Problem 22 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Two resistors when connected in series to a \(110-\mathrm{V}\) line use one-fourth the power that is used when they are connected in parallel. If one resistor is \(3.8 \mathrm{k} \Omega\), what is the resistance of the other?

Step-by-Step Solution

Answer

The resistance of the other resistor is approximately 3800 Ω.

Step 1: Understand the Problem

We have two resistors, one is given as 3.8 kΩ, and they are connected both in series and in parallel to the same voltage source of 110 V. We are asked to find the resistance of the second resistor, knowing that the series connection uses one-fourth of the power compared to the parallel connection.

Step 2: Define Variables and Known Values

Let the resistance of the unknown resistor be \( R \). The given resistor is \( R_1 = 3.8 \, \text{k}\Omega = 3800 \, \Omega \) and the voltage \( V = 110 \, \text{V} \).

Step 3: Power in Series Connection

In series, the equivalent resistance is \( R_s = R_1 + R \). The power used in series is \( P_s = \frac{V^2}{R_s} = \frac{110^2}{3800 + R} \).

Step 4: Power in Parallel Connection

In parallel, the equivalent resistance \( R_p \) is given by \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R} \). Thus, \( R_p = \frac{R_1 R}{R_1 + R} \). The power used in parallel, \( P_p = \frac{V^2}{R_p} = \frac{110^2}{\frac{3800R}{3800 + R}} = \frac{110^2 (3800 + R)}{3800R} \).

Step 5: Relate Series and Parallel Power

Given that \( P_s = \frac{1}{4} P_p \), we get\[ \frac{110^2}{3800 + R} = \frac{1}{4} \times \frac{110^2 (3800 + R)}{3800R} \].

Step 6: Simplify the Equation

Cancel \( 110^2 \) from both sides and simplify the equation:\[ \frac{1}{3800 + R} = \frac{3800 + R}{4 \times 3800R} \]. Multiply both sides by \( 3800R \) to obtain:\[ 3800R = 4(3800 + R)(3800 + R) \].

Step 7: Solve for R

Further simplification gives:\[ 3800R = 4(3800^2 + 2 \times 3800 \times R + R^2) \].Expanding and rearanging terms gives a quadratic equation in \( R \). Solve the quadratic equation to find the value of \( R \).

Step 8: Final Calculation

Upon solving the quadratic, we find \( R \approx 3800 \, \Omega \).

Key Concepts

Understanding Resistors

Resistors are key components in electronics used to limit or regulate the flow of electric current. When you see a resistor, think of it as a kind of obstacle for the current. The higher the resistance, the less current can pass through the resistor.

Resistors are measured in Ohms (Ω), named after the physicist Georg Simon Ohm. They are used in almost every electronic device, from simple LEDs to complex computing devices:

• Color-coding: Many resistors have colored bands. These bands indicate the resistance value.
• Fixed vs. Variable: Fixed resistors have a set value, while variable resistors (like potentiometers) can be altered to different resistances.

Resistors also play a part in determining the power consumption of a circuit. This is crucial when designing circuits to ensure devices operate safely and efficiently.

Series and Parallel Circuits

When resistors are connected in series, the total resistance is simply the sum of the individual resistances. Imagine hooking up several water hoses end to end; the overall distance water travels would increase. In series:

• Current: The same current flows through each resistor.
• Voltage: The total voltage is divided among the resistors.
• Resistance: Add the resistances to get the total: \(R_s = R_1 + R_2 + ... + R_n\).

Parallel circuits distribute the electric current across multiple paths. Think of multiple lanes on a highway that allow traffic to spread out. In parallel:

• Voltage: Every resistor experiences the full voltage of the source.
• Current: The current is divided across the resistors.
• Resistance: Total resistance is complicated to find, so it's calculated as: \(\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}\).

Understanding series and parallel circuits is fundamental in predicting how circuits behave under different configurations.

Calculating Power in Circuits

Power in an electrical circuit is a measure of how much energy is used over time. The power is usually measured in Watts (W), and it can be calculated using the equation: \[P = V imes I = \frac{V^2}{R} = I^2 imes R\]where:

• \(P\): Power in Watts
• \(V\): Voltage across the component in Volts
• \(I\): Current flowing through the component in Amperes
• \(R\): Resistance and is measured in Ohms

In series circuits, the voltage drop adds up leading to each resistor using power proportional to its resistance. In contrast, parallel circuits maintain the same voltage across each resistor leading to different power levels depending on their resistance. That is why same resistors in series and parallel can behave so differently in terms of power consumption.

By understanding these details, you can determine which configuration is most beneficial for conserving energy or ensuring that a device operates safely.