Problem 2 In a simple circuit diagram we have a power suppl

Problem # 2

In a simple circuit diagram we have a power supply that is connected in series with a resistor and a light bulb. The resistor is 1k with a tolerance of 10%(Uncertainty of 100). The k value for the light bulb is a well known constant to be 882.328.
You measure the current across the resistor with your ammeter five times with values of: 112 mA, 98 mA, 103 mA, 109 mA, 99 mA.

a) What is the Mean and estimated error of the mean for Current?

b) Kirchhoff’s law states for our simple circuit diagram: V = IR + kI^2 What is the average value of V?

c) If we express Kirchhoff’s law in terms of V = a + b, where a = IR and b = kI^2 What is the uncertainty of a (delta a), and b (delta b)?

d) What is the uncertainty of V (delta V)? Express your final answer as Mean ± Variance

Solution

a) mean = ( 112 + 98 + 103 + 109 + 99 ) / 5 = 104.2 mA

Uncertainty in a measurement ( x)

Uncertainty in a single measurement of x. You determine this uncertainty by making multiple measurements. You know from your data that x lies somewhere between x _maxand x _min .

x = R / 2 = (x_max - x _min) / 2 = (112 - 98 ) / 2 = 3.5

Uncertainty in the Mean ( x_avg)

Uncertainty in the mean value of x. The actual value of x will be somewher /e in a neighborhood around x_avg . This neighborhood of values is the uncertainty in the mean.

x_avg = x / N = 3.5 / 5 = 1.565

b) (11068034.432 + 8473976.112 + 9360720.752 + 10483041.968 + 8647795.728) / 5 = 9606713.7984