Understanding electrical circuits is fundamental to grasping how modern technology functions, from simple household devices to complex electronic systems. Activity 1.2.Even so, 3 provides a practical exploration into the behavior of circuits, specifically focusing on the differences between series and parallel configurations. Practically speaking, this activity is crucial for students and enthusiasts alike, as it lays the groundwork for more advanced concepts in electronics and physics. The provided answer key offers a structured approach to verifying solutions and deepening comprehension of circuit principles.
Activity 1.2.3: Exploring Series and Parallel Circuits - Answer Key
Introduction This activity involves constructing simple electrical circuits using batteries, wires, resistors, and light bulbs to observe how voltage, current, and resistance behave differently in series versus parallel setups. The core objective is to identify the key characteristics that distinguish these fundamental circuit types and to calculate equivalent resistances and predict current flow accurately. Mastery of these concepts is essential for troubleshooting real-world circuits and designing new ones.
Steps and Solutions
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Series Circuit Construction (Question 1):
- Step: Connect one battery, one resistor, and one light bulb in a single continuous loop. Ensure all components are connected end-to-end.
- Solution: The circuit diagram should show components connected sequentially: Battery (+) terminal connected to Resistor, Resistor connected to Light Bulb, Light Bulb connected to Battery (-) terminal. This forms a single path for current flow.
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Parallel Circuit Construction (Question 2):
- Step: Connect one battery, two resistors, and two light bulbs such that both resistors (and bulbs) are connected directly across the battery terminals, creating multiple branches.
- Solution: The circuit diagram should show the battery connected to two distinct paths. One path contains Resistor A and Bulb A in series, the other path contains Resistor B and Bulb B in series. Both paths start and end at the same battery points.
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Current Calculation (Question 3):
- Step: Given a circuit with a 9V battery and a single resistor of 10Ω, calculate the current flowing through the circuit.
- Solution: Using Ohm's Law (I = V / R), where V = 9V and R = 10Ω, the current I = 9V / 10Ω = 0.9 Amperes (A). This current is the same at every point in a series circuit.
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Equivalent Resistance - Series (Question 4):
- Step: Calculate the equivalent resistance of three resistors (R1 = 5Ω, R2 = 10Ω, R3 = 15Ω) connected end-to-end in a series circuit.
- Solution: In a series circuit, total resistance (R_total) is the sum of individual resistances. R_total = R1 + R2 + R3 = 5Ω + 10Ω + 15Ω = 30Ω. The equivalent resistance is always greater than the largest individual resistor in a series circuit.
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Equivalent Resistance - Parallel (Question 5):
- Step: Calculate the equivalent resistance of two resistors (R1 = 6Ω, R2 = 3Ω) connected side-by-side across the same battery terminals.
- Solution: In a parallel circuit, the equivalent resistance (R_eq) is calculated using the reciprocal formula: 1/R_eq = 1/R1 + 1/R2. 1/R_eq = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2. Which means, R_eq = 2Ω. The equivalent resistance is always less than the smallest individual resistor in a parallel circuit.
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Current Distribution (Question 6):
- Step: Explain why the current through each resistor is different in a parallel circuit with R1 = 4Ω and R2 = 12Ω connected to a 12V battery.
- Solution: In a parallel circuit, the voltage across each branch is the same (12V). On the flip side, current is determined by Ohm's Law (I = V / R). So, the current through R1 is I1 = 12V / 4Ω = 3A, and the current through R2 is I2 = 12V / 12Ω = 1A. The total current from the battery is the sum (I_total = I1 + I2 = 4A). The current splits according to the resistance of each branch.
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Brightness Comparison (Question 7):
- Step: Compare the brightness of a single light bulb connected to a 9V battery with the brightness of the same bulb connected in series with another identical bulb to the same 9V battery.
- Solution: When two identical bulbs are connected in series to the same 9V battery, each bulb experiences only half the voltage (4.5V) compared to the single bulb (9V). Since brightness is proportional to power (P = V² / R), the voltage drop across each series bulb is halved, resulting in each bulb being significantly dimmer than the single bulb. Brightness is reduced in series due to voltage division.
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Voltage Drop (Question 8):
Calculate the voltage drop across a 20Ω resistor in a series circuit with a 12V battery and a total resistance of 60Ω.
Solution: In a series circuit, the voltage drop across a resistor is proportional to its resistance. First, find the total current: I = V / R_total = 12V / 60Ω = 0.2A. Then, apply Ohm's Law to the 20Ω resistor: V_R = I × R = 0.2A × 20Ω = 4V. Alternatively, use the voltage divider rule: V_R = (R / R_total) × V = (20Ω / 60Ω) × 12V = 4V. The voltage drop is directly proportional to the resistor's share of the total resistance.
- Power Dissipation (Question 9): Determine the power dissipated by a 10Ω resistor carrying a current of 2A.
Solution: Power can be calculated using P = I²R. Substituting the values: P = (2A)² × 10Ω = 4 × 10 = 40W. Alternatively, if the voltage across the resistor is known, P = V × I or P = V²/R could be used. Power dissipation increases with the square of the current.
- Short Circuit Analysis (Question 10): Explain what happens to the current in a parallel circuit when one branch is accidentally shorted (resistance approaches zero).
Solution: When a branch is shorted, its resistance becomes nearly zero, creating a path of least resistance. According to Ohm's Law, the current through that branch becomes extremely large (I = V / 0 ≈ ∞), limited only by the internal resistance of the power source and wiring. This can cause overheating, damage to the circuit, or tripping of protective devices like fuses. A short circuit diverts most of the current through the low-resistance path, potentially causing hazardous conditions.
Conclusion
Understanding the behavior of resistors in series and parallel configurations is fundamental to analyzing electrical circuits. Whether calculating equivalent resistance, current distribution, or power dissipation, mastering these concepts enables precise control over circuit performance. Series connections increase total resistance and divide voltage, while parallel connections decrease total resistance and divide current. These principles, governed by Ohm's Law and Kirchhoff's Laws, are essential for designing safe and efficient circuits. Practical applications range from simple lighting systems to complex electronic devices, highlighting the importance of these foundational principles in electrical engineering and everyday technology.
