College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Ohm's law
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Path of Least Resistance
In a series of alternate pathways, the direction of least resistance is the actual or metaphorical route that offers the least resistance to forwarding motion by a given individual or body.
Question
![**Question:**
What is the resistance of a \(2.00 \times 10^2 \, \Omega\), and a \(4.00 \times 10^3 \, \Omega\) resistor connected in parallel?
---
**Explanation:**
To find the total resistance \( R_t \) of resistors connected in parallel, use the formula:
\[
\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}
\]
In this scenario, \( R_1 = 2.00 \times 10^2 \, \Omega \) and \( R_2 = 4.00 \times 10^3 \, \Omega \).
**Calculation:**
1. **Calculate the reciprocal of each resistor:**
\[
\frac{1}{R_1} = \frac{1}{200} = 0.005 \, \text{S}
\]
\[
\frac{1}{R_2} = \frac{1}{4000} = 0.00025 \, \text{S}
\]
2. **Add the reciprocals:**
\[
\frac{1}{R_t} = 0.005 + 0.00025 = 0.00525 \, \text{S}
\]
3. **Find the reciprocal to get \( R_t \):**
\[
R_t = \frac{1}{0.00525} \approx 190.48 \, \Omega
\]
**Conclusion:**
The equivalent resistance of a \(2.00 \times 10^2 \, \Omega\) and a \(4.00 \times 10^3 \, \Omega\) resistor connected in parallel is approximately \(190.48 \, \Omega\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F470a9dcb-02ee-4386-831b-818341ed1d34%2F47c30b83-2c28-4796-9894-9dda4d378673%2Fj6wixcn_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
What is the resistance of a \(2.00 \times 10^2 \, \Omega\), and a \(4.00 \times 10^3 \, \Omega\) resistor connected in parallel?
---
**Explanation:**
To find the total resistance \( R_t \) of resistors connected in parallel, use the formula:
\[
\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}
\]
In this scenario, \( R_1 = 2.00 \times 10^2 \, \Omega \) and \( R_2 = 4.00 \times 10^3 \, \Omega \).
**Calculation:**
1. **Calculate the reciprocal of each resistor:**
\[
\frac{1}{R_1} = \frac{1}{200} = 0.005 \, \text{S}
\]
\[
\frac{1}{R_2} = \frac{1}{4000} = 0.00025 \, \text{S}
\]
2. **Add the reciprocals:**
\[
\frac{1}{R_t} = 0.005 + 0.00025 = 0.00525 \, \text{S}
\]
3. **Find the reciprocal to get \( R_t \):**
\[
R_t = \frac{1}{0.00525} \approx 190.48 \, \Omega
\]
**Conclusion:**
The equivalent resistance of a \(2.00 \times 10^2 \, \Omega\) and a \(4.00 \times 10^3 \, \Omega\) resistor connected in parallel is approximately \(190.48 \, \Omega\).
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