Wheatstone bridge?

Question

The value of R at which balace was obtained (V=0) was measured to be 2205 ohms. R1= 1000 ohms, R2= 2200 ohms and I calculated Ru to be 1002 ohms from the equation:

Ru/R = R1/R2. The power suppy was V= 5V. If the power supply were changed to 15 volts, how would it affect the value of Ru? Why?

Answer 1

It would have been helpful to describe the locations of the resistors in the circuit, but most people asking Wheatstone bridge questions love to leave that stuff out. Anyway, if it's a typical setup, changing the power supply voltage doesn't change the balance point or the calculated value of Ru; it changes the voltages at either side of the meter but they remain equal because the power supply voltage is equally proportioned by the two voltage dividers R1 & R2 and Ru & R.