1. Complete the following circuit diagram to show the two light bulbs in parallel to each other, connected to the 12 V battery. 2. Marty placed a voltmeter.

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Presentation transcript:

1. Complete the following circuit diagram to show the two light bulbs in parallel to each other, connected to the 12 V battery. 2. Marty placed a voltmeter across one of the bulbs in the circuit. What would be the voltage reading?

3. Discuss the advantages of wiring the bulbs in parallel rather than in series. 4. Marty decides to put a fuse into the circuit. What is the purpose for adding the fuse?

1. Parallel

2. 12 V 3. If one bulb blows the other will keep going because there is still a complete conducting pathway All bulbs (in parallel) are the same brightness because the voltage is the same across each bulb.

4. The fuse will protect the circuit components. Instead of potentially blowing an expensive component during a power surge, a fuse will blow and can be easily replaced.

 Electrical resistance (R) is a measure of the degree to which an object opposes an electric current through it.  Measured in ohms ( Ω ).  Calculated using the formula:

 When electric current flows in a circuit with resistance, it does work.  Devices convert this work into many useful forms, such as heat, light motion and sound.

 Electric power is defined as the rate at which electrical energy is transformed by an electric circuit.  The unit of power is the watt (W).  Calculated using the formula:

R V I V P I Resistance Power

 Study the circuit diagram below. The two lamps are NOT identical. When an ammeter is placed in this circuit a reading of 1.2 A is recorded.

a) Using the equation V = IR, calculate voltage across the 4  lamp. b) Calculate the power output of the 4  lamp. Give an appropriate unit. c) In terms of energy, explain what the power output in (b) means. d) Calculate the resistance of lamp B. e) Describe how the brightness of lamp B compares with the 4  lamp.

a) Correct solution V = IR = 1.2  4 = 4.8 v b) Correct solution AND unit P = VI = 4.8  1.2 = 5.76 W c) Tells you the rate at which electrical energy is converted into light and / or heat.

d) Correct solution V lamp A = IR = 1.2 x 4 = 4.8 V V lamp B = 12 – 4.8 = 7.2 V R = V ÷ I = 7.2 ÷ 1.2 = 6 

 Pg. 58 #2, 4 and 6