In the Wheatstone bridge circuit shown, \( R_1 = 1.5 \, k\Omega \) and \( R_2 = R_3 = R_4 = 1 \, k\Omega \). The switch is initially open and the voltage between the points C and D is \( V_{CD} \). Upon closing the switch at \( t = 0 \), the resistance in the arm AD changes by an amount \( \delta R_1 \), and the voltage between C and D changes by \( \delta V_{CD} \). The sensitivity of the bridge in volt/kilohm, defined as \( \left| \frac{\delta V_{CD}}{\delta R_1} \right| \), is _________.
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In a Wheatstone bridge, the sensitivity is the rate of change of the output voltage with respect to the change in resistance in one arm of the bridge.
Given the following resistances:
- \( R_1 = 1.5 \, k\Omega \)
- \( R_2 = R_3 = R_4 = 1 \, k\Omega \)
- The voltage \( V_{CD} \) is initially measured across the bridge when the switch is open.
- \( \delta R_1 \) is the change in the resistance \( R_1 \), and \( \delta V_{CD} \) is the change in voltage across points C and D. Step 1: Wheatstone Bridge Equation
For a Wheatstone bridge, the output voltage \( V_{CD} \) is given by the equation:
\[
V_{CD} = V_{in} \times \frac{R_2}{R_1 + R_2} - \frac{R_4}{R_3 + R_4}.
\]
Step 2: Sensitivity Definition
The sensitivity of the bridge is defined as the rate of change of the output voltage with respect to the change in the resistance of the arm AD:
\[
\text{Sensitivity} = \left| \frac{\delta V_{CD}}{\delta R_1} \right|.
\]
Step 3: Apply Small Change Approximation
To calculate the sensitivity, we differentiate \( V_{CD} \) with respect to \( R_1 \). Taking the derivative with respect to \( R_1 \) gives:
\[
\frac{\delta V_{CD}}{\delta R_1} = \frac{V_{in}}{(R_1 + R_2)^2}.
\]
Step 4: Substituting the values
Substituting the given values \( R_1 = 1.5 \, k\Omega \), \( R_2 = 1 \, k\Omega \), and \( V_{in} = 10 \, V \), we get:
\[
\frac{\delta V_{CD}}{\delta R_1} = \frac{10}{(1.5 + 1)^2} = \frac{10}{(2.5)^2} = \frac{10}{6.25} = 1.6 \, \text{V/k}\Omega.
\]
Thus, the sensitivity of the bridge is \( 1.6 \, \text{V/k}\Omega \).
