Physics Lab 2b

Physics Lab 2B Methodology 1. Explain the methodology you used to determine the value of the unknown resistor. Also describe an alternative approach you could have used instead. (Assuming you could not remove the unknown resistor from the circuit.) Method 1: In method 1, by measuring the current and voltage values across the unknown resistor, we can calculate the resistance value using the formula R=V/I. We just measured the voltage using the multimeter by attaching the measuring sticks to each end of each resistor (voltmeter in parallel). For the current, we measured this by attaching the measuring sticks to one component’s leg and to the neghbouring component, ie. attached the ammeter in series. Method 2: An alternate method to determine the value of the unknown resistor would be to find the equivalent resistance of the circuit and then solve for the unknown resistor. The equivalent resistance of the circuit can be measured using a multimeter once the power supply is disconnected. The other values of the resistors are known, having already been measured previously. The unknown resistor can then be solved using the resistor values and appropriate equations of equivalent resistance of circuits in parallel and in series. Data & Analysis ● A table of all measured & calculated data used in your analysis, with uncertainties. Show sample calculations for your calculated values and calculated uncertainties. (Remember significant digits!) ● The equivalent resistance of both circuits. (Hint: Recall Lab 2A protocol step 5.) ● The value of the unknown resistor. (Depending on your methodology, it may be very difficult to find the uncertainty of the unknown resistor. You will not be graded on this and may omit an uncertainty for your unknown resistor value.) Interpretation of Results 2. Do the resistances you calculated in Protocol steps 1- 4 agree with the theoretical (pre- lab) resistances, within uncertainty? Pre-lab resistances: R1 = 51 Ω R2 = 100 Ω R3 =10 Ω R4 = 10 Ω The theoretical value is in agreement within uncertainty if the value falls within the acceptable range of the experimental value (the calculated uncertainty). Measured resistances: R1 = 50 Ω R2 = 100 Ω R3 = 10 Ω R4 =10 Ω with an uncertainty of 1 for each resistance.

The resistances measured individually with the multimeter all lie within the theoretical resistances as R2, R3 and R4 have exactly the same values and R1 lies within the theoretical resistance with its uncertainty of 1 as it is 1 less than the theoretical value. Calculated resistances: R1 = 50 ± 3 R2 = 100 ± 20 R3 = 10 ± 2 R4 = 10 ± 1 The resistances found through measurements of voltage and current with the multimeter then calculated through ohm’s law also agree with the theoretical resistances as the values all lie within the acceptable ranges with their calculated uncertainties. 3. Did the voltage and current across R1 increase or decrease when it was exchanged for the unknown resistor? Why? Voltage R1 = 0.74 V Current R1 = 0.015 A Voltage U = 0.83 V Current U = 0.0045 A The voltage increased for the Unknown in comparison to R1, this is expected as R4 is fixed so the loop will get the same amount of voltage as a whole in both scenarios which allows us to look at the breakdown of individual resistances and their effects only within that loop. If we increase the value of resistor 1, then Req for the loop will require more voltage, which ultimately means that the voltage will increase for the unknown if the value of the resistor increases. For the current in the unknown resistor, the current decreased, which is expected as seen by Ohm’s law V=IR in which an increased resistance will produce a decreased current as resistance and current are inversely proportional to each other. Alternatively, the current is the flow of electrons thus as the resistance in the material increases, it increases the interactions between the electrons and the material, thus slowing the electrons down. 4. Did the equivalent resistance of the circuit increase or decrease when R1 was exchanged for the unknown resistor? Why? Equivalent resistance with R1: 45 Ω Equivalent resistance with unknown resistor: 80 Ω The equivalent resistance of the circuit increased with the unknown resistance. R1/ the unknown resistor are in parallel with R2 and R3 therefore to find the equivalent resistance the following equation must be used 1/Req = 1/(R2+R3) + 1/R1 then plus R4 because the

Circuit 1 values Measured resistance (Ω) Uncertainty of resistance Measured voltage (V) Uncertainty of voltage Measured current (A) Uncertainty of current Calculated resistance (Ω) Uncertainty of resistance R1 50 1 0.74 0.01 0.015 0.001 R=V/I = 0.739/0.0149 =50 𝛅 R = R ( δ𝑉 𝑉 ) 2 + ( δ𝐴 𝐴 ) 2 =(49.5973154)(SQRT((0.01/0.7 39)^2 + (0.001/0.0149)^2)) = 3 R2 100 1 0.67 0.01 0.007 0.001 100 20 R3 10 1 0.066 0.001 0.007 0.001 10 2 R4 10 1 0.2 0.01 0.021 0.001 10 1 Unknown resistance Measured resistance (Ω) Uncertainty of resistance Measured voltage (V) Uncertainty of voltage Measured current (A) Uncertainty of current Calculated resistance (Ω) Uncertainty of resistance R unkown 197 1 0.83 0.01 0.004 0.001 180 40 Equivalent resistance: Circuit 1 Measured resistance (Ω) Uncertainty of resistance Calculated resistance (Ω) Equivalent resistance 45 1 44.4 Equivalent resistance Circuit 2 Measured resistance (Ω) Uncertainty of resistance Calculated resistance (Ω) Equivalent resistance 80 1 80.3