SAT Physics Circuit Elements and DC Circuits - Principal Components Of A DC Circuit

SAT Physics Circuit Elements and DC Circuits - Principal Components Of A DC Circuit

PRINCIPAL COMPONENTS OF A DC CIRCUIT
 
Battery

The purpose of a battery is to create electric potential. Think of a battery as a pump that creates the electrical pressure to push charges through an electric circuit. When a wire is connected to the positive terminal of a battery, charges flow as a current through the wire on their way to the negative terminal of the battery. Potential (voltage) is analogous to height in mechanics. When a mass is released from rest, it moves from a high height to a low height while moving through a distance Ah. If a positive charge is released from rest, it moves from a high potential to a low potential while moving through a potential difference AV. Positive charges match the force and energy characteristics of masses in gravity fields. Thus positive charges are the default charge in electricity. As a result, the convention is to visualize posi­tive charges leaving the positive terminal of the battery (high potential) and flowing through the circuit on the way to the negative terminal (low potential). Although this is how circuits are analyzed, it is not in truth what is really happening. Only electrons are free to move in circuits, and the actual motion of charges is composed of electrons moving in the opposite direction. Mathematically it makes no difference, as protons and electrons have the same charge. This just affects direction of charge flow. Even though it is not what is actually hap­pening, the convention is to visualize the flow charges in a circuit as a positive flow.
The potential difference (voltage) between the positive and negative terminals of the bat­tery provides the electrical pressure to the circuit. The overall resistance of the components in the circuit dictates the amount of charge allowed to flow from the battery. The flow of charge is known as the current, 7, and is measured in units of amperes (A) or coulombs per second.
In circuit diagrams, a long line and a short line represent batteries, as shown in Figure 12.1(a). Sometimes a battery is represented by a series of long and short lines as seen in Figure 12.1(b). This represents multiple battery cells connected in series, much like a series of batteries lined up end to end in a flashlight
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Figure 12.1. Batteries

The side of the battery drawn with a long line is the positive terminal. The side drawn with a short line is the negative terminal. When batteries are connected into circuits, positive charges depart from the positive terminal, travel through the circuit, and arrive back at the negative terminal of the battery.

Resistors
A resistor is a device with a known amount of resistance. As a result, resistors can be used to control current flow in the various branches of a circuit. The resistance, R, to current flow has many applications in how a circuit works. As current flow is resisted, some of the electrical energy is dissipated into heat energy in units of joules. Figure 12.2 shows the symbol for a resistor used in circuit diagrams.
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Figure 12.2. Resistor symbol

Resistors are rated by how much they can resist the flow of current. The actual measure­ment of resistance is in units of Ohms, Ω. Ohm’s law can be used to determine the resistance of ohmic resistors:
V=IR
Using the voltage (V) applied to the circuit and the current (I) flowing through the circuit, you can determine the resistance (R). Ohmic resistors are resistors that obey Ohm’s law. Resistors that do not obey Ohm’s law are said to be nonohmic. Problems on the SAT Subject Test will most likely involve resistors, and other devices having resistance, that are ohmic. Unless otherwise stated in the exam, assume devices are ohmic.
Lightbulbs are essentially resistors that give off light. The principal difference between a plain resistor and a lightbulb is that a lightbulb uses some of the energy dissipated through resistance to produce light. In circuit diagrams, lightbulbs may be represented as resistors. However, the symbol for lightbulbs often varies, as shown in Figure 12.3.
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Figure 12.3. Lightbulb symbols

Switch
The purpose of a switch is to direct current flow around a circuit. In a schematic diagram of a circuit, a switch is represented as a line at an angle to the circuit. Figure 12.4 shows a switch in the open position. When open, a switch blocks the flow of current. When a switch is closed, the circuit becomes complete and a current is able to flow.
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Figure 12.4. Symbol for an open switch

DC CIRCUITS
The SAT Subject Test in Physics will most likely have series and parallel resistor circuit problems for you to solve. Typically, these problems can be categorized as one of three types: series only, parallel only, or series-parallel. Each category has specific problem-solving strat-egies involved. Those strategies will be discussed in this section.
When multiple resistors are used in a circuit, they work together and have a combined resistance known as equivalent resistance. Adding resistors results in a single mathematical resistance value, which describes a single resistor that is equivalent to the resistors working together. A single resistor possessing this equivalent resistance can replace all the resistors added together. The method of finding equivalent resistance differs for series and parallel circuits.

Series Circuits
In a series circuit, the electrical components are arranged so there is only one path through the circuit. Figure 12.5 shows three resistors arranged in series with a 12-volt battery.
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Figure 12.5. Series circuit

The equivalent resistance for resistors in series, Rs, is simply the sum of the individual resistances.
R1 + R2 + R3 + ...................
If all of the resistors in a circuit are added together, the equivalent resistance is also the total resistance, Rs that the battery must push against.
As more resistors are added in series, the total resistance increases. According to Ohm’s law, V = IR, current flow is inversely proportional to resistance. Therefore, increasing resistance decreases the total amount of current.
In the complete series circuit shown in Figure 12.5, there is only one pathway. This means the current must flow through every resistor in turn. Think of current as water and the wires and resistors as the pipes carrying the water from the top of a hill (the positive terminal of the battery) to the bottom of a hill (the negative terminal of the battery). In a series circuit, there is a single path that all the current must follow. If the current flows through R1 it must also flow through R2 and R3. With only one pathway available, each of these resistors has the same amount of current flowing through them. Therefore, the total current leaving the battery,R1  is equal to the current in every series resistor.
IT = I1 = I2 = I3 = ................................
Although resistors in series all receive the same current, they do not have the same voltage. Voltage (electric potential) is similar to height. The 12-volt potential difference of the battery in Figure 12.5 can be thought of as a 12-volt hill. The positive terminal of the battery is at the top of a hill with a value of 12 volts. The negative terminal of the battery is at the bottom of the hill with a value of 0 volts. Charges flow as a current and drop down the voltage hill, essentially falling through the 12-volt potential difference. On their way to the bottom, the charges pass through circuit components such as resistors. However, no matter which path the charges follow, they must lose all the voltage given to them by the battery. The exact volt­age drop in each resistor depends on its resistance and how it is connected to the battery.
In a series circuit, the voltage drops must add up to the potential of the battery. There is only one path through the circuit, and the charges drop through each resistor in turn. The voltage drops across each resistor are summed, adding up to the total voltage produced by the battery.
VT = v1 + v2 + v3 + .............
Solving Series Circuits
1-6
Determine the voltage drop across each resistor.
 
WHAT'S THE TRICK?

You can use a table like the one below to organize circuit problems. The column headings are the variables for Ohm’s law, and any row in the table can be cal­culated using this law. The components used in the circuit are listed on the left. The top row contains the total values for the entire circuit and is labeled as the battery. The battery produces the total potential for the circuit, pushes against the total resistance in the circuit, and supplies the total current to the circuit. The values shown in bold type are the values given in the original problem. The values in regular type are the values determined during the course of the problem. They are preceded by a number enclosed in parentheses. This is the order in which the problem is solved. Below the table, the four steps are shown in detail.
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1. Add the resistors to find total resistance.
RT = R1 + R2 + R3
RT = 1Ω + 2Ω + 3Ω = 6Ω
2. Use Ohm’s law to find the total current.
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3. The current remains the same in series.
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4. Use Ohm’s law for each resistor.
V = IR
V1 =  l1R1  = (2 A)(1 Ω) = 2 V
V2 = l2R2 = (2 A)(2 Ω) = 4 V
V2 = l2R2 = (2 A)(3 Ω) = 6 V
You can use the voltage values of the individual resistors to double-check your solution. In series, the voltage drops should sum to equal the voltage of the battery.
VT = V1 + V2 + V3
VT = 2 V + 4 V + 6 V = 12 V
Using a table such as the one shown in Example 12.1 makes it easy to organize the values that may be required in a particular problem. This table illustrates the voltage, current, and resistance for individual resistors and the entire circuit. You can answer many general ques­tions about a circuit simply by reading the completed table.

Parallel Circuits
In a parallel circuit, the electrical components are arranged so there is more than one path through the circuit.Figure 12.6 shows three resistor arranged in parallel with a battery.
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Figure 12.6. Parallel circuits

The equivalent resistance for resistors in parallel, Rp, is more complicated than for resistors in series. Parallel resistors are added together so that the sum of their individual reciprocals equals the reciprocal of the total resistance.
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­Since the above formula solves for the reciprocal of parallel resistance, 1-12 you must remember to invert the answer to determine the equivalent resistance in parallel, Rp. In Figure 12.6, all the resistors are in parallel. The equivalent resistance is also the total resis­tance that the battery must push against. As more resistors are added in parallel, the total resistance decreases, causing the total current in the circuit to increase.
In a parallel circuit, the current moves through multiple pathways. The current flowing through each resistor is a portion of the total current that is produced by the battery. Think of the current as a fixed amount of water leaving the battery. When the water arrives at the junctions, where the paths in the circuit separate, the water will split. Some will flow through resistor 1, some through resistor 2, and the remainder through resistor 3. Then the water will reunite and continue back toward the battery. The total amount of water in the circuit is always the same. When it splits, it must still add up to the total amount that leaves the battery and that later returns to the battery. Therefore, the total current produced by the battery, IT, is equal to the sum of the current in the parallel resistors.
IT = I1 + I2 + I3 + .........
The potential in a parallel circuit is the same for all resistors. Again, potential (voltage) can be thought of as height and current can be represented by the flow of water. The battery pushes the water up the hill through a set height. In Figure 12.6, there are three paths, one through each resistor. If water leaves the top of the hill (positive terminal of the battery) and flows through the circuit, it may follow different paths. However, the water must always drop the same amount of height in order to return to the bottom of the hill (negative terminal of the battery). The drop in height must also be equal to the height created by the pumping action of the battery. In a parallel circuit, the voltage drops across each resistor are equal to each other and they are equal to the total voltage produced by the battery.
VT = V1 = V2 = V3 = ...........
Most electrical wiring for home use is wired in parallel. Each time a switch is closed, the same voltage (120 V for most homes) is applied across a resistor (lightbulb, television set, computer, and so on). The effect of having one appliance turned on or off does not affect the voltage applied to other appliances that are turned on. A home wired in series would require all appliances to be on all the time in order for the circuit to be completed.
             Solving parallel circuits
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Determine the current in each resistor.
 
WHAT'S THE TRICK?

As before, use a table to organize the problem.
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1. To find the total resistance, add the inverse of each resistor.
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2. Use Ohm’s law to find the total current.
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3. Voltage remains the same in parallel.
VT = V1 = V2 = V3 = 6V
4. Use Ohm’s law for each resistor.
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Use the current values of the individual resistors to double-check the solution. In parallel, the current in each pathway should sum to equal the current produced by the battery.
IT = I1 = I2 = 6A + 3A = 9A

Series-Parallel Circuits
A series-parallel circuit is somewhat more complicated. However, you can employ the same table. Before using the table, however, it is helpful to find the total equivalent resistance, Rv of the circuit.
Series-Parallel Circuits
1-18
(A) Determine the equivalent resistance.
 
WHAT'S THE TRICK?

Group resistors and add them together to simplify the circuit progressively. In the diagram above, R2 and R3 are in series and can be consolidated as R23.
R23 : R2 + R3 : 2 Ω + 2 Ω = 4 Ω
Now redraw the circuit.
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R23 and R4 can now be added in parallel to obtain equivalent resistor R234:
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Invert to solve for resistance in parallel.
R234 = 2 Ω.
Redraw the circuit as shown below.
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R234 can now be added to R1 to find the total equivalent resistance in the circuit.
RT = R1234 = R1 + R234 = 1 Ω + 2 Ω = 3 Ω
(B) Determine the voltage drops and current through each resistor.
 
WHAT'S THE TRICK?

The values shown in bold type below are the values given in the original problem and the values of the equivalent resistances for R23, R234, and R1234. The total resistance, R1234, is the resistance the battery must push against. It is recorded in the table below on the row labeled “Battery.” The values in regular type are the values determined during the course of the problem. They are preceded by a number enclosed in parentheses. This is the order in which the problem is solved.
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1. Use Ohm’s law to solve for the total current
1-23
2. Current remains the same in each resistor in series.
IT = I1 = I234 = 4A
3. Use Ohm's law to find the voltage in each resistor.
 V = IR
V1 = l1R1 = (4 A)(1 Ω)= 4 V
V234 = l234R234 = (4 A)(2 Ω)= 8 V
Voltage adds in series.you can use this to double-check the values so far.
VT = V1 + V234 = 4 V + 8 V= 12 V   
4. There is a voltage drop of 4 V across the R1 resistor.Since the battery has a total voltage of 12 V, this 4 V drop leaves 8 V to go across parallel resistor R23 and R4.Since they are in parallel,R23 and R4 both receive 8 V.
5. Use Ohm's law to find the current in each resistor.
1-24
current adds in parallel.you can use this to double- check the values.
I234 = I23 + I4 = 2 A + 2 A= 4 A
6. It has been established that 4 A flows through R234. of this total, 2 A flows through R4. Therefore, the remaining 2 A must flow through R2 and R3.
7. Use Ohm’s law for to find the voltage in each resistor.
V2 = l2R2 = (2 A)(2 Ω) = 4 V
V3 = I3R3 = (2 A)(2 Ω) = 4V
Voltage adds in series. You can use this to double-check the values.
V23 = V2 + V3 = 4 V + 4 V =8 V

HEAT AND POWER DISSIPATION
As current flows through resistors, the frictional resistance causes the wires to heat up. The amount of heat, Q, dissipated per second is known as power dissipation. Heat is similar to work in that heat is also a transfer of energy into or out of a system. Heat is the transfer of thermal energy. Be careful when working with heat in electricity problems. The variable Q is used both for heat and for charge. So it is possible to mistake them for each other. Power is quantified in units of watts (joules per second).

Joule’s Law
Joule’s law states that the heat dissipated in a circuit is equal to the current squared multi­plied by the resistance and the time that the current flows through the resistor.
Heat = Q = I2Rt
In this case, the variable Q represents heat measured in joules. Joule’s law shows that heat is directly proportional to the square of current. If current doubles, heat quadruples.

Power
Power is the rate of change in energy. It is measured in joules per second (J/s), which is known as watts (W).
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In circuits, the change in energy may involve the electric energy generated (created) by power sources or dissipated (lost) in circuit components, ΔUE = QV. Keep in mind that cur­rent is the amount of charge divided by time. Combining these equations creates a useful power equation for use in analyzing electric circuits.
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The resulting equation, P= IV, can be combined with Ohm’s law, V = IR, to create other useful power equation variations.
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Most problems can be answered using both P = IV and V = IR as separate equations. However, being aware of all the power equation variations does allow for faster problem solving.

Brightness of Lightbulbs
The brightness of a lightbulb is related to the amount of power it dissipates. If given a circuit with different lightbulbs and asked to determine which lightbulb is brighter, simply treat the lightbulbs as resistors and solve for the current and voltage of each lightbulb, as shown in the previous circuit example problems. Then multiply the currents and voltages (P = IV) to determine the power dissipated by each bulb. The bulb dissipating the most power will be the brightest.
Lightbulbs are sold by the amount of power, in watts, that they dissipate. The greater the power printed on the bulb, the brighter it burns. However, the power that lightbulbs are rated as dissipating at is true only if the bulbs are used in typical household circuits. Homes in the United States are wired in parallel with voltages of 110 V to 120 V. In purely parallel circuits, voltage in all branches is the same as the voltage of the power source. Every lightbulb should then have the same voltage. Under these exact conditions, the printed power on a lightbulb is the power that will be dissipated. However, if the voltage supplied to a parallel circuit is not 110 V to 120 V or if the lightbulbs are moved to a series circuit, then the power dissipated will differ from the power rating printed on the lightbulb.
Remember that lightbulbs are essentially resistors that glow. Although the power rating on a bulb is not really constant, the resistance of a lightbulb is constant no matter which circuit it is used in. For lightbulbs, the relationship between power, potential, and resistance is impor­tant. This makes the following power equation very useful when working with lightbulbs.
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In purely parallel circuits composed of lightbulbs, all branches and all bulbs have the same voltage as the power supply. This means they are all exposed to the maximum voltage, and they all dissipate the greatest amount of power. Therefore, lightbulbs in parallel circuits are at their brightest. Changing the voltage of the power supply changes the brightness of bulbs in parallel. According to the above equation, power is directly proportional to the square of voltage. Doubling voltage quadruples power dissipation.
If lightbulbs are wired in series, they use less power than they are rated for, and they bum dimmer. In series, voltage adds. Each bulb in series receives only a portion of the total voltage supplied by the power source. A lower voltage means less power and dimmer bulbs.

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