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Grade 9 Academic Science – Unit 4 Electricity Series and Parallel Circuits Section 13.10 Pages 571-575 SERIES CIRCUIT One path for current to flow When two lights are connected in series, the brightness of each lamp decreases. Why? As more loads (…in this case, lights…) are added, more RESISTANCE to the flow of current is built up. With more resistance, less current flows into each light. Thus, brightness goes down as more loads are added. See Sample Problem #1 on Page 571. Example A circuit diagram has two lights in series. The total voltage in the circuit is 50 V. The total resistance (R) of the two lights is 10 (Recall: = ohms). What is the current (I) in the circuit? HINT: Use Ohm’s Law formula I = V/R Solution I = 50 / 10 I=5A The current through each light is 5 amps Another circuit diagram has four lights arranged in a series circuit. The total voltage is the same as the first circuit (i.e., V = 50 V). Total resistance has increased to 25 . What is the current on this circuit? Solution I = 50 / 25 I=2A The current through each light is 2 amps. Thus, the lights in the second circuit would not burn as brightly. Task Draw a circuit diagram with five lights in series. The Potential Difference (i.e., total voltage) in the circuit is 120 V. The total resistance (R) of the lights is 40 (Recall: = ohms). What is the current through each lamp? HINT: Use Ohm’s Law formula I = V/R Voltage across a Load in Series …Not enough? To add more, look at the POTENTIAL ENERGY of the circuit. A battery converts chemical energy into electric potential energy. As the electron leave the battery, they establish flow (i.e., they push each other along). As the electrons move, the electric potential energy becomes kinetic energy. When the electron passes through one light, the electric potential energy is converted into heat and light (…kinetic energy…). When the electron re-enters the battery, it is recharged (i.e., given more energy) which keeps the movement of electrons going (inand-out). Since voltage is related to potential energy, the voltage drop across one load (light) equal the voltage drop across the battery. If two lights occur on the series circuit, only half the potential electric energy is converted into heat and light in each light. Why? The battery can only supply so much energy, and to keep the current flowing, the energy must be distributed across all the loads. The voltage drop across the light is half the voltage drop across the battery. If more loads are added, less potential energy is converted to light and heat at each load. The voltage drop across each load decreases. The formula is Vload = Vsource / Number of Loads Example See Sample Problem #2 on Page 572 A series circuit contains six identical lamps. The potential difference of the battery is 120 V. What is the potential difference across each lamp? Solution Vload = Vsource / # of loads Vload = 120 / 6 Vload = 20 V The potential difference across each lamp is 20 volts Practice Twenty identical lights are connected in series. A 10 V battery is hooked up as the power supply. What is the potential difference across each light? Potential difference at the source is 200 V. Five very large and identical floodlights are connected in a series circuit to the voltage source. What is the potential difference across each light? PARALLEL CIRCUIT More than one path for current to flow. The number of paths is determined by the number of loads that are connected in parallel. When two lights are connected in parallel, the brightness of each lamp remains the same. Why? The electrons leaving the battery are split between the paths. Thus, the quantity of electrons leaving the battery is greater than the quantity of the electrons moving through each load. The formula is Iload = Isource / Number of Loads where Iload is the current (amps) through each identical load, Isource is the current coming from the energy source, and number of loads is the total number of identical loads in the entire circuit Example See Sample Problem #3 on Page 573 There are five loads. Each load has a separate path in a parallel circuit. Total Resistance is 100 . The potential difference of the energy sources is 20V. What is the current through each light? Solution Isource =V/R = 20 / 100 = 0.2 A Iload = Isource / # of loads = 0.2 / 5 = 0.04 A The current through each lamp is 0.04 amps Practice Four identical lights are connected in parallel. Total resistance is 10 . A 10 V battery is hooked up as the power supply. What is the current across each light? Ten identical lights are connected in parallel. Total resistance is 100 . A 5000 V battery is hooked up as the power supply. What is the current across each light? HOMEWORK Page 575 Questions 2-7