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?Wheatstone bridge?
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 %26amp; R2 and Ru %26amp; R.
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