Consider the circuit shown in the figure below

Summary

This educational video tutorial guides viewers through solving a multi-part circuit problem, covering how to calculate equivalent resistances in series and parallel configurations. The presenter demonstrates step-by-step solutions, including using Ohm's law to determine current and voltage drops across various resistors. The tutorial covers the combination of 10 ohm, 5 ohm, 4 ohm, 3 ohm, and 2 ohm resistors, ultimately calculating the total current and voltage across specific resistors. The video provides a clear, structured approach to analyzing and solving complex circuit problems, ideal for those learning electrical engineering concepts.

Takeaways

• 😀 The problem involves calculating the equivalent resistance of resistors in various configurations (parallel and series).
• 😀 For Part A, the equivalent resistance of a 10 ohm and 5 ohm resistor in parallel is calculated using the formula 1/R_eq = 1/R1 + 1/R2.
• 😀 The equivalent resistance of the 10 ohm and 5 ohm resistors in parallel is 3.33 ohms.
• 😀 In Part B, the equivalent resistance of the 10 ohm, 5 ohm, and 4 ohm resistors is calculated. The 10 ohm and 5 ohm resistors are first combined in parallel, resulting in 3.33 ohms, which is then added to the 4 ohm resistor in series, giving a total resistance of 7.33 ohms.
• 😀 Part C involves combining the 7.33 ohm and 3 ohm resistors in parallel, resulting in an equivalent resistance of 2.13 ohms.
• 😀 In Part D, the final equivalent resistance is obtained by combining the 2.13 ohm resistance from Part C with the 2 ohm resistor in series, resulting in a total of 4.13 ohms.
• 😀 Part II asks for the total current in the circuit, which is calculated using Ohm's law (I = V/R). With a total resistance of 4.13 ohms and a battery voltage of 8V, the total current is 1.94 amps.
• 😀 In Part F, the voltage drop across the 2 ohm resistor is calculated as 3.88V, using Ohm's law with a current of 1.94 amps.
• 😀 Part G calculates the voltage across the 3 ohm resistor by subtracting the voltage drop across the 2 ohm resistor from the total battery voltage, resulting in 4.12V.
• 😀 Part H calculates the current flowing through the 3 ohm resistor, which is 1.37 amps, using the voltage across it (4.12V) and Ohm's law.

Q & A

What is the formula used to calculate the equivalent resistance of two resistors in parallel?

The formula for calculating the equivalent resistance (R_eq) of two resistors in parallel is: 1/R_eq = 1/R1 + 1/R2, where R1 and R2 are the resistances of the two resistors.

How is the equivalent resistance of 10Ω and 5Ω resistors calculated in the first part of the script?

In Part A, the equivalent resistance of the 10Ω and 5Ω resistors in parallel is calculated using the formula 1/R_eq = 1/10 + 1/5. This simplifies to 1/R_eq = 0.3, and inverting both sides gives R_eq = 10/3 = 3.33Ω.

What happens after combining the 10Ω and 5Ω resistors into a single equivalent resistor?

After combining the 10Ω and 5Ω resistors into a single 3.33Ω equivalent resistor, this combined resistance is then in series with the 4Ω resistor for further calculation in Part B.

How do you calculate the equivalent resistance when two resistors are in series?

When resistors are in series, the equivalent resistance is simply the sum of their individual resistances. The formula is R_eq = R1 + R2.

How is the combined resistance of the 3.33Ω (from Part A) and 4Ω resistors calculated?

In Part B, the 3.33Ω resistor (from Part A) and the 4Ω resistor are in series. The combined equivalent resistance is calculated as R_eq = 3.33Ω + 4Ω = 7.33Ω.

What formula is used to calculate the equivalent resistance of resistors in parallel, and how is it applied in Part C?

The formula used for resistors in parallel is 1/R_eq = 1/R1 + 1/R2. In Part C, the 7.33Ω resistor and the 3Ω resistor are combined in parallel. The calculation gives 1/R_eq = 1/7.33 + 1/3, which results in R_eq ≈ 2.13Ω.

What happens when the 2.13Ω (from Part C) and 2Ω resistors are combined?

In Part D, the 2.13Ω (from Part C) and the 2Ω resistors are combined in series. The equivalent resistance is calculated as R_eq = 2.13Ω + 2Ω = 4.13Ω.

What is the formula for calculating the total current in a circuit, and how is it applied in Part E?

The formula for calculating total current is Ohm's Law: I = V/R, where I is the current, V is the voltage, and R is the total resistance. In Part E, using a battery voltage of 8V and a total resistance of 4.13Ω, the current is calculated as I = 8V / 4.13Ω ≈ 1.94A.

How is the voltage drop across the 2Ω resistor calculated in Part F?

In Part F, the voltage drop across the 2Ω resistor is calculated using Ohm's Law: ΔV = I × R. With a current of 1.94A and a resistance of 2Ω, the voltage drop is ΔV = 1.94A × 2Ω = 3.88V.

How is the voltage across the 3Ω resistor found in Part G?

In Part G, the voltage across the 3Ω resistor is calculated by subtracting the voltage drop across the 2Ω resistor from the total battery voltage. The calculation is 8V - 3.88V = 4.12V.

How is the current through the 3Ω resistor calculated in Part H?

In Part H, the current through the 3Ω resistor is calculated using Ohm's Law: I = V/R. With a voltage of 4.12V across the 3Ω resistor, the current is I = 4.12V / 3Ω ≈ 1.37A.