Experiment 2c Class: Name: ( ) Date: 2c Effect of length and thickness on resistance Objective To find out how the resistance of a metal wire is affected by its length and thickness (cross-sectional area). Background information 1 The resistance of a conductor is defined as: voltage across conductor Resistance = current through conductor or 2 3 R= V I By measuring the current I through a conductor when a known voltage V is applied across it, the resistance of the conductor V can be determined from the formula R = . I The graph in Figure 2c-1 is a straight line passing through the origin, which means x and y are directly proportional, i.e. y y0 slope = x = k 0 y0 y∝x or y = kx (k is a constant) origin x0 x Fig 2c-1 Apparatus ❏ 2 eureka wires of different thickness (0.5 m, mounted on a wooden strip) ❏ 1 voltmeter ❏ 1 ammeter ❏ 1 battery box ❏ 1 switch ❏ several connecting leads (one with a crocodile clip at one end) 24 New Physics at Work (Second Edition) © Oxford University Press 2007 Class: Name: ( Procedure ✐ Before the experiment, discuss with Ss the need for control of variable—keep thickness of the wire constant when investigating the effect of length on resistance and keep length constant when investigating the effect of thickness of wire. Experiment 2c ) Date: Effect of length 1 Set up the apparatus as shown in Figure 2c-2: (a) Connect one of the eureka wires in series with an ammeter, a battery box and a switch. (b) Connect a voltmeter across the wire using a crocodile clip as a sliding contact at one end. Be careful! Do not touch the hot wire! Disconnect the circuit once the readings are taken. Otherwise, the eureka wire may overheat. battery box switch ammeter voltmeter crocodile clip two 0.5 m-eureka wires of different thickness mounted on wooden strip + eureka wire A length of wire + V Fig 2c-2 2 (a) Vary the length of the wire connected by sliding the crocodile clip along the wire. (b) Take the ammeter and voltmeter readings. Record the results in Table 2c-1 on p.26. New Physics at Work (Second Edition) © Oxford University Press 2007 25 Experiment 2c Class: Name: ( ) Date: 3 Repeat several times with other lengths. Record the results in Table 2c-1 and calculate the resistance of different lengths of the wire. ✎ Results: Length of wire l / m 0.10 0.20 0.30 0.40 0.48 Voltage across wire V / V 0.48 0.90 1.30 1.70 2.10 Current through wire I / A 0.50 0.50 0.50 0.50 0.50 Resistance of wire R / Ω 0.96 1.80 2.60 3.40 4.20 Table 2c-1 4 Plot a graph of the resistance R of wire against the length l of wire in Figure 2c-3. resistance of wire R / 8 4 3 2 1 0 0.1 0.2 0.3 0.4 0.5 length of wire l / m Fig 2c-3 26 New Physics at Work (Second Edition) © Oxford University Press 2007 Class: Name: ( Experiment 2c ) Date: Effect of thickness (cross-sectional area) 5 Measure the current and voltage of equal length of the two wires of different thickness. Record the results in Table 2c-2 and calculate the resistance of the two wires. ✎ Results: Thin wire Thick wire Voltage across wire V / V 1.00 1.00 Current through wire I / A 0.24 1.00 Resistance of wire R / Ω 4.20 1.00 Table 2c-2 Discussion ✎ How is the resistance of the wire related to its length? The resistance of the wire is directly proportional to its length. ✎ How does the resistance of the wire change with its cross-sectional area? The resistance of the wire decreases as its cross-sectional area increases. 1For metal wires of the same thickness and material, the directly proportional resistance is ___________________________________ to the length of the wires. 2For metal wires of the same length and material, the larger the smaller cross-sectional area of the wires, the ________________________ the resistance. New Physics at Work (Second Edition) © Oxford University Press 2007 27 Experiment 2c Class: Name: ( ) Date: Further thinking ✎ Design an experiment to show whether the resistance of a metal wire is inversely proportional to its cross-sectional area. Write down the necessary procedures. Use several eureka wires of different diameters. Measure the current and voltage of equal length of the wires to calculate their corresponding resistances. Plot a graph • • 1 4 = . If the of the resistance of the wire against cross-sectional area π × (diameter)2 graph is a straight line passing through the origin, the resistance of a metal wire is inversely proportional to its cross-sectional area. 28 New Physics at Work (Second Edition) © Oxford University Press 2007