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What we are Doing Today ? 1) Question(s) of Day 2) Kirchhoff’s Laws & Magnetism Notes 3) Continue Lab 41 & 42 at the back 4) Make sure you finished the HW p.173 of the textbook questions 10-18 5) Lab is due on Thursday 6) Tutorial after school today. Room 308 Circuits Series and Parallel Circuits Kirchoff’s Laws Kirchoff’s Laws show the relationship between current and voltage Kirchoff’s First Law…current (junction rule) At any junction, the sum of the currents entering the junction equals the sum of the currents leaving Current in a series circuit: Any junction has one path in and one path out…..current is the same everywhere Current in a parallel circuit: Sum of the currents entering equals the sum of the currents leaving Kirchoff’s Second Law…potential difference (loop rule) The sum of the drops in potential (voltages) equals the potential (voltage) at the source Potential difference in a series circuit: What are the missing voltages in the above circuits? Voltages add up to the voltage at the source Potential difference in a parallel circuit: What are the voltages in bulb 1 and bulb 2? Voltage drop in each loop is the same as the voltage at the source. Series Circuits Series Circuits Current One path - all current flows through each resistor (remember the junction rule?) Equation: I = I1 = I2 = I3 .... Voltage The sum of the drops in potential difference equals the potential difference at the source (Remember the loop rule?) Equation: V = V1 + V2 + V3 ... Resistance V = V1 + V2 + V3 ... But V=IR, so….. IR = I1R1 + I2R2 + I3R3 .... Since I = I1 = I2 = I3 .... Current cancels in the equation, and….. Req = R1 + R2 + R3 .... Series Circuit Example: In the circuit below: R1=4Ω, R2=2Ω, and R3=6Ω. If the current in the circuit is 0.5 A, what is the voltage at the source? Method 1 Find total resistance in the circuit: R = R1+R2+R3 = 4Ω + 2Ω + 6Ω = 12Ω Then use Ohm’s Law to find total voltage: V = IR = 0.5 A x 12 Ω = 6V Method 2 Use Ohm’s Law to find voltage drop in each resistor Then add all the voltages: V = V1 + V2 + V3 = 2V + 1V + 3V = 6V Parallel Circuits Parallel Circuits Current Sum of currents entering a junction equals the sum of the currents leaving (remember the junction rule?) Equation: I = I1 + I2 + I3 .... Example: I1=2A I2=1A I4=6A What is I3? I4 = I1 + I2 + I3 6A = 2A + 1A + ? I3 = 3A Voltage The sum of the drops in potential difference equals the potential difference at the source (Remember the loop rule?) The voltage in each loop is the same as the source of potential: Equation: V = V1 = V2 = V3 ... Resistance I = I1 + I2 + I3 .... But I=V/R, so….. Since V V1 V2 V3 = + + ... R R1 R2 R3 V = V1 = V2 = V3 ... Voltage cancels in the equation, and….. 1 1 1 1 = + + ... Req R1 R2 R3 Parallel Circuit Example: R1=2Ω, R2=2Ω, and R3=4Ω. If the voltage at the source is 16V, what is the total current in the circuit? Add a voltmeter to show the voltage drop across R3. Method 1 Find total resistance in the circuit: 1 1 1 1 = + + Req R1 R2 R3 Need a common denominator Then use Ohm’s Law to find total voltage: Method 2 Use Ohm’s Law to find current in each resistor Then add all the currents: I = I1 + I2 + I3 = 8A + 8A + 4A = 20A What happens to total current in a parallel circuit when you add more appliances? Additional appliances - more resistance More paths…total resistance DECREASES Less resistance, more current More current, more thermal energy (heat) Short circuits, fires……. Solution? What devices do you have in your home to protect you from too much current? 1. Fuses - short wire melts, breaks circuit (thickness of wire determines current rating) 2. Circuit breakers - switch opens when current reaches a preset value 3. GFI (ground fault interrupter) - found in kitchen and bathroom outlets - detects differences in current caused by extra current paths - works like a circuit breaker For a Summary Look at table 5.28 on page 163 of the textbook What to do now? 1) 2) 3) Homework p. 173-174 questions 19-20 Finish lab 41 & 42 at the back Homework check p.173 questions 10-18