15.0 V 500 N 900 Ω www 200 Ω 1500 N www ) (a) Calculate the equivalent resistance of the circuit. 5) (b) If 50.0 mA of current moves through the 500 $2 resistor what is the potential difference across the resistor?
15.0 V 500 N 900 Ω www 200 Ω 1500 N www ) (a) Calculate the equivalent resistance of the circuit. 5) (b) If 50.0 mA of current moves through the 500 $2 resistor what is the potential difference across the resistor?
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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Question
Use the circuit equation to solve the practice problem.
![### Circuit Analysis: Parallel Resistors
#### Diagram Explanation
The image shows a circuit with a 15.0 V battery connected in series to a 500 Ω resistor. Following the 500 Ω resistor, three resistors in parallel are connected to the circuit: 900 Ω, 200 Ω, and 1500 Ω.
#### Problem Set
**(a)** Calculate the equivalent resistance of the circuit.
**(b)** If 50.0 mA of current moves through the 500 Ω resistor, what is the potential difference across the resistor?
#### Solution Steps:
1. **Calculating Equivalent Resistance (a):**
- Start with the resistors in parallel: 900 Ω, 200 Ω, and 1500 Ω.
- Use the formula for parallel resistors:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{900} + \frac{1}{200} + \frac{1}{1500}
\]
- Calculate the reciprocal sum:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{900} + \frac{1}{200} + \frac{1}{1500} = \frac{1}{R_{\text{total}}}
\]
- Calculate the equivalent parallel resistance \( R_{\text{parallel}} \).
2. **Combine with the Series Resistor:**
- The equivalent resistance \( R_{\text{eq}} \) of the entire circuit is:
\[
R_{\text{eq}} = R_{\text{series}} + R_{\text{parallel}}
\]
3. **Potential Difference Calculation (b):**
- Use Ohm’s Law for the 500 Ω resistor given a current of 50.0 mA:
\[
V = IR
\]
- Calculate \( V \) using the given current and resistance:
\[
V = 0.050 \, \text{A} \times 500 \, \Omega
\]
This information facilitates educational engagement by providing a detailed breakdown of how to calculate equivalent resistance in a series-parallel circuit and applying Ohm's Law to find potential differences.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faaba32b4-cc3a-4364-adc8-ff237e8eebae%2Fff1a33e3-e703-4307-a5b6-61d4bbd74fda%2F88amaxp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Circuit Analysis: Parallel Resistors
#### Diagram Explanation
The image shows a circuit with a 15.0 V battery connected in series to a 500 Ω resistor. Following the 500 Ω resistor, three resistors in parallel are connected to the circuit: 900 Ω, 200 Ω, and 1500 Ω.
#### Problem Set
**(a)** Calculate the equivalent resistance of the circuit.
**(b)** If 50.0 mA of current moves through the 500 Ω resistor, what is the potential difference across the resistor?
#### Solution Steps:
1. **Calculating Equivalent Resistance (a):**
- Start with the resistors in parallel: 900 Ω, 200 Ω, and 1500 Ω.
- Use the formula for parallel resistors:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{900} + \frac{1}{200} + \frac{1}{1500}
\]
- Calculate the reciprocal sum:
\[
\frac{1}{R_{\text{parallel}}} = \frac{1}{900} + \frac{1}{200} + \frac{1}{1500} = \frac{1}{R_{\text{total}}}
\]
- Calculate the equivalent parallel resistance \( R_{\text{parallel}} \).
2. **Combine with the Series Resistor:**
- The equivalent resistance \( R_{\text{eq}} \) of the entire circuit is:
\[
R_{\text{eq}} = R_{\text{series}} + R_{\text{parallel}}
\]
3. **Potential Difference Calculation (b):**
- Use Ohm’s Law for the 500 Ω resistor given a current of 50.0 mA:
\[
V = IR
\]
- Calculate \( V \) using the given current and resistance:
\[
V = 0.050 \, \text{A} \times 500 \, \Omega
\]
This information facilitates educational engagement by providing a detailed breakdown of how to calculate equivalent resistance in a series-parallel circuit and applying Ohm's Law to find potential differences.
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