Four resistors A, B, C, and D form a Wheatstone bridge. The bridge is balanced when C = 100 Ω. If A and B are interchanged, the bridge balances for C = 121 Ω. The value of D is

Explanation

In a balanced Wheatstone bridge, the voltage across the resistors C and D is zero. This means that the ratio of the resistances of the two legs of the bridge must be equal.Let R_A and R_B(underscore value represents subscript) be the resistances of resistors A and B, respectively. When the bridge is balanced with C = 100 Ω, we have the following equation:R_A / R_B = D / 100When A and B are interchanged, the bridge balances for C = 121 Ω, so we have the following equation:R_B / R_A = D / 121We can solve these two equations simultaneously to find the values of R_A, R_B, and D.First, we can rearrange the first equation to get an expression for R\_B:R_B = R_A (D / 100)Next, we can substitute this expression into the second equation to get an equation that only involves R_A and D:R_A / R_A (D / 100) = D / 121Simplifying this equation, we get:1 / (D / 100) = D / 121Cross-multiplying and solving for D, we get:D^2 = 100 x 121D = sqrt(100 x 121)D ≈ 110 ΩTherefore, the value of resistor D is approximately 110 Ω.Note: This answer assumes that the resistances of the resistors are exact values and that the bridge is perfectly balanced. In practice, there may be small errors due to tolerances in the resistors and imperfections in the bridge, so the actual value of D may be slightly different