Problem 62 - Physics

Question

A \(60.0-\Omega\) resistor is connected in parallel with a \(120-\Omega\) resistor. This parallel group is connected in series with a \(20.0-\Omega\) resistor. The total combination is connected across a \(15.0\) - \(\mathrm{V}\) battery. Find (a) the current and (b) the power delivered to the \(120.0-\Omega\) resistor.

Step-by-Step Solution

Answer

The current through the 120Ω resistor is 0.0833 A, and the power is 0.833 W.

Step 1: Find Equivalent Resistance of Parallel Resistors

The first step is to find the equivalent resistance of the resistors in parallel. The formula for two resistors in parallel is \( R_{parallel} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} \). Plug in the values: \( R_1 = 60 \; \Omega \) and \( R_2 = 120 \; \Omega \). This gives \( R_{parallel} = \frac{1}{\frac{1}{60} + \frac{1}{120}} = 40 \; \Omega \).

Step 2: Calculate Total Resistance of the Circuit

Now that we have the equivalent resistance of the parallel resistors, add the series resistor. The total resistance \( R_{total} \) is calculated by \( R_{total} = R_{parallel} + R_{series} \). Here, \( R_{series} = 20 \; \Omega \), so \( R_{total} = 40 \; \Omega + 20 \; \Omega = 60 \; \Omega \).

Step 3: Calculate the Total Current from the Battery

Using Ohm's Law, the total current \( I_{total} \) flowing in the circuit can be calculated by \( I_{total} = \frac{V}{R_{total}} \). The voltage \( V = 15 \; V \), and \( R_{total} = 60 \; \Omega \), so \( I_{total} = \frac{15}{60} = 0.25 \; A \).

Step 4: Determine the Current through the 120Ω Resistor

For parallel resistors, the current divides inversely proportional to their resistance. Current through the \( 120 \; \Omega \) resistor \( I_{120} \) is given by \( I_{120} = I_{total} \frac{R_{parallel}}{R_2} \). Thus, \( I_{120} = 0.25 \times \frac{40}{120} = 0.0833 \; A \).

Step 5: Calculate Power Delivered to 120Ω Resistor

The power delivered to a resistor is given by \( P = I^2 \times R \). Substitute \( I_{120} = 0.0833 \; A \) and \( R = 120 \; \Omega \) into the formula, so \( P = (0.0833)^2 \times 120 \approx 0.833 \; W \).

Key Concepts

Parallel Resistance

When dealing with circuits that have parallel resistors, understanding how to calculate the equivalent resistance is crucial. In a parallel circuit, each resistor is connected across the same two points, resulting in multiple pathways for the current. This configuration allows the total current to be divided among the different paths. The formula to find the equivalent resistance of two resistors in parallel is given by:

• \( R_{parallel} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}} \)

For example, if you have a 60-ohm resistor and a 120-ohm resistor in parallel, you calculate their equivalent resistance by plugging these values into the formula. You get:

• \( R_{parallel} = \frac{1}{\frac{1}{60} + \frac{1}{120}} \)
• \( R_{parallel} = 40 \, \Omega \)

This equivalent resistance becomes a single resistor's resistance that represents the entire parallel network, simplifying the circuit analysis.

Series Resistance

Series resistance is a straightforward concept where resistors are connected end-to-end, and the same current flows through each resistor. The total resistance in a series circuit is simply the sum of the individual resistances, because there is only one path for the current to flow.

• The formula is \( R_{total} = R_1 + R_2 + R_3 + \ldots \)

In our circuit example, once we have found the equivalent resistance of the parallel components as 40-ohm, this can be seen as a single resistor. It is then added to any resistors in series, like the 20-ohm resistor:

• \( R_{total} = 40 \, \Omega + 20 \, \Omega = 60 \, \Omega \)

This total resistance can then be used to calculate other aspects of the circuit, such as the total current or voltage drop across specific points.

Ohm's Law

Ohm's Law is pivotal in understanding the relationship between voltage, current, and resistance in electric circuits. The formula is represented as \( V = I \times R \), where

• \( V \) is the voltage across the circuit (in volts)
• \( I \) is the current through the circuit (in amperes)
• \( R \) is the resistance of the circuit (in ohms)

This relationship tells us that current is directly proportional to voltage and inversely proportional to resistance. For example, if you have a total resistance of 60 ohms and a battery providing 15 volts, you can find the current using the formula:

• \( I_{total} = \frac{15}{60} = 0.25 \, A \)

This means a quarter of an ampere flows through the entire circuit. Ohm's Law is fundamental for calculating the different variables within any circuit.

Current Division Rule

In parallel circuits, the current division rule is essential for determining how the total current splits across different branches. The current through a given branch is inversely proportional to the resistance of that branch, assuming other branches that share the total current. The formula is:

• \( I_1 = I_{total} \times \frac{R_{total}}{R_1} \)

In our exercise, to find the current through the 120-ohm resistor using the total current of 0.25 A, we apply:

• \( I_{120} = 0.25 \times \frac{40}{120} = 0.0833 \, A \)

This rule helps to predict how much current will go through each resistor, helping in further calculations such as power in each branch.

Power in Resistor

Power in a resistor is determined by how much energy is consumed as electrical current flows through it. The power can be calculated using the formula:

• \( P = I^2 \times R \)

Where:

• \( P \) is the power (in watts)
• \( I \) is the current through the resistor (in amperes)
• \( R \) is the resistance (in ohms)

In our example, after finding the current through the 120-ohm resistor as 0.0833 A, the power consumed by this resistor can be found by:

• \( P = (0.0833)^2 \times 120 \approx 0.833 \, W \)

This gives the energy rate at which the resistor is converting electrical energy into heat and possibly light, revealing crucial details about the circuit's performance.