Laboratory activity about the Electric circuit Fundamental., Lab Reports of Electrical Circuit Analysis

Download Laboratory activity about the Electric circuit Fundamental. and more Lab Reports Electrical Circuit Analysis in PDF only on Docsity! LABORATORY ACTIVITY 1: Electric Circuit Fundamentals PART 1: Voltage and Current in a DC Circuit I. Objectives 1. Investigate how to use a voltmeter to measure voltage across a circuit component. 2. Investigate how to use an ammeter to measure current flow in a circuit component. 3. Investigate what happens if the rated value of a component is exceeded (only simulation will be used to investigate this step). II. Materials ❖ One dc power supply ❖ One dc 0-20 V voltmeter ❖ One dc 0-100 mA ammeter ❖ One Multimeter ❖ One 5 V, 1-W lamp III. Discussion Ohm's law, or V=IR, is the most fundamental law in electricity. The V stands for voltage, which is the potential difference between two charges. To put it another way, it's a measurement of how much work it takes to transfer a unit charge between two places. Common ground is used to describe how each device in a system uses the same zero-point reference (or ground) to ensure that the same potential difference (or voltage) is applied all across the system. Common ground is used to describe how each device in a system uses the same zero-point reference (or ground) to ensure that the same potential difference (or voltage) is applied all across the system. The next aspect of Ohm's law is current, which is measured in amperes and commonly denoted by I. Current moves through conductors with a certain amount of friction, or opposition to motion. The quantity of current in a circuit is determined by the voltage and resistance in the circuit that opposes current flow. Resistance, like voltage, is a quantity that exists between two points. As a result, voltage and resistance are frequently expressed as being “between” or “across” two points in a circuit. Electrical resistance is a unit of measurement for the amount of current repulsion in a circuit and measured in Ohms. The light bulb is the most typical use for resistance in a circuit. The light bulb provides enough resistance to a circuit to allow the filament inside to heat up and generate light. When a resistance is added to a series circuit, the overall resistance increases, causing the current to decline. And due to the decline of current, electrons that pass in the bulb do not lose much voltage than before. The amount of voltage lost in each bulb will be determined by the resistances, but one thing is certain and it is that the total of the voltages lost in each resistance will always equal the voltage of the battery. IV. Procedures 1. Construct the circuit shown in figure 1 on the Multisim software. Use a 5 V, 1 W lamp for simulation. (In the Components Toolbar, select the Indicator icon; now find Voltmeter, Ammeter, and Lamp in this window.) 2. Set the dc supply voltage to 3 V and click the Simulate Switch and verify that the battery voltage is 3 V as measured by the voltmeter. Record the voltage across the lamp terminals “V” and the current “I” flowing through it in Table 1. Calculate the power dissipation in the lamp using the relationship P = VI and note it down. 3. Change the dc supply voltage to 5 V. Run the analysis again. Record voltage and current in Table 1 and calculate the power dissipation in the lamp. 4. Change the dc supply voltage to 7 V. Run the analysis and see the effect on the intensity of light. Record voltage and current in Table 1 and calculate the power dissipation. 5. Change the dc supply voltage to 8 V. Run the analysis and observe the value of current, also observe the glow of the lamp. What happened? Explain. Computed Manually by Ohms Law Experiment 6. Build the circuit of figure 1 with manually computed components by Ohm’s law. Before switching on the dc power supply, set its voltage to 0 V. for the voltage. In addition, the value of the current dramatically decreases when the voltage source changes into 8V due to the overflow of the current to the load. The glow of the lamp in the circuit turned off because of the excess power supplied. Since our lamp is only capable of 1W, it automatically shut down when the voltage raised into 8V because it will supply a 7𝑥104−6W which is higher than the capability of the lamp. Below are the computations for power using the acquired values of voltage and current flowing in the circuit. The formula used to determine the power is voltage multiplied by current or simply P=VI. V=3 V I=0.12 A P=VI P=(3V) (0.12A) P=0.36W V=5 V I=0.2 A P=VI P=(5V) (0.2A) P=1W V=7 V I=0.28 A P=VI P=(7V) (0.28A) P=1.96W Source Voltage(V) Lamp Voltage (V) Current (A) Power, VI (W) 3 3V 0.12A 0.36W 5 5V 0.2A 1W 7 7V 0.28A 1.96W Table 1: Simulation Results using Multisim electronics workbench Table 1 shows the lamp voltage, current, and power across the circuit given with different voltages. It can be seen that if the voltage of the power source is 3V the voltage across the lamp is also 3V, the current measured in Ampere would be 0.12A and the power would be 0.36W. When the voltage from the power source was increased to 5V the current became 0.2A and the voltage across the lamp became 5V, also the power changed and became 1W. Furthermore, when the power source was increased to 7V the lamp voltage became 7V as well and the current became 0.28A and the power became 1.96W. Computed Manually by Ohm's Law Experiment Below are computations done to determine the value of current and power. The formula used to obtain the value of current is the division of voltage to the given resistance or I=V/R. While the formula that was used to get the power is the product of voltage and current or P=VI. V=3V V=5V R=25Ω R=25Ω I = I = I = I = I = 0.12A I =0.2A P=VI P=VI P=(3V) (0.12A) P=(5V) (0.2A) P= 0.36W P= 1W Source Voltage(V) Lamp Voltage(V) Current(A) Power , VI(W) 3 3V 0.12A 0.36W 5 5V 0.2A 1W Table 2: Computed Manually by Ohm’s Law Table 2 shows the values of the computed manually by Ohm’s Law. The formula of power which is voltage multiplying the current is used to get the power in the lamp. It can be seen that the voltage of the power source is 3V the voltage across the lamp is also 3V, to get the current; the value of the voltage will be divided to the value of resistance and the measured current is 0.12A and the power would be 0.36W. Furthermore, when the power source was increased to 5V the lamp voltage became 5V as well and there’s a current of 0.2A and it produced a power of 1W. VI. Question and Analysis Question 1: Why is the lamp damaged when the voltage across it goes to 7V? Explain by comparing the power dissipation with the rated value. The lamp was damaged when the voltage across goes to 7V because the power dissipated by the 7V-source is greater than the rated value which is 1W. As the voltage across the circuit goes beyond 7V, the light bulb breaks. Question 2: Compare the Multisim results with the hardwired laboratory results. Comment on the results. The results on the Multisim and manually computed using Ohm’s Law are not different from each other. The results are exactly the same from the both setups. Analysis 1 After the simulations we have done in the Multisim software, we have determined that voltmeter is used when you have to determine the voltage across the circuit and it will always vary depending on the given values. Analysis 2 We have determined that ammeter is used to give the amount of current flowing in the circuit. Using it is simple and just like the other testing tool, the result it will give will vary depending on the given data. 3. Vary the dc supply voltage Vs in steps of 2 V and record current in each case. Enter your result in Table 4. 4. Plot “I” vs. “V” in the graph of Table 5. 5. Calculate the value of resistor based on the slope of the V-I characteristic curve plotted in step 6. Repeat steps 1 to 5 with the manually computed components by Ohm’s law. Note: Use different color pens to plot the experimental results in Table 5. V. Data and Results, Computation, Drawings and Tables Figure 2 displays the circuit diagram with resistor, multimeter, and wire for a resistance measurement. The circuit has a given 100 Ω resistor that is being checked by the use of a multimeter to measure resistance. Figure 2. Resistance Measurement Figure 3 shows the circuit diagram with the voltage source of 10V. Moreover, the value of the voltage is calculated by connecting a multimeter in parallel with the resistor. Figure 3.1 shows the circuit diagram with the voltage source of 10V. The current recorded 100mA is flowing across the circuit with the resistance of 100Ω and the voltage of 10V. The value of the voltage “V1” is calculated across resistor R1 by connecting a multimeter in parallel to it. In addition, the value of current “I” flowing through R1 will be calculated by connecting another multimeter in series to R1. Figure 3. Verifying Ohm’s law Figure 3.1 Verifying Ohm’s law Computed Manually by Ohm’s Law Below are computations to determine the resistance value and percent error using the given data. V= 10V R= 𝑉 𝐼 I= 100mA R= 0.1A R= 100Ω % Error Computation % Error = × 100% % Error = × 100% % Error = × 100% % Error = 0Ω× 100% % Error = 0% Ohmmeter Reading Multisim R1 =100Ω Manually computed by Ohm’s Law R1 =100Ω % Error 0% Table 3. Resistance Measurement The parameter that is given in Figure 4.4 in a circuit is being measured by the multimeter. With the resistance of 100 Ω and charge of 8V, after these were simulated it gave the current amount of 80 mA flowing the given circuit. Figure 4.5 V-I Measurements In the last given circuit in Figure 4.5, the simulation to test the amount of current flowing was done. With the given charge of 10V and 100 Ω of resistance, after placing the multimeter in the panel, it resulted to totality of 100mA current flowing in the circuit. Figure 4.4 V - I Measurements Multisim Manually computed by Ohm’s law V(volt) 1(mA) V(volt) 1(mA) 0V 0mA 0V 0mA 2V 20mA 2V 20mA 4V 40mA 4V 40mA 6V 60mA 6V 60mA 8V 80mA 8V 80mA 10V 100mA 10V 100mA Table 4. V-I Measurements Table 4 indicates the data and results gathered in the Multisim application and computed manually with the help of Ohm’s law. With the given values of voltage and resistance used to determine the current flowing through the circuit we can notice that the result derived from two different ways are the same. Below are the computations done manually to determine the current flow with the help of Ohm’s Law(V=I/R). To find the current flow, we will use the formula I=V/R, where I is the value of current to determine, V is for the voltage and R is for the resistance which is 100 ohms. Note that the unit used in current is ampere and 1 A = 1000 mA. V= 0V R= 100Ω V= 2V R= 100Ω V= 4V R= 100Ω I= I= I= I= I= I= I=0mA I=0.02A or 20mA I=0.04A or 40mA V= 6V R= 100Ω V= 8V R= 100Ω V= 10V R= 100Ω I= I= I= I= I= I= I=0.06A or 60mA I=0.08A or 80mA I=0.1A or 100mA