Chapter Review
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Chapter 21: Electric Currents and Direct-Current Circuits Chapter Review |
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Chapter Review
21 - 4 Resistors in Series and Parallel
A resistor in a circuit, indicated by a jagged line, represents a circuit element (such as a light bulb or a heater) that contains resistance. Circuit elements can be connected in different ways and the overall resistance, or equivalent resistance, of the combination depends on how they are connected. In this section we learn how to determine the equivalent resistance for combinations of resistors connected in series and in parallel.
Resistors in series are connected one after the other (or end to end). Connecting resistors in series has the same effect as making one resistor longer. It is important to remember that
the same current flows through resistors that are in series with each other.
The result is that the equivalent resistance Req of N resistors connected in series is the sum of the individual resistances
Consequently, Req is greater than any of the individual resistances that contribute to it.
Resistors in parallel are connected across the same potential difference.
Connecting resistors in parallel has the same effect as making one resistor wider. The result is that the equivalent resistance of N resistors connected in parallel can be found by the following relation:
Once 1/Req is determined, we can invert this result this result to get the value of Req in ohms. As a consequence of this result, we can see that for resistors in parallel, Req is smaller than the smallest of the individual resistances that contribute to it.
Physlet Illustration: Resistors in Parallel | |
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| In this simulation, a 10 Ohm, 15 Ohm, and 30 Ohm resistor are connected in parallel. Use the voltmeters and ammeters to measure the voltages and currents in the circuit. Are the values consistent with the rule for adding resistors in parallel? Can you verify Kirchoff's node rule? | |
Hints:
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Example 21.3 Equivalent Resistance: Determine the equivalent resistance of the circuit shown below given that R1 = 3.3 W, R2 = 4.5 W, R3 = 6.1 W, and R4 = 2.8 W.
Picture the Problem The picture shows the circuit in question.
Strategy We will first get the equivalent resistance of each parallel combination, then combine those results in series.
Solution
| 1. The equivalent resistance of R1 and R2 is: |  |
| 2. The equivalent resistance of R3 and R4 is: | |
| 3. The overall equivalent resistance then is: | Req = R12 + R34 = 1.904 W + 1.919 W = 3.8 W |
Insight When finding the equivalent resistance of many resistors it is often useful to take an "inside-out" approach as shown here.
Practice Quiz
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21-5 Kirchhoff's Rules
As circuitry becomes more complicated it becomes difficult to analyze the circuit using simple applications of Ohm's law. For these we use Kirchhoff's rules which apply the conservation of electric charge and the conservation of energy to circuit analysis.
Kirchhoff's first rule (for the conservation of charge) is called the junction rule:
The algebraic sum of all currents meeting at any junction in a circuit must equal zero.
A junction is a point in a circuit where three or more wires meet so that the current in the circuit may take different paths into or out of this point. In applying the junction rule as stated above, currents entering the junction are given a positive sign and current leaving the junction are given a negative sign. Application of the junction rule seems to imply that the directions of the currents must be known. However, it is not always obvious what the direction of every current will be. Fortunately, in these cases it is sufficient to just guess a direction and solve the equations. If the value of the current whose direction you guessed works out to be negative then you guessed wrong and you therefore know that it actually flows in the opposite direction from what you guessed.
Kirchhoff's second rule (for the conservation of energy) is called the loop rule:
The algebraic sum of all potential differences around any closed loop in a circuit must equal zero.
Applying these two rules involves several choices. These choices include the following:
For step 3 there are several things you should remember when writing your loop rule equations.
Applying these rules carefully will give you a system of equations that can be solved for currents, resistances, and potential differences in circuits.
Exercise 21.4 Kirchhoff's Rules: Use Kirchhoff's rules to determine the overall current, and the current through each of the resistors, in the circuit below. The relevent data is that R1 = 2.5 W, R2 = 3.3 W, R3 = 1.8 W, E1 = 20 V, and E2 = 6.0 V.
Solution
First I will choose the directions for the currents in the circuit realizing that some or all of these choices may not be correct. The figure below also labels the junctions in the circuit A - D.
By examining the circuit, we see that I3 can be determined by applying the loop rule to the small loop involving R3 and E2 shown below.
Moving around this loop counterclockwise from junction D back to D produces the loop rule equation
E2 - I3R3 = 0,
which gives
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The fact that we got a positive value for I3 shows that we chose the correct direction.
Further examination suggests that I2 can be determined by applying the loop rule to the large loop that includes E1 and E2 and R2.
Moving completely around this loop counterclockwise produces the loop rule equation
E1 + E2 + I2R2 = 0.
This gives
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The minus sign means that we chose the wrong direction for I2.
Let's pause for a moment to notice that I1 can be found using three different loops. (a) One loop would be the parallel combination involving R1 and R2 only, (b) another loop would be the larger path including E1, E2, and R1, and (c) the third loop would come from the path including E1, R3 , and R1. The choice with the least chance for error is (b). Why?
Moving around this loop counterclockwise produces the loop rule equation
E1 + E2 + I1R1 = 0.
This gives
You can see that the wrong direction was also chosen for I1.
All that remains is to find the overall current I. Inspection of the circuit suggests that we can determine I by applying the junction rule at junction A.
The junction rule equation at this junction becomes
-I - I1 - I2 = 0.
This gives
I = -(I1 + I2) = -(-10.4 A - 7.88 A) = 18 A.
So the final results are: I1 = 10 A, I2 = 7.9 A, I3 = 3.3 A, and I = 18 A.
Insight Make sure you understand the sign associated with each term in the loop rule and junction rule equations; it is very important that these be done consistently and correctly. Also, keep in mind that when you get a negative result for a current, that negative result must be used in any subsequent mathematical step; only drop the negative when stating the final result for that current. It is always better if you can choose the correct current directions in the beginning. In the circuit for this problem the polarities of the two emfs would immediately suggest that the directions of I1 and I2 should be opposite to what was chosen. Try to see this fact. The choice made for this example was done only to show you what happens when the wrong choice is made.
Physlet Illustration: Resistors in Series | |
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| In this simulation, a 10 Ohm, 15 Ohm, and 25 Ohm resistor are connected in series. Use the voltmeters and ammeters to measure the voltages and currents in the circuit. Are the values consistent with the rule for adding resistors in series? Can you verify Kirchoff's loop rule? | |
Hints:
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Practice Quiz
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