it desirable for an ammeter to have a high resistance or low resistance? Why? ➢ Yes‚ it is desirable for an ammeter to have a high resistance or low resistance because the lower the resistance‚ the better the ammeter as it is connected in series and so its resistance will affect the current flow and thereby affect any calculations to be obtained. 2. An ammeter of resistance Ra is connected through a rheostat to a cell of negligible internal resistance. When the rheostat is adjusted
Premium Electric current Electrical resistance Battery
MAGNETISM‚ SOUND‚ AND LIGHT SERIES CIRCUIT In this unit you’ll begin learning how to analyze circuits. Circuit analysis means looking at a schematic diagram for a circuit and computing the voltage‚ current‚ or power for any component in that circuit. Closely related to the task of circuit analysis is the task of troubleshooting‚ which means figuring out what is wrong in a circuit that is not working correctly. Analyzing and troubleshooting go hand in hand; when a circuit is not working correctly‚
Free Resistor Electrical resistance Series and parallel circuits
currents entering a node is equal to zero. 2. What is the equation for the total current in a parallel circuit? It= I1+I2+I3…. 3. Determine the total current in the following nodes. 1) It= 2mA+4mA+6mA= 0.012x10^-3=12mA 2) I + I1+I3= I2 I= 2mA+4mA= 1mA I= 1mA- 6mA I= -5mA 4. Determine the total current in the following circuit. It= I1+I2+I3= 10mA+20mA+40mA= 0.07 x 10^-3= 70mA 5. If VS = 10 V in the circuit above‚ what are the following voltages? R1= V/I= 10V/10mA= 1000 x 10^3= 1kΩ
Free Resistor Series and parallel circuits Electric current
Discussion Through this experiment we have learnt about the basic concept of simulation and circuit design using TINA Sofware in the computer lab. TINA is a software that used to designing‚ simulating and analyzing analog‚ digital and mixed electronic circuits. Analysis result can be displayed as sophisticated diagrams or an range of virtual instruments. Any circuit can be design easily by using schematic editor. The components can be chosen from the large library. The parameters of each components
Premium Resistor Electrical impedance Voltage divider
Week 5 iLab #1 Parts: Breadboard DC Power Supply Hand Held DMM Test leads Wire 100 Ω resistor‚ 120 Ω resistor‚ 220 Ω resistor‚ 330 Ω resistor 1. Identify each of these circuits. 2. Given the circuit below‚ calculate the values listed below. IT: _64mA__ IA: __18.18mA____ IB: __45.45mA__ VR1: _3.99V___ VR2: __5.99V__ VR3: _5.45V___ VR4: __4.54V___ Ra= R1+R2= 220Ω+330Ω= 550Ω Rb=R3+R4= 120Ω+100Ω= 220Ω Ia= Vs/Ra= 10V/550Ω= 0.01818
Premium Resistor Measurement Electric current
ECET110 Professor: Laboratory Number: 1 Laboratory Title: Analysis of a Series Circuit using Simulation and Actual Construction Submittal Date: 3/8/2014 Objectives: 1. To construct a series circuit and measure its equivalent resistance. 2. To predict and verify electrical characteristics of a series circuit using Ohm’s Law and Kirchhoff’s Voltage Law. 3. Determine the voltages across each resistor in the series circuit using the voltage divider rule. Measure and verify the same using simulation
Premium Ohm's law Electric current Resistor
resistance of a circuit with three parallel resistors with the values below? Gt= G1+G2+G3= 0.5mS+0.25mS+0.75mS= 0.0015 x 10^-3= 1.5mS Rt= 1/Gt= 1/1.5mS= 667Ω 5. Use the reciprocal method to calculate the total resistance of the circuit below. Rt= 1/(1/R1+1/R2)= 1/(1/2.2kΩ + 1/3.3kΩ)= 1320 x 10^3= 1.320kΩ 6. Calculate the total resistance of the circuit above using the product-over-sum rule. Rt= R1(R2)/R1+R2= 2.2kΩ(3.3kΩ)/2.2kΩ+3.3kΩ= 1320 x 10^3= 1.320kΩ 7. Construct the circuit above on
Premium Measurement Resistor Series and parallel circuits
ECT 122 Week 3 iLab #2 Parts: Breadboard DC power supply Handheld DMM Test leads Wire 2 - 1.0kΩ resistor 2.2kΩ resistor 3.3kΩ resistor 1. Using the circuit below‚ calculate the following values. On the circuit drawing‚ indicate the polarities for each voltage and show the current direction. 0.00133 x = 1.33mA R1 + R2 + R3 + R4 = 1kΩ + 2.2kΩ + 3.3kΩ + 1kΩ = 7500 x 10^3 = 7.5kΩ 1.33mA x 1kΩ = 1.33V 1.33mA x 2.2kΩ = 2.93V 1.33mA x 3.3kΩ = 4.39V 1.33mA x 1kΩ
Premium Ohm's law Resistor Power
EC1 VI-1 The results of experiment EC1 part V1 are summarized as below. V‚ I‚ Rm are directly measured Resistor A Resistor B Resistor C Resistor D V±ΔV(V) 6.748±0.001 7.759±0.001 9.061±0.001 10.383±0.001 I±ΔI(mA) 44±0.5 35±0.5 23±0.5 10±0.5 V/I±ΔR(Ω) 153.36±1.7 221.69±3.2 393.96±8.61 1038.30±52.02 Rm±ΔRm(Ω) 145.21±0.01 214.90±0.01 385.60±0.01 907.50±0.01 VI-2 (V‚ I) pairs for Resistor A using a DC power supply V-2 f = 0 Hz # V(Volts) I(mA) 1
Free Resistor Electrical resistance Series and parallel circuits
DC Circuits: Electric Power PHY 114 Lab Report 10/09/2013 Abstract: The purpose of this experiment was to look at electric power dissipated in a load resistor in a circuit with a real power source‚ which has a finite internal resistance. We discover a surprising result about how to deliver the most power to a device from a source. Consider a simple circuit in which a battery of voltage drives a current through a resistor of resistance. As we have seen‚ the battery is continuously doing
Free Ohm's law Electric current Resistor