Understanding the Voltage Drops and Power Dissipation in Series Resistors

Introduction to Circuit Analysis

Circuit analysis is a fundamental aspect of electrical engineering, essential for understanding how electrical current behaves when passing through different components. This article will delve into the behavior of resistors connected in series, specifically exploring how voltage drops and power dissipation occur in such a configuration. We will use a practical example involving a 12V battery and three resistors with different resistance values to illustrate these concepts.

Given resistors of 10 ohms, 20 ohms, and 30 ohms, connected in series with a 12V battery, we will calculate the voltage drops across each resistor and the power dissipated in each component.

Total Resistance Calculation

The first step in analyzing any series circuit is to determine the total resistance. In a series circuit, resistors are added together to obtain the total resistance. Here, the resistors are 10 ohms, 20 ohms, and 30 ohms. Therefore, the total resistance is:

$$text{Total resistance} R10 R20 R30 10 20 30 60 , text{ohms}$$

Calculating the Current in the Circuit

To find the current flowing through the circuit, we can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). Hence, the current in this circuit is:

$$text{Current} I frac{E}{R} frac{12V}{60 , text{ohms}} 0.2 , text{A} 200 , text{mA}$$

Voltage Drop Across Each Resistor

Once the current is known, we can calculate the voltage drop across each resistor using the formula: Voltage Drop Current × Resistance.

For the 10-ohm resistor:

$$text{Voltage Drop} E10 IR10 0.2 , text{A} times 10 , text{ohms} 2 , text{V}$$

For the 20-ohm resistor:

$$text{Voltage Drop} E20 IR20 0.2 , text{A} times 20 , text{ohms} 4 , text{V}$$

For the 30-ohm resistor:

$$text{Voltage Drop} E30 IR30 0.2 , text{A} times 30 , text{ohms} 6 , text{V}$$

Verifying with Kirchoff's Voltage Law

Kirchoff's Voltage Law (KVL) states that the sum of the voltage drops around any closed loop in a circuit must equal the total supply voltage. Therefore, we can verify our calculations:

$$12 , text{V} 2 , text{V} 4 , text{V} 6 , text{V} 12 , text{V}$$

Power Dissipation in Each Resistor

The power dissipated in a resistor is given by the formula: Power Voltage × Current. Alternatively, it can also be calculated as Power Current^2 × Resistance.

For the 10-ohm resistor:

$$text{Power} P10 (0.2 , text{A})^2 times 10 , text{ohms} 0.04 , text{W} times 10 , text{ohms} 0.4 , text{W}$$

For the 20-ohm resistor:

$$text{Power} P20 (0.2 , text{A})^2 times 20 , text{ohms} 0.04 , text{W} times 20 , text{ohms} 0.8 , text{W}$$

For the 30-ohm resistor:

$$text{Power} P30 (0.2 , text{A})^2 times 30 , text{ohms} 0.04 , text{W} times 30 , text{ohms} 1.2 , text{W}$$

A Note on Homework and Learning

While it is natural to seek answers quickly, self-learning and problem-solving are crucial skills in any academic field. Attempting to solve problems independently can provide deeper understanding and retention of information. If you find these calculations challenging, it might help to review your course materials or seek guidance from your instructor.

Furthermore, practicing circuit analysis is essential for any aspiring electrical engineer. Tools like circuit simulation software can also aid in visualizing and understanding complex circuit behaviors.

Remember, it is always beneficial to understand the underlying principles and equations rather than just grasping the final numbers. Happy learning!