4) For a balanced Wheatstone bridge (ie. when null condition is met, Eo = 0) R2= 5k Ohm, R3= 27 k Ohm, R4 = 9 kOhm, find the value of R1. A R1 R3 Es Eo Ec in R4 R2 B
4) For a balanced Wheatstone bridge (ie. when null condition is met, Eo = 0) R2= 5k Ohm, R3= 27 k Ohm, R4 = 9 kOhm, find the value of R1. A R1 R3 Es Eo Ec in R4 R2 B
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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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![**Problem Statement:**
For a balanced Wheatstone bridge (i.e., when the null condition is met, \( E_0 = 0 \)), with the given resistances \( R_2 = 5 \text{ kOhm} \), \( R_3 = 27 \text{ kOhm} \), and \( R_4 = 9 \text{ kOhm} \), find the value of \( R_1 \).
**Diagram Explanation:**
The diagram is a schematic representation of a Wheatstone bridge circuit. The bridge is composed of four resistors:
- \( R_1 \) is connected between points \( A \) and \( E_C \).
- \( R_2 = 5 \text{ kOhm} \) is connected between points \( B \) and \( E_C \).
- \( R_3 = 27 \text{ kOhm} \) is connected between points \( A \) and \( E_D \).
- \( R_4 = 9 \text{ kOhm} \) is connected between points \( B \) and \( E_D \).
The bridge is powered by a voltage source \( E_s \) across points \( E_C \) and \( B \).
In the balanced condition, the potential difference \( E_0 \) between the junction points \( E_C \) and \( E_D \) is zero.
**Solution:**
For a Wheatstone bridge to be balanced, the following condition must be satisfied:
\[
\frac{R_1}{R_2} = \frac{R_3}{R_4}
\]
Substitute the known values:
\[
\frac{R_1}{5 \text{ kOhm}} = \frac{27 \text{ kOhm}}{9 \text{ kOhm}}
\]
Solving for \( R_1 \):
\[
R_1 = \frac{5 \times 27}{9} \text{ kOhm}
\]
Simplify to find:
\[
R_1 = 15 \text{ kOhm}
\]
Thus, the value of \( R_1 \) is \( 15 \text{ kOhm} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2ac1715-6ccd-4f91-87eb-b09a0d0b46c6%2F7beee1b9-9ee8-469c-b453-806964f66c2b%2Fn1h68h8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
For a balanced Wheatstone bridge (i.e., when the null condition is met, \( E_0 = 0 \)), with the given resistances \( R_2 = 5 \text{ kOhm} \), \( R_3 = 27 \text{ kOhm} \), and \( R_4 = 9 \text{ kOhm} \), find the value of \( R_1 \).
**Diagram Explanation:**
The diagram is a schematic representation of a Wheatstone bridge circuit. The bridge is composed of four resistors:
- \( R_1 \) is connected between points \( A \) and \( E_C \).
- \( R_2 = 5 \text{ kOhm} \) is connected between points \( B \) and \( E_C \).
- \( R_3 = 27 \text{ kOhm} \) is connected between points \( A \) and \( E_D \).
- \( R_4 = 9 \text{ kOhm} \) is connected between points \( B \) and \( E_D \).
The bridge is powered by a voltage source \( E_s \) across points \( E_C \) and \( B \).
In the balanced condition, the potential difference \( E_0 \) between the junction points \( E_C \) and \( E_D \) is zero.
**Solution:**
For a Wheatstone bridge to be balanced, the following condition must be satisfied:
\[
\frac{R_1}{R_2} = \frac{R_3}{R_4}
\]
Substitute the known values:
\[
\frac{R_1}{5 \text{ kOhm}} = \frac{27 \text{ kOhm}}{9 \text{ kOhm}}
\]
Solving for \( R_1 \):
\[
R_1 = \frac{5 \times 27}{9} \text{ kOhm}
\]
Simplify to find:
\[
R_1 = 15 \text{ kOhm}
\]
Thus, the value of \( R_1 \) is \( 15 \text{ kOhm} \).
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