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Question:
Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:1a)
OOO £83622
ol rl 3223s a ©
83 cl 3°2238
a Ee
SU ERs: %8
SH Ethel
: vEgas
Sl Bie
:::2 $35:8%3
sa BRE
Te S48338 2
ttc B 737d sa fl
BEES cErsziis
ERE 2ZLERoE
SE BEE
Ent ee fogs
Lal SxnE2 SR
11 TfNESEs°38
BB BIB Serco
Bt Eg szigesd
$:i4 pSfEEis
Sl EERE
53 36M 3587552
gg fis
$3338:
8-8 2
§5Fs TE
sEzagiE
Sod salts
ME
igsatns
2833z9a
28855329
e ang TE
352538°%%
3852535
| iigscgs
3g8gigs
Epatege
385233
1223s
ll
Fs
This is AI generated Response.
Asked by: manish kumar
Created at: 2025-10-07 07:46:48
Asked by: manish kumar
Created at: 2025-10-07 07:46:48
Final Answer:
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Let's break down the problem and solve it step by step.
---
## **Step 1: Given Data**
- **Wheatstone bridge resistance (all four arms):** \( R = 120\, \Omega \)
- **Supply voltage to bridge (\( V_{in} \))**: \( 75\, V \)
- **Difference amplifier gain:** \( G = 10 \)
- **Strain gauge GF (gauge factor):** \( K = 2.16 \)
- **Amplified output voltage under load (\( V_{out,amp} \)):** \( 2.2\, mV \)
- **Specimen Young's modulus (\( E \)):** \( 200\, GPa = 2 \times 10^{11}\, Pa \)
---
## **Step 2: Find Output Voltage (Wheatstone Bridge)**
The amplifier output is \( V_{out,amp} = 2.2\, mV \).
Since the amplifier has a gain of 10:
\[
V_{out,bridge} = \frac{V_{out,amp}}{G} = \frac{2.2\, mV}{10} = .22\, mV = 2.2 \times 10^{-4}\, V
\]
---
## **Step 3: Wheatstone Bridge Output (For Small Changes in R)**
For a quarter-bridge (one arm changes):
\[
V_{out,bridge} = \left( \frac{1}{4} \frac{\Delta R}{R} \right) V_{in}
\]
---
## **Step 4: Calculate \(\Delta R\) (Change in Resistance)**
Rearrange the above equation:
\[
\frac{V_{out,bridge}}{V_{in}} = \frac{1}{4} \frac{\Delta R}{R}
\]
\[
\frac{\Delta R}{R} = 4 \cdot \frac{V_{out,bridge}}{V_{in}}
\]
Plug in values:
\[
\frac{\Delta R}{R} = 4 \cdot \frac{2.2 \times 10^{-4}}{7.5}
\]
\[
= 4 \cdot 2.9333 \times 10^{-5}
\]
\[
= 1.1733 \times 10^{-4}
\]
\[
\Delta R = R \cdot 1.1733 \times 10^{-4}
\]
\[
= 120 \cdot 1.1733 \times 10^{-4}
\]
\[
= .01408\, \Omega
\]
---
## **Step 5: Strain (\( \epsilon \))**
The strain gauge equation:
\[
\frac{\Delta R}{R} = K \cdot \epsilon
\]
\[
\epsilon = \frac{\Delta R / R}{K}
\]
\[
= \frac{1.1733 \times 10^{-4}}{2.16}
\]
\[
= 5.432 \times 10^{-5}
\]
---
## **Step 6: Stress (\( \sigma \))**
\[
\sigma = E \cdot \epsilon
\]
\[
= 2 \times 10^{11} \cdot 5.432 \times 10^{-5}
\]
\[
= 1.0864 \times 10^7\, Pa
\]
---
## **Step 7: Sensitivity (\( S \))**
\[
S = \frac{V_{out,amp}}{\sigma} = \frac{2.2 \times 10^{-3}\, V}{1.0864 \times 10^7\, Pa}
\]
\[
= 2.025 \times 10^{-10}\, V/Pa = .2025\, mV/MPa
\]
---
## **Final Answers**
- \( \Delta R = .01408\, \Omega \)
- \( \epsilon = 5.432 \times 10^{-5} \)
- \( \sigma = 1.086 \times 10^7\, Pa \)
- **Sensitivity:** \( S = .2025\, mV/MPa \)
**Correct Option (as highlighted):**
\[
\boxed{\Delta R = .01408\, \Omega,\, \epsilon = 5.432 \times 10^{-5},\, S = .2025\, mV/MPa}
\]
Let me know if you want detailed explanation for any step!
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