Inductance and Capacitance - Lab Report Example

Inductance and capacitance Name Date Inductance and capacitance This task has three experiments that include task 1, task 2, and task 3 that explores the operation of resistors, inductors, and capacitors. Task 1: Resistance and Resistivity In any conductor there are certain factors that influence its resistance, which is its cross-sectional area, length, temperature and resistivity. Cross-sectional area can be equated to cross sectional area of a hallway. If the hallway is wide high current will be allowed to pass and vice versa. A short conductor allows the current to pass fast, as compared to a hallway a short hallway will allow people to move past at a high rate. Resistivity of a material depends on material electronic structure and temperature. As the temperature of a material increases its temperature also increases. Material with less resistance values allows the charge to pass fast. The resistance of a conductor is given by the equation Where R = resistance (Ω) l = length (m) ρ = resistivity (Ω m) A = area (m2) In this experiment, pencil lead of a set of HB1 pencils was used. The resistance of the lead was measured and classified whether it was a conductor, a semiconductor, or an insulator. The objective of this experiment was to investigate the equation that relates to resistance of a material with its length, cross-sectional area and resistivity. Equipments used. HB grade pencils and pencil sharpener. Multimeter that can measure inductance, resistance, and capacitance Crocodile clips connectors Ruler, vernier and micrometer Part A: Resistivity A group of HP pencils was obtained and both ends were sharpened so as to expose the lead. The resistance meter was used to measure the resistance of the each pencil lead. For the effects of random variation in graphite/clay to be reduced, the average resistance and average length of the pencils was calculated. Part B: Variation of Resistance with Length. The resistances of a group of pencils with varying lengths were measured by use of resistance meter. The values obtained were recorded on a table. Results We used a Vernier caliper to measure the diameter of the pencil leads. The diameter of the pencil lead was 2.23mm. The table below shows the values of length of the pencil lead verses their resistances. Table for a group of pencils and their resistances Length (m) Resistance (Ω) 0.084 13.4 0.131 18.6 0.145 20.4 0.155 24.7 0.176 28.4 A graph of resistance verses length is shown below. A = πr2 = π x 0.0011052 From the graph, the gradient = 161.5 Thus, 161.5 The resistivity of the pencil, Ωm2 This shows that the pencil is a conductor because it has low resistivity Task 2: Inductance of a Solenoid A solenoid is a straight coil of wire in which when voltage is applied through, it generates a uniform magnetic field similar to that of a bar magnetic. The objective of this experiment was to investigate the equation which describes the inductance of a long solenoid. A solenoid This was done by wrapping the wire around the nail and counting the number of turns. Then the inductance was measured and recorded on a table. The inductance is obtained from the following equation. Where N= number of turns A= average cross-sectional area (m2) l = length in meters µo= permeability of free space µr= relative permeability Method use i. The ruler was used to measure the length of the coil and average diameter ii. LCR meter was connected and the inductance was measured. The LCR meter is shown below. LCR meter iii. The inductance equation was rearranged to make µr the subject of the formula and the value of relative permeability was then calculated. The results were then compared. iv. The length of the inductor was then varied by stretching or compressing the wire coil. v. The results showing the variation of inductance with length was recorded in a table and the results were plotted in a graph. The results were compared with the expected. Results The cross sectional area of the solenoid, The table for length and inductance of the solenoid Length (m) Inductance (mH) 1/length (m-1) 0.024 0.000112 41.67 0.03 0.000096 33.33 0.045 0.000083 22.22 0.07 0.000056 14.29 0.088 0.000044 11.36 The graph of length verse inductance is shown below. From the graph The gradient = 2 x 10-6 The gradient is equivalent to µoµrN2A. Thus, The calculated is and the experimental is 0.05067. The difference is due to errors encountered in the experiment. This can be due error in measurement or temperature change. Task 3: Capacitance of a parallel plate capacitor This is a simple type of capacitor which is made up of two conducting plates of area (A) separated by a distance (d). The material between the plates can be air, vacuum, or insulator, an insulator being put between the spaces increases the capacitance of the capacitor. Capacitance is the property of the capacitor to store charges between the parallel plates in form of electrostatic field. When an electric current pass through a capacitor, it becomes charged and the electrostatic field stores energy. The objective of this experiment was to investigate the equation which describes the capacitance of a parallel plate capacitor. This was done by gluing two sheets of aluminum foil to either sides of paper as shown below. The capacitance is calculated from the following equation. Whereby = C is the capacitance (F) A = the area of plates (m2) Ɛo= permeability of free space () µr = relative permittivity d = distance between plates (m) Apparatus used i. Aluminum foil for conducting plates ii. Dielectric material (a sheet of A4 paper was used) iii. A glue stick iv. LCR meter or multimeter that can measure capacitance v. Ruler vi. Pair of scissors Method used i. A parallel plate capacitor was constructed with the width and height of aluminum plates. ii. The capacitance meter was used to measure its capacitance and the results recorded. iii. The micrometer was used to measure the thickness of the 5 layer sandwich, and the paper and foil separately. Then the length between capacitance plates was then estimated. iv. Capacitance equation was rearranged and relative permittivity of the paper used as dielectric was calculated. v. The results were compared with those expected from dielectric material. vi. A selection of representative capacitance dimensions was obtained and capacitance values were measured. vii. The obtained capacitance values together with those from classmates were used to plot the graph of capacitance versus area. Results The thickness of: The capacitor = 0.11 mm Paper = 0.09mm Foil = 0.0070 mm The distance between the plates = 0.09mm The capacitance = 3.84uf A= 0.186 x 0.168 = 0.0312m2, Ɛo = 8.85 x 10-12 The relative permittivity of the paper = 1.2516. According to Bird (2014), the relative permittivity of the paper is 2.3. The relative permittivity of the paper obtained from this experiment is less due to the variation in temperature and errors introduce during the measurement. The errors were minimized by taking accurate measurements and ensuring that the measurement were at room temperature. References Bird J. (2014). Electrical Circuit Theory and Technology, Routledge Kubala, T. S. (2009). Electricity 2: Devices, circuits, and materials. Australia: Delmar Cengage Learning. Read More