The output voltage is given by the voltage divider equation: v_2 = v_1 r_2/r_1 + r_2. suppose the voltage source produces a voltage of v_1 = 5v with an accuracy of plusminus 10mv, the resistor r_1 has a nominal value of 1k ohm with a tolerance of plusminus 5%, and we measure the voltage v_2 with our voltmeter to be 3.5v and we know the error in our voltmeter reading to be plusminus 20mv. if we use the above voltage divider equation to calculate the value of the resistor, r_2, include the following derivations in your prelab: (1) what is the calculated value of r_2? (2) what kind of error should we expect in our calculated value of r_2? that is, what is the maximum and minimum amount we could expect to be off by?

R_2 = 2333.333 ohms

Δ R_2 = 148 ohms , R_2 = (2185.333 , 2481.333) ohms

Explanation:

Given:

                                 V_2 = V_1 * (R_2 / (R_1 + R_2))

V_1 = 5 +/- 0.01 V

V_2 = 3.5 +/- 0.02 V

R_1 = 1000 +/- 50 ohm

Find:

(1) What is the calculated value of R_2?

(2) What kind of error should we expect in our calculated value of R_2? That is, what is the maximum and minimum amount we could expect to be off by?

Solution:

- Re-arrange the given expression:

                             (V_2 / V_1) = R_2 / (R_1 + R_2)

                            R_1 * V_2 + R_2*V_2 = R_2*V_1

                               R_2 = R_1*V_2 / (V_1 - V_2)

- Now plug in the mean values:

                             R_2 = 1000*3.5 / (5 - 3.5)

                             R_2 = 2333.333 ohms

- We will use weighted uncertainty method for tolerance of R_2

                             Δ R_2 / R_2 = Δ R_1 / R_1 + 2* (Δ V_2 / V_2) + Δ V_1 / V_1

                             Δ R_2 / 2333.333 = 50/1000 + 2*(0.02/3.5) + 0.01/5

                             Δ R_2 / 2333.333 = 111 / 1750

                             Δ R_2 = 148 ohms

- The value of R_2 will be off by 148 ohms, the minimum and maximum values are:

                            R_2 = (2185.333 , 2481.333) ohms

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