10) If the voltage, V, in an electric circuit is held constant, the current, I, is inversely proportional to the resistance, R. If the current is 350 milliamperes when the resistance is 5 ohms, find the current when the resistance is 35 ohms.
10) If the voltage, V, in an electric circuit is held constant, the current, I, is inversely proportional to the resistance, R. If the current is 350 milliamperes when the resistance is 5 ohms, find the current when the resistance is 35 ohms.
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Please assist in this intermediate algebra problem. This chapter is on polynomials and rationals
![### Problem 10: Electrical Circuit Analysis
**Question:**
If the voltage, \( V \), in an electric circuit is held constant, the current, \( I \), is inversely proportional to the resistance, \( R \). If the current is 350 milliamperes when the resistance is 5 ohms, find the current when the resistance is 35 ohms.
**Solution:**
To solve this problem, we need to use the concept of Ohm's Law and the relationship between current and resistance, given that the voltage is constant.
#### Step-by-step Solution:
1. **Understand that Current is Inversely Proportional to Resistance:**
The relationship can be expressed as:
\[
I \propto \frac{1}{R}
\]
Which means:
\[
I = \frac{k}{R}
\]
where \( k \) is a constant.
2. **Determine the Constant \( k \):**
We know from the problem statement that:
\[
I = 350 \text{ mA} \quad \text{when} \quad R = 5 \text{ ohms}
\]
So we can substitute these values into the equation to find \( k \):
\[
350 = \frac{k}{5}
\]
Solving for \( k \), we get:
\[
k = 350 \times 5 = 1750
\]
3. **Find the Current for \( R = 35 \text{ ohms} \):**
Using the relationship from step 1 and the value of \( k \) we calculated, we substitute \( R = 35 \):
\[
I = \frac{1750}{35}
\]
Simplifying, we get:
\[
I = 50 \text{ mA}
\]
So, the current when the resistance is 35 ohms is **50 milliamperes**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a79e915-3d5c-4c79-9ba5-7a5ef81967b7%2F9eb91051-d31f-4b50-980c-577e997f3fe7%2Fz8fk0gl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 10: Electrical Circuit Analysis
**Question:**
If the voltage, \( V \), in an electric circuit is held constant, the current, \( I \), is inversely proportional to the resistance, \( R \). If the current is 350 milliamperes when the resistance is 5 ohms, find the current when the resistance is 35 ohms.
**Solution:**
To solve this problem, we need to use the concept of Ohm's Law and the relationship between current and resistance, given that the voltage is constant.
#### Step-by-step Solution:
1. **Understand that Current is Inversely Proportional to Resistance:**
The relationship can be expressed as:
\[
I \propto \frac{1}{R}
\]
Which means:
\[
I = \frac{k}{R}
\]
where \( k \) is a constant.
2. **Determine the Constant \( k \):**
We know from the problem statement that:
\[
I = 350 \text{ mA} \quad \text{when} \quad R = 5 \text{ ohms}
\]
So we can substitute these values into the equation to find \( k \):
\[
350 = \frac{k}{5}
\]
Solving for \( k \), we get:
\[
k = 350 \times 5 = 1750
\]
3. **Find the Current for \( R = 35 \text{ ohms} \):**
Using the relationship from step 1 and the value of \( k \) we calculated, we substitute \( R = 35 \):
\[
I = \frac{1750}{35}
\]
Simplifying, we get:
\[
I = 50 \text{ mA}
\]
So, the current when the resistance is 35 ohms is **50 milliamperes**.
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