Resistors In Series And Parallel Practice Problems

Let's delve into the fascinating world of resistors arranged in series and parallel configurations, focusing on practical problem-solving to solidify your understanding. Mastering these concepts is fundamental to understanding basic circuit analysis and is crucial for anyone working with electronics.

Resistors in Series and Parallel: A Practical Guide

Resistors, fundamental components in electronic circuits, control the flow of electrical current. Understanding how they behave in series and parallel arrangements is essential for designing and analyzing circuits effectively. We'll explore the core principles, formulas, and step-by-step solutions to common resistor network problems.

Series Resistors: A Single Path

When resistors are connected in series, they form a single path for current to flow. Imagine a single lane road; all cars must travel the same route.

Key Characteristics of Series Resistors:

• Same Current: The current flowing through each resistor in a series circuit is the same.
• Voltage Division: The total voltage across the series combination is divided among the resistors, proportional to their resistance values.
• Total Resistance: The total resistance (equivalent resistance) of resistors in series is the sum of individual resistances.

Formula for Total Resistance in Series:

R_total = R₁ + R₂ + R₃ + ... + R_n

Where:

• R_total is the total resistance.
• R₁, R₂, R₃, ..., R_n are the individual resistances.

Practical Problem 1: Calculating Total Resistance

Problem: Three resistors with values of 100 ohms, 220 ohms, and 330 ohms are connected in series. Calculate the total resistance.

Solution:

R_total = R₁ + R₂ + R₃ R_total = 100 ohms + 220 ohms + 330 ohms R_total = 650 ohms

Therefore, the total resistance of the series combination is 650 ohms.

Practical Problem 2: Calculating Voltage Drop Across Each Resistor

Problem: Two resistors, R₁ = 1k ohms and R₂ = 2k ohms, are connected in series with a 9V battery. Calculate the voltage drop across each resistor.

Solution:

1. Calculate Total Resistance:

   R_total = R₁ + R₂ R_total = 1000 ohms + 2000 ohms R_total = 3000 ohms

2. Calculate the Current (I) flowing through the circuit using Ohm's Law:

   Ohm's Law: V = I * R (Voltage = Current * Resistance) Therefore: I = V / R I = 9V / 3000 ohms I = 0.003 A or 3 mA (milliamperes)

3. Calculate Voltage Drop Across R₁ (V₁):

   V₁ = I * R₁ V₁ = 0.003 A * 1000 ohms V₁ = 3V

4. Calculate Voltage Drop Across R₂ (V₂):

   V₂ = I * R₂ V₂ = 0.003 A * 2000 ohms V₂ = 6V

Therefore, the voltage drop across R₁ is 3V, and the voltage drop across R₂ is 6V. Notice that the sum of the voltage drops (3V + 6V) equals the total voltage (9V), as expected in a series circuit.

Practical Problem 3: Determining Resistor Value for a Specific Voltage Drop

Problem: You have a 12V power supply and need to drop the voltage to 5V for a particular component. You have a 1k ohm resistor available. What value of resistor should you place in series with the 1k ohm resistor to achieve the desired 5V drop across the component (which is in parallel with the 1k ohm resistor)?

Solution:

1. Determine the required voltage drop across the added resistor: Since the total voltage is 12V and we want 5V across the 1k ohm resistor, the added resistor must drop 12V - 5V = 7V.
2. Calculate the current flowing through the 1k ohm resistor (and therefore the entire series circuit): Using Ohm's Law: I = V / R = 5V / 1000 ohms = 0.005A or 5mA.
3. Calculate the required resistance of the added resistor: Using Ohm's Law again, but this time solving for R: R = V / I = 7V / 0.005A = 1400 ohms or 1.4k ohms.

Therefore, you should place a 1.4k ohm resistor in series with the 1k ohm resistor to achieve a 5V drop across the 1k ohm resistor (and the component in parallel with it).

Parallel Resistors: Multiple Paths

When resistors are connected in parallel, they provide multiple paths for current to flow. Think of a multi-lane highway; cars can choose different routes.

Key Characteristics of Parallel Resistors:

• Same Voltage: The voltage across each resistor in a parallel circuit is the same.
• Current Division: The total current entering the parallel combination divides among the resistors, inversely proportional to their resistance values.
• Total Resistance: The total resistance (equivalent resistance) of resistors in parallel is less than the smallest individual resistance.

Formula for Total Resistance in Parallel (General Case):

1 / R_total = 1 / R₁ + 1 / R₂ + 1 / R₃ + ... + 1 / R_n

Formula for Total Resistance in Parallel (Two Resistors):

R_total = (R₁ * R₂) / (R₁ + R₂)

Practical Problem 4: Calculating Total Resistance (Parallel)

Problem: Two resistors with values of 470 ohms and 1k ohms are connected in parallel. Calculate the total resistance.

Solution:

Using the two-resistor formula:

R_total = (R₁ * R₂) / (R₁ + R₂) R_total = (470 ohms * 1000 ohms) / (470 ohms + 1000 ohms) R_total = 470000 / 1470 R_total ≈ 319.73 ohms

Therefore, the total resistance of the parallel combination is approximately 319.73 ohms.

Practical Problem 5: Calculating Current Through Each Resistor (Parallel)

Problem: A 6V battery is connected across two resistors in parallel: R₁ = 200 ohms and R₂ = 300 ohms. Calculate the current flowing through each resistor.

Solution:

1. Voltage across each resistor is the same: In a parallel circuit, the voltage across each resistor is equal to the source voltage, which is 6V.

2. Calculate Current through R₁ (I₁) using Ohm's Law:

   I₁ = V / R₁ I₁ = 6V / 200 ohms I₁ = 0.03 A or 30 mA

3. Calculate Current through R₂ (I₂) using Ohm's Law:

   I₂ = V / R₂ I₂ = 6V / 300 ohms I₂ = 0.02 A or 20 mA

Therefore, the current flowing through R₁ is 30 mA, and the current flowing through R₂ is 20 mA.

Practical Problem 6: Calculating Total Current and Total Resistance (Parallel)

Problem: Resistors of 10 ohms, 20 ohms, and 30 ohms are connected in parallel to a 12V source. Find the total current supplied by the source and the total equivalent resistance of the parallel combination.

Solution:

1. Calculate the current through each resistor:

   • I₁ = V / R₁ = 12V / 10 ohms = 1.2A
   • I₂ = V / R₂ = 12V / 20 ohms = 0.6A
   • I₃ = V / R₃ = 12V / 30 ohms = 0.4A

2. Calculate the total current:

   I_total = I₁ + I₂ + I₃ = 1.2A + 0.6A + 0.4A = 2.2A

3. Calculate the total equivalent resistance: You can use either the reciprocal formula or Ohm's Law:

   • Reciprocal Formula: 1 / R_total = 1/10 + 1/20 + 1/30 = 6/60 + 3/60 + 2/60 = 11/60. Therefore, R_total = 60/11 ≈ 5.45 ohms.
   • Ohm's Law: R_total = V / I_total = 12V / 2.2A ≈ 5.45 ohms.

Therefore, the total current supplied by the source is 2.2A, and the total equivalent resistance is approximately 5.45 ohms.

Series-Parallel Combinations: Putting It All Together

Many circuits involve a combination of series and parallel resistors. Solving these requires breaking down the circuit into smaller, manageable sections.

General Strategy:

1. Identify series and parallel sections: Look for resistors that are clearly in series or parallel with each other.
2. Simplify: Calculate the equivalent resistance of the series and parallel sections.
3. Redraw the circuit: Replace the simplified sections with their equivalent resistances.
4. Repeat: Continue simplifying until you have a single equivalent resistance for the entire network.
5. Solve for currents and voltages: Use Ohm's Law and Kirchhoff's Laws to determine the currents and voltages in the circuit.

Practical Problem 7: Analyzing a Series-Parallel Circuit

Problem: Consider a circuit with the following configuration:

• A 12V voltage source.
• R₁ = 100 ohms in series with the source.
• A parallel combination of R₂ = 200 ohms and R₃ = 300 ohms connected after R₁.

Calculate:

• The total resistance of the circuit.
• The total current supplied by the source.
• The current through R₂ and R₃.
• The voltage drop across each resistor.

Solution:

1. Simplify the parallel combination (R₂ and R₃):

   R₂₃ = (R₂ * R₃) / (R₂ + R₃) = (200 ohms * 300 ohms) / (200 ohms + 300 ohms) = 60000 / 500 = 120 ohms.

2. Redraw the circuit: Now you have a simplified circuit with a 12V source, R₁ = 100 ohms in series with R₂₃ = 120 ohms.

3. Calculate the total resistance (R_total):

   R_total = R₁ + R₂₃ = 100 ohms + 120 ohms = 220 ohms.

4. Calculate the total current (I_total) supplied by the source:

   I_total = V / R_total = 12V / 220 ohms ≈ 0.0545 A or 54.5 mA.

5. Calculate the voltage drop across R₁ (V₁):

   V₁ = I_total * R₁ = 0.0545 A * 100 ohms ≈ 5.45V.

6. Calculate the voltage drop across the parallel combination (R₂₃): Since R₂₃ is equivalent to the parallel combination of R₂ and R₃, the voltage drop across them is the same. We can find it by subtracting V₁ from the source voltage:

   V₂₃ = V - V₁ = 12V - 5.45V ≈ 6.55V.

7. Calculate the current through R₂ (I₂):

   I₂ = V₂₃ / R₂ = 6.55V / 200 ohms ≈ 0.0328 A or 32.8 mA.

8. Calculate the current through R₃ (I₃):

   I₃ = V₂₃ / R₃ = 6.55V / 300 ohms ≈ 0.0218 A or 21.8 mA.

Summary of Results:

• Total Resistance: 220 ohms
• Total Current: 54.5 mA
• Voltage Drop across R₁: 5.45V
• Voltage Drop across R₂: 6.55V
• Voltage Drop across R₃: 6.55V
• Current through R₂: 32.8 mA
• Current through R₃: 21.8 mA

Notice that I₂ + I₃ ≈ I_total (32.8 mA + 21.8 mA ≈ 54.6 mA, which is very close to 54.5 mA, the small difference due to rounding). This confirms that the total current divides between the parallel branches.

Practical Problem 8: A More Complex Series-Parallel Circuit

Problem: A circuit consists of a 24V source connected to the following:

• R1 = 20 ohms in series with the source.
• After R1, the circuit splits into two parallel branches:
  • Branch 1: R2 = 30 ohms in series with R3 = 40 ohms.
  • Branch 2: R4 = 50 ohms.

Find:

• The total equivalent resistance of the circuit.
• The total current supplied by the source.
• The current through each resistor.
• The voltage drop across each resistor.

Solution:

1. Simplify Branch 1 (R2 and R3): Since R2 and R3 are in series, their equivalent resistance is R23 = R2 + R3 = 30 ohms + 40 ohms = 70 ohms.
2. Simplify the parallel combination of Branch 1 (R23) and Branch 2 (R4): Use the parallel resistance formula: R234 = (R23 * R4) / (R23 + R4) = (70 ohms * 50 ohms) / (70 ohms + 50 ohms) = 3500 / 120 ≈ 29.17 ohms.
3. Redraw the circuit: You now have a simplified circuit with a 24V source, R1 = 20 ohms in series with R234 ≈ 29.17 ohms.
4. Calculate the total equivalent resistance (Rtotal): Rtotal = R1 + R234 = 20 ohms + 29.17 ohms ≈ 49.17 ohms.
5. Calculate the total current (Itotal) supplied by the source: Itotal = V / Rtotal = 24V / 49.17 ohms ≈ 0.488 A.
6. Calculate the voltage drop across R1 (V1): V1 = Itotal * R1 = 0.488 A * 20 ohms ≈ 9.76V.
7. Calculate the voltage drop across the parallel combination (R234): This is the same as the voltage remaining after the drop across R1: V234 = V - V1 = 24V - 9.76V ≈ 14.24V.
8. Calculate the current through Branch 1 (I23): Since R2 and R3 are in series in Branch 1, they have the same current. I23 = V234 / R23 = 14.24V / 70 ohms ≈ 0.203 A.
9. Calculate the current through Branch 2 (I4): I4 = V234 / R4 = 14.24V / 50 ohms ≈ 0.285 A.
10. Calculate the voltage drop across R2 (V2): V2 = I23 * R2 = 0.203 A * 30 ohms ≈ 6.09V.
11. Calculate the voltage drop across R3 (V3): V3 = I23 * R3 = 0.203 A * 40 ohms ≈ 8.12V.
12. Calculate the voltage drop across R4 (V4): V4 = V234 ≈ 14.24V (since it's in parallel with the combination of R2 and R3).

Summary of Results:

• Total Equivalent Resistance: ≈ 49.17 ohms
• Total Current: ≈ 0.488 A
• Current through R1: ≈ 0.488 A
• Current through R2: ≈ 0.203 A
• Current through R3: ≈ 0.203 A
• Current through R4: ≈ 0.285 A
• Voltage Drop across R1: ≈ 9.76V
• Voltage Drop across R2: ≈ 6.09V
• Voltage Drop across R3: ≈ 8.12V
• Voltage Drop across R4: ≈ 14.24V

Important Verification:

• Kirchhoff's Current Law (KCL) at the junction after R1: Itotal ≈ I23 + I4 (0.488 A ≈ 0.203 A + 0.285 A, which is true).
• Kirchhoff's Voltage Law (KVL) around the loop containing the source, R1, R2, and R3: V ≈ V1 + V2 + V3 (24V ≈ 9.76V + 6.09V + 8.12V, which is approximately true, discrepancies due to rounding).

Key Takeaways and Practical Considerations

• Series circuits: Current is constant, voltage divides, resistance adds.
• Parallel circuits: Voltage is constant, current divides, resistance decreases.
• Series-Parallel circuits: Break down the circuit systematically to simplify.
• Ohm's Law (V = I * R) is your best friend!
• Kirchhoff's Laws (KCL and KVL) are essential for verifying your results.
• Power Dissipation: Each resistor dissipates power as heat. The power dissipated by a resistor is given by P = I² * R = V² / R. Be mindful of the power rating of resistors, especially in high-current circuits.
• Tolerance: Resistors have tolerance values (e.g., 5%, 10%), meaning their actual resistance may vary slightly from the stated value. This can affect circuit performance, especially in precision applications.
• Real-World Applications: Resistors are used everywhere – from simple LED circuits to complex amplifier designs. Understanding series and parallel combinations is crucial for analyzing and troubleshooting these circuits.

By working through these problems and understanding the underlying principles, you'll be well-equipped to tackle a wide range of resistor network challenges. Remember to practice consistently and always verify your results! Good luck!