Understanding Electricity: Current, Voltage, Power and Parallel Circuits in Physics Lab, Lab Reports of Physics

Download Understanding Electricity: Current, Voltage, Power and Parallel Circuits in Physics Lab and more Lab Reports Physics in PDF only on Docsity! PHYSICS 162 Name: . Winter 2006 Lab/Lecture Activity #3 EXPERIMENT #1 What's in a Watt? Objectives: 1. To better understand the relationship between current, voltage and power. 2. To analyze series and parallel connections of electric circuits and determine why your home is connected as a parallel circuit. PART A : RESISTANCE, CURRENT, VOLTAGE AND POWER Ohm's law defines the amount of current that will flow through device and how it depends on the resistance. : RESISTANCE = VOLTAGE / CURRENT or R = V/I. CURRENT = VOLTAGE / RESISTANCE or I = V/R The operating power of a device is given by POWER = CURRENT * VOLTAGE or P = I*V. Here we will examine the current, voltage, resistance and power for several devices operating at their normal voltage, 120 V. An ammeter ( a device that measures current flow) is connected to an outlet box. When a lamp is plugged into the outlet box the ammeter will register the current flowing to that light bulb. Plug the lamp with the specified bulb into the outlet box. Turn the power on and record the current in amperes (A) in the table below. Calculate the resistance in ohms (Ω) and the power in watts (W). DATA BULB VOLT (V) CURRENT (A) RESISTANCE (Ω) POWER(W) (V) (I) (R= V I ) P = I*V 40 Watt 120 60 Watt 120 100 Watt 120 Question: How does the current relate to the rated wattage of the bulb? PART B: CURRENT AND VOLTAGE IN PARALLEL CIRCUITS A parallel circuit is one where the devices are wired in such a way that current can independently flow through any branch. To observe the effect of the total current of outlets connected in parallel you will plug in two lamps with different wattage light bulbs and measure the current. 1. Plug a 40 watt light bulb into one of the outlets. Apply power and measure the current. 2. Plug a 60 watt bulb into a second outlet and record the current. DATA BULB(s) VOLT (V) CURRENT (A) RESISTANCE (Ω) POWER(W) (V) (I) (R= V I ) P = I*V 40W 120 60W+40W 120 Question: Compare the results to the currents in Part 1 and determine if the currents measured for the individual bulbs add up to the measured currents in Part 2. Observation: When you have the two lights plugged in, what happens to the one light bulb when you unscrew the other light bulb? PHYSICS 162 Name: . Winter 2005 Lab/Lecture Activity #3 EXPERIMENT 2 Energy Conversion: Gravitational Potential Energy to Electrical Energy Purpose: The purpose of this part of the lab is to examine the conversion of potential energy to usable electrical energy. The setup consists of a pulley wheel attached to a small generator. A string is wound around the pulley wheel. Dropping a mass attached to the string will turn the wheel and generate electricity, lighting up a bulb. A current and voltage probe are connected to a computer and can monitor the current and voltage generated by the falling mass. Measuring the current and voltage produced, and the time it takes the mass to fall a distance h, will enable you to calculate how much of the potential energy of the falling mass is being converted into light and heat. EXPERIMENT: PART A: First connect a light bulb to the leads of the generator and connect a 0.7kg mass to the string. Determine and measure the height you will drop the mass from. Drop the mass a couple of times and observe what happens to the bulb as the mass falls. h = ____________________ meters Drop the mass with the leads disconnected from the light bulb. Does the mass fall faster or slower...Why? PART B: With the falling 0.7kg mass on the string. measure the average current produced, the average voltage across the resistor, and the time it takes for the mass to fall the distance h. V = voltage (in Volts) P = power = V x I (in Watts) I = current (in Amps) t = time (in seconds) E = energy = P x t (in Joules) DATA: 0.7 kg mass V (volts) I (Amperes) P = I*V (Watts) t (seconds) Average CALCULATIONS: Average Energy Out (E = Paverage*taverage) = Joules Repeat with the 1 kg mass. Movie located at website DATA: 1 kg mass trial # V (volts) I (Amperes) P = I*V (Watts) t (seconds) 1 CALCULATIONS: Light Bulb Energy : (E = P*t) Joules PART C: Calculate the efficiency of this energy conversion process. Efficiency = Energy out / Energy in. EFFICIENCY OF 0.7 kg MASS: Energy in = gravitational potential energy = mgh = Joules Energy out = P*t = Joules Efficiency: = Energy Out Energy In x100 = % EFFICIENCY OF 1 kg MASS: Energy in gravitational potential energy = mgh = Joules Energy out: = P*t = Joules Efficiency: = Energy Out Energy In x100 = % Where did the rest of the gravitational potential energy go?