Recall that R is the total resistance of the circuit. Hence a proper treatment o
ID: 1599228 • Letter: R
Question
Recall that R is the total resistance of the circuit. Hence a proper treatment of this RL circuit Would include the resistance of the coil (and other wires) as well as as the resistance of the resistor. If we did not consider the coil resistance, about how much error do you think this omission would cause?
4. Now click on the XY cursor to togste it off Reseale the plot the window. Find your tool 63.2% of this maximum voltage from the data box on plot. Calculate the vol maximum voltage. the 63.2% (0.632)v CooC volts max 5. Use the Magnifier tool to expand the section of s near this y 63.2% voltage. Y ou migh have to rescale as well to fit this selected section in the full window. click on the XY cursor tool (3 and place the crosshairs on the plot where your y-value corresponds to your V voltage. Read the x-value to get the time 63.2% at this point. time at V voltage tz 0.0214 seconds. 63.2% 6. Take the difference of these two times to obtain the time constant, t. Record the value here. At- t2-ti "T" 3- xlo seconds 7. It should be true that T L R. From this, calculate the inductance OL of the inductor L TR mH 8. What is the marked value for the Inductance? 3.2 Lmarked 9. What is the difference between the marked value of the inductor (the "accepted'' value) and the measured value of the indcutor (from experiment? differenceExplanation / Answer
Here since the experimentally observed value of inductance is 9.9 mH while the marked value is 8.2 mH this implies that total resistance of the circuit is greater than that expected by taking marked value
Since,
L/R =t (t is time constant)
Thus using marked value the resistance of the circuit is
R= 8.2/0.3
R=27.33 ohm
While using experimental value it is
R = 9.9/0.3
R =33 ohm
Thus it is clear that this extra resistance is resistance of coil
i.e. 5.67 ohm is coil resistance. So there will be 20.74 increase in the time constant of the circuit.
Error= ((9.9/27.33)-0.3/0.3)*100
Error= 20.746
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