Question

A Wheatstone bridge is constructed using a \(1.00-\mathrm{m}\) -long Nichrome wire (the purple line in the figure) with a conducting contact that can slide along the wire. A resistor, \(R_{1}=100 . \Omega\), is placed on one side of the bridge, and another resistor, \(R\), of unknown resistance, is placed on the other side. The contact is moved along the Nichrome wire, and it is found that the ammeter reading is zero for \(L=25.0 \mathrm{~cm} .\) Knowing that the wire has a uniform cross section throughout its length, determine the unknown resistance.

Short answer

Answer: The unknown resistance in the Wheatstone bridge is 300Ω.

Step-by-step solution

Write the balanced condition for the Wheatstone Bridge

The balanced condition of the Wheatstone bridge states that the ratio of the resistances in the two branches of the bridge is equal. That is: \(\dfrac{R_{A}}{R_{B}}=\dfrac{R_{1}}{R}\) Here, \(R_A\) and \(R_B\) are the resistances of the segments of the Nichrome wire having lengths \(L\) and \((1.00 - L)\) respectively.

Find resistance of Nichrome wire segments

Now, let's write the resistance formula for a wire and find the resistances of the two segments of the Nichrome wire in terms of their lengths: \(R_{A}=\frac{\rho L}{A}\) and \(R_{B}=\frac{\rho(1.00-L)}{A}\) Here, \(\rho\) is the resistivity of the Nichrome wire, which remains constant throughout the wire, and \(A\) is the cross-sectional area of the wire. Since the entire wire has a uniform cross section, \(A\) is the same for both segments.

Substitute expressions for wire resistances

Now, substitute the expressions for \(R_A\) and \(R_B\) found in step 2 back into the balanced condition equation found in step 1: \(\frac{\frac{\rho L}{A}}{\frac{\rho(1.00-L)}{A}} = \frac{100\Omega}{R}\)

Simplify the equation and solve for the unknown resistance

Simplify the previous equation and solve for the unknown resistance \(R\): \(\frac{L}{1.00-L} = \frac{100\Omega}{R}\) We're given that \(L = 25.0\ \text{cm} = 0.250\ \text{m}\). Substitute this value into the equation and solve for \(R\): \(\frac{0.250}{1.00 - 0.250} = \frac{100\Omega}{R}\) Rearrange the equation to isolate \(R\): \(100\Omega \cdot \frac{1.00 - 0.250}{0.250} = R\) Calculate R: \(100\Omega \cdot \frac{0.75}{0.250} = R\) \(R = 300\Omega\) The unknown resistance in the Wheatstone bridge is \(300\Omega\).