows through you (bad). When drawing a circuit, it is important to keep track of which voltages and which currents you re talking about. The current must be the same in a single wire: electrons are not accumulating anywhere, so they must be moving by at the same rate. If a wire splits into two, then some of the current goes one way, some goes the other. A voltage, on the other hand, is defined between two points. We say a voltage drop, or a potential difference is present between these two points. What this means that there is a difference in potential energy between these two points. A voltage drop in electricity is exactly like dropping a ball in a gravitational field: depending on how far it falls Power I keep emphasizing how a voltage difference is related to the difference in potential energy. When a wire has resistance, the energy goes to getting the charges through. It s like pushing your way through a crowd. By the time you ve made it through the crowd, you ve run out of energy, and need another push. Energy overall is conserved: this energy goes to heating the resistor. In a light bulb, some of the energy is converted to light as well, although incandescent bulbs are very inefficient: most of the energy goes to heating. When there is energy flow, there is power. Power is as always P = E t Here we know that the change in energy between the two terminals is due to a change in voltage. Thus E = qv. This energy is converted to kinetic energy, but because of resistance, it is converted to other forms as well. Plugging this into the formula for power gives P = ( q)v t = q t V = IV The MKS units of power are, as we ve seen, the Watt, which is a J/s. Here, we see that one watt is one volt-ampere as well. You can check that this is consistent with the definitions of the units. Since V = IR as well, we can rewrite the power formula in the forms P = V I = I R = V R
Problem What is the resistance of a 50 W light bulb? Answer The 50 W is based on the 0 V in the US. To find the resistance, we use P = V /R. R = V P = (0 V ) 50 W = 88 Ω Parallel and Series Resistances Last time I pointed out that the resistance of a wire increases with length, but decreases with cross-sectional area. The same ideas hold for circuits with multiple resistors. Say put a resistor of resistance R after one of resistance R, so that the current through the first one I is the same as the second I. This is known as putting the resistors in series. Say we connect these resistors in series to a batter of fixed voltage V. The current is not have I = V/R and I = V/R. Ohm s Law applies between any two points, so we have I = V /R, and I = V /R. V is the voltage drop across the first resistor, while V is the voltage drop across the second. Since the resistors are in series, I = I, so V R = V R Since voltage is defined as between two points, the total voltage V = V + V. You can use these two relations to solve for V and V and hence I, giving I = I = V R + R Note that this means you can view the two resistors as being a single resistor with a resistance R series = R + R In other words, if I put the series resistors together in a box, the box has the total resistance R + R. Putting resistors in series is like making a wire longer and increasing its resistance. To lower resistance, we can put resistors in parallel. Two resistors R and R in parallel have the same voltage across each: V = V = V. This requires splitting the current: the current coming out of the battery I is split into two. Since overall charge is conserved, I = I + I.Thus resistors in series have the same current through them, resistors in parallel have the same voltage across them. As always, we have I = V /R = V/R and I = V /R = V/R. Plugging this in gives I = I + I = V R + V R 3
Thus ( V = I + ) R R Thus if you were to put the two resistors in parallel in a box, the box has total resistance = + R parallel R R Problem A simple example of a parallel circuit is a three-way light bulb. It consists of two resistors in parallel. In the dimmest position, the current goes through the resistor R. In the middle position, the current goes through R, and in the brightest position it is going through both resistors. Say we have a 50/00/50 W bulb. What is the total current in the 50 W position? Answer First, let s figure out the individual resistances. We ve already done this for the 50 W setting, because this consists of a single resistor with 0 V across it. This leads to R = 88Ω. The value of R is found from the same P = V /R we used before. Since the power is twice as large, the resistance must be twice as small, so R = 44Ω. We can get the current in the 50 W setting in two equivalent ways. Note that because the resistors are in parallel, V = V = 0 V. We have I tot = I + I = V R + V R = 0 V 88Ω + 0 V 44Ω Equivalently, we can use the formula for parallel resistances. This gives R parallel = 96Ω. Then = R parallel 88Ω + 44Ω =.004/Ω I tot = V = 0 V R parallel 96Ω =.5 A =.47 A +.833 A =.5 A Alternating Current All the formulas above apply for DC current. For AC current, the physics is the same, but the formulas need to be treated a little more carefully. The point is that the voltage and the current are changing with time. In the US, the wall current is V = V 0 sin ωt 4
Most of the formulas are the same. However, the formula for power is different. It is changed to P average = V 0I 0 The reason for the / is that because I 0 and V 0 are the peak values of the voltage and current. Thus to get the power dissipated in the circuit, you need to average over all the values. This results in the factor of / in the power. To compensate for this, people often define the rootmean-squared voltage and current: V rms = V 0 I rms = I 0 Then Ohm s law still applies, so V rms = I rms R, but now P average = V rms I rms The upshot is that if you use the rms voltage and current, all the old formulas imply. In fact, we ve already used this fact. I ve said that wall voltage is 0 V. This is rms voltage, which is why all the formulas I ve used are correct. 5