Here, we note the equivalent resistance as. Notice that resistors and are in series. They can be combined into a single equivalent resistance. One method of keeping track of the process is to include the resistors as subscripts. Here the equivalent resistance of and is. The circuit now reduces to three resistors, shown in Figure 6.
Redrawing, we now see that resistors and constitute a parallel circuit. Those two resistors can be reduced to an equivalent resistance:. This step of the process reduces the circuit to two resistors, shown in in Figure 6. Here, the circuit reduces to two resistors, which in this case are in series. These two resistors can be reduced to an equivalent resistance, which is the equivalent resistance of the circuit:.
The main goal of this circuit analysis is reached, and the circuit is now reduced to a single resistor and single voltage source. Now we can analyze the circuit. The current provided by the voltage source is. This current runs through resistor and is designated as.
Looking at Figure 6. The resistors and are in series so the currents and are equal to. The potential drops are and. The final analysis is to look at the power supplied by the voltage source and the power dissipated by the resistors.
The power dissipated by the resistors is. The total energy is constant in any process. Therefore, the power supplied by the voltage source is. Analyzing the power supplied to the circuit and the power dissipated by the resistors is a good check for the validity of the analysis; they should be equal. We can consider to be the resistance of wires leading to and a Find the equivalent resistance of the circuit. Then use this result to find the equivalent resistance of the series connection with.
The current through is equal to the current from the battery. The voltage across can be found using. To find the equivalent resistance of the circuit, notice that the parallel connection of R 2 R2 and R 3 R3 is in series with R 1 R1 , so the equivalent resistance is. The total resistance of this combination is intermediate between the pure series and pure parallel values and , respectively. The current through is equal to the current supplied by the battery:. The voltage across is.
The voltage applied to and is less than the voltage supplied by the battery by an amount. When wire resistance is large, it can significantly affect the operation of the devices represented by and. To find the current through , we must first find the voltage applied to it. The voltage across the two resistors in parallel is the same:. The current is less than the that flowed through when it was connected in parallel to the battery in the previous parallel circuit example.
The power dissipated by is given by. The analysis of complex circuits can often be simplified by reducing the circuit to a voltage source and an equivalent resistance. Even if the entire circuit cannot be reduced to a single voltage source and a single equivalent resistance, portions of the circuit may be reduced, greatly simplifying the analysis. Consider the electrical circuits in your home. Give at least two examples of circuits that must use a combination of series and parallel circuits to operate efficiently.
One implication of this last example is that resistance in wires reduces the current and power delivered to a resistor. If wire resistance is relatively large, as in a worn or a very long extension cord, then this loss can be significant.
If a large current is drawn, the drop in the wires can also be significant and may become apparent from the heat generated in the cord. For example, when you are rummaging in the refrigerator and the motor comes on, the refrigerator light dims momentarily. Similarly, you can see the passenger compartment light dim when you start the engine of your car although this may be due to resistance inside the battery itself.
What is happening in these high-current situations is illustrated in Figure 6. The device represented by has a very low resistance, so when it is switched on, a large current flows.
This increased current causes a larger drop in the wires represented by , reducing the voltage across the light bulb which is , which then dims noticeably. Two resistors connected in series are connected to two resistors that are connected in parallel. The series-parallel combination is connected to a battery.
Each resistor has a resistance of. The wires connecting the resistors and battery have negligible resistance. A current of runs through resistor. What is the voltage supplied by the voltage source? Since they are in series, the current through equals the current through. Since , the current through each will be. The power dissipated by the resistors is equal to the sum of the power dissipated by each resistor:.
Since the power dissipated by the resistors equals the power supplied by the battery, our solution seems consistent. Significance If a problem has a combination of series and parallel, as in this example, it can be reduced in steps by using the preceding problem-solving strategy and by considering individual groups of series or parallel connections.
Why do lights dim when a large appliance is switched on? The answer is that the large current the appliance motor draws causes a significant drop in the wires and reduces the voltage across the light. A switch has a variable resistance that is nearly zero when closed and extremely large when open, and it is placed in series with the device it controls.
Explain the effect the switch in Figure 7 has on current when open and when closed. Figure 7. A switch is ordinarily in series with a resistance and voltage source. Ideally, the switch has nearly zero resistance when closed but has an extremely large resistance when open.
Note that in this diagram, the script E represents the voltage or electromotive force of the battery. There is a voltage across an open switch, such as in Figure 7. Why, then, is the power dissipated by the open switch small?
A student in a physics lab mistakenly wired a light bulb, battery, and switch as shown in Figure 8. Explain why the bulb is on when the switch is open, and off when the switch is closed. Do not try this—it is hard on the battery! Figure 8. Knowing that the severity of a shock depends on the magnitude of the current through your body, would you prefer to be in series or parallel with a resistance, such as the heating element of a toaster, if shocked by it? Some strings of holiday lights are wired in series to save wiring costs.
An old version utilized bulbs that break the electrical connection, like an open switch, when they burn out. If one such bulb burns out, what happens to the others? If such a string operates on V and has 40 identical bulbs, what is the normal operating voltage of each? Newer versions use bulbs that short circuit, like a closed switch, when they burn out.
If such a string operates on V and has 39 remaining identical bulbs, what is then the operating voltage of each? If two household lightbulbs rated 60 W and W are connected in series to household power, which will be brighter? Suppose you are doing a physics lab that asks you to put a resistor into a circuit, but all the resistors supplied have a larger resistance than the requested value. How would you connect the available resistances to attempt to get the smaller value asked for?
Explain why resistance cords become warm and waste energy when the radio is on. Some light bulbs have three power settings not including zero , obtained from multiple filaments that are individually switched and wired in parallel. What is the minimum number of filaments needed for three power settings? Note: Data taken from figures can be assumed to be accurate to three significant digits. What are the largest and smallest resistances you can obtain by connecting a An W toaster, a W electric frying pan, and a W lamp are plugged into the same outlet in a A, V circuit.
The three devices are in parallel when plugged into the same socket. What power would one headlight and the starter consume if connected in series to a Neglect any other resistance in the circuit and any change in resistance in the two devices. In both parts explicitly show how you follow the steps in the Problem-Solving Strategies for Series and Parallel Resistors above. Referring to Figure 5: a Calculate P 3 and note how it compares with P 3 found in the first two example problems in this module.
Refer to Figure 6 and the discussion of lights dimming when a heavy appliance comes on. Assume negligible change in bulb resistance. A kV power transmission line carrying 5. What is the resistance to ground of of these insulators? Figure 9. High-voltage kV transmission line carrying 5. The row of ceramic insulators provide 1. Skip to main content. Circuits and DC Instruments. Search for:. Resistors in Series and Parallel Learning Objectives By the end of this section, you will be able to: Draw a circuit with resistors in parallel and in series.
Contrast the way total resistance is calculated for resistors in series and in parallel. Explain why total resistance of a parallel circuit is less than the smallest resistance of any of the resistors in that circuit.
Calculate total resistance of a circuit that contains a mixture of resistors connected in series and in parallel. Making Connections: Conservation Laws The derivations of the expressions for series and parallel resistance are based on the laws of conservation of energy and conservation of charge, which state that total charge and total energy are constant in any process.
These two laws are directly involved in all electrical phenomena and will be invoked repeatedly to explain both specific effects and the general behavior of electricity. Example 1. The same current flows through each resistor in series. Individual resistors in series do not get the total source voltage, but divide it. Example 2.
Strategy and Solution for a The total resistance for a parallel combination of resistors is found using the equation below. Discussion for b Current I for each device is much larger than for the same devices connected in series see the previous example. This is consistent with conservation of charge. Strategy and Solution for d The power dissipated by each resistor can be found using any of the equations relating power to current, voltage, and resistance, since all three are known.
Each resistor in parallel has the same full voltage of the source applied to it. Charge does NOT become used up by resistors such that there is less of it at one location compared to another.
The charges can be thought of as marching together through the wires of an electric circuit, everywhere marching at the same rate. Current - the rate at which charge flows - is everywhere the same. It is the same at the first resistor as it is at the last resistor as it is in the battery. Mathematically, one might write. These current values are easily calculated if the battery voltage is known and the individual resistance values are known. Using the individual resistor values and the equation above, the equivalent resistance can be calculated.
As discussed in Lesson 1 , the electrochemical cell of a circuit supplies energy to the charge to move it through the cell and to establish an electric potential difference across the two ends of the external circuit. This is to say that the electric potential at the positive terminal is 1.
As charge moves through the external circuit, it encounters a loss of 1. This loss in electric potential is referred to as a voltage drop. It occurs as the electrical energy of the charge is transformed to other forms of energy thermal, light, mechanical, etc.
If an electric circuit powered by a 1. There is a voltage drop for each resistor, but the sum of these voltage drops is 1. This concept can be expressed mathematically by the following equation:. To illustrate this mathematical principle in action, consider the two circuits shown below in Diagrams A and B. Suppose that you were to asked to determine the two unknown values of the electric potential difference across the light bulbs in each circuit.
To determine their values, you would have to use the equation above. The battery is depicted by its customary schematic symbol and its voltage is listed next to it. Determine the voltage drop for the two light bulbs and then click the Check Answers button to see if you are correct. Earlier in Lesson 1, the use of an electric potential diagram was discussed.
An electric potential diagram is a conceptual tool for representing the electric potential difference between several points on an electric circuit. Consider the circuit diagram below and its corresponding electric potential diagram. The circuit shown in the diagram above is powered by a volt energy source.
There are three resistors in the circuit connected in series, each having its own voltage drop. The negative sign for the electric potential difference simply denotes that there is a loss in electric potential when passing through the resistor. Conventional current is directed through the external circuit from the positive terminal to the negative terminal. Since the schematic symbol for a voltage source uses a long bar to represent the positive terminal, location A in the diagram is at the positive terminal or the high potential terminal.
Location A is at 12 volts of electric potential and location H the negative terminal is at 0 volts. In passing through the battery, the charge gains 12 volts of electric potential. And in passing through the external circuit, the charge loses 12 volts of electric potential as depicted by the electric potential diagram shown to the right of the schematic diagram. This 12 volts of electric potential is lost in three steps with each step corresponding to the flow through a resistor.
In passing through the connecting wires between resistors, there is little loss in electric potential due to the fact that a wire offers relatively little resistance to the flow of charge. Since locations A and B are separated by a wire, they are at virtually the same electric potential of 12 V. When a charge passes through its first resistor, it loses 3 V of electric potential and drops down to 9 V at location C. Since location D is separated from location C by a mere wire, it is at virtually the same 9 V electric potential as C.
When a charge passes through its second resistor, it loses 7 V of electric potential and drops down to 2 V at location E. Since location F is separated from location E by a mere wire, it is at virtually the same 2 V electric potential as E. Finally, as a charge passes through its last resistor, it loses 2 V of electric potential and drops down to 0 V at G.
At locations G and H, the charge is out of energy and needs an energy boost in order to traverse the external circuit again. The energy boost is provided by the battery as the charge is moved from H to A.
The Ohm's law equation can be used for any individual resistor in a series circuit. When combining Ohm's law with some of the principles already discussed on this page, a big idea emerges. Wherever the resistance is greatest, the voltage drop will be greatest about that resistor.
The Ohm's law equation can be used to not only predict that resistor in a series circuit will have the greatest voltage drop, it can also be used to calculate the actual voltage drop values.
The above principles and formulae can be used to analyze a series circuit and determine the values of the current at and electric potential difference across each of the resistors in a series circuit. Their use will be demonstrated by the mathematical analysis of the circuit shown below. The goal is to use the formulae to determine the equivalent resistance of the circuit R eq , the current at the battery I tot , and the voltage drops and current for each of the three resistors. The analysis begins by using the resistance values for the individual resistors in order to determine the equivalent resistance of the circuit.
Now that the equivalent resistance is known, the current at the battery can be determined using the Ohm's law equation.
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