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Kirchhoff's Voltage and Current Laws AND Ohm’s and Watt’s Laws Kirchhoff’s Voltage Law The sum of the voltage drops around a circuit is equal to the sum of the voltage source. 10V 10V 10V 6V 4V Kirchhoff’s Current Law The total amount of current coming out of a power source must equal the total amount going back into the power source. 1A 10V 1A Remember: Current is the flow of electricity in a circuit. The unit is the Ampere (A) 2A 10V 2A Kirchhoff’s Current Law (Applied to series and parallel circuits con’t) Series/Parallel Circuits In a circuit that is comprised of both series and parallel, the current remains the same through a series path, divides up through the parallel path depending on the resistance of each path, then goes back together. 4A 4A 3A 1A Electrical Components and Their Characteristics Disclaimer All electrical/electronic components require voltage and current to make them work. When working with electrical/electronic components it is VERY IMPORTANT to understand that these components will only work properly when the correct voltage and current are supplied to them. All electrical/electronic components are RATED for a specific range of voltage and current. Exceeding these ratings results in decreased life expectancy or failure of the component. Ohm’s Law Ohm’s Law • Ohm’s Law is the relationship between voltage, current and resistance. V The law states: One ohm is the resistance value through which one volt will maintain a current of one ampere. V = Voltage I = Current (in Amps) R = Resistance in Ohms (Ω) Example: V=IxR V = 1A x 1Ω V = 1volt I R V=IxR Ohm’s Law Continued • In a circuit you know the voltage and therefore V = I x R isn’t the most useful. However, you can change the formula around to solve for resistance and current . V V=IxR can be changed to I=V/R to solve for current Example: Supply of 10volts and a 100Ω resistor, what is the current in the circuit? I=V/R I = 10V / 100Ω I = .1A The current running through the circuit is .1A. I R V = Voltage, I = Current, R = Resistance Ohm’s Law Continued • All components require a specific amount of current to work. Therefore, the most useful use of Ohm’s Law in electronics is to solve for resistance. V V=IxR can be changed to R=V/I Example: You have a 10volt supply and require a .01A (10mA) current to run an electrical component, what resistance is needed? R=V/I R = 10V / .01A R = 1000Ω The resistor needed is 1000Ω. I R V = Voltage, I = Current, R = Resistance Ohm’s Law Examples Solve for the missing value Voltage (volts) Current (Amps) Resistance (Ohms) 20 2 10 10 2 5 5 .05A (50mA) 100 10 .02 500 5 .005A 5mA 1000 V I R V = Voltage, I = Current, R = Resistance V=IxR I =V/R R=V/I Voltage Dividers (Applying Ohm’s Law) When you have two resistors in series they create a voltage divider. Voltage dividers are used to create SPECIFIC voltages for use in circuits (ie – USB power supply, sensors with the Arduino) How it works • If the voltage of the circuit is 5 volts and the resistors are 2.5ohms each, the current would be 1amp (I = V/R) • Since the total current will never change and the resistors are the same, the voltage between the resistors is half the supply voltage. V = I x R 1A x 2.5 Ω = 2.5V • IF the resistors are different such as 4ohm ad 1 ohms, the current is still 1amp. However, the voltage between the two is NOT 5volts, now. The voltage drop across the top resistor will be greater than the bottom one. TOP = 1A x 4Ω = 4V Bottom = 1A x 1 Ω = 1V 5v 0v Divided voltage Voltage Divider in USB Project • The supply voltage is 5 volts and the resistors first voltage divider is 150K and 100K (I = V/R) 5v Total resistance = 150k + 100k = 250k Current through resistors (I=V/R) I = 5/250k I = .00002A or .02mA Top resistor voltage drop (V = I x R) .00002A x 150000 (150K) = 3V Bottom resistor voltage drop .00002A x 100000 (100k) = 2V 0v LED In Circuit Example (Putting it all together) • ALL components consume voltage as current electricity passes through them. • The first lab proves this when you measure the voltages around the circuit. The last question in the lab is “How many volts is the LED using up?” The answer will be between 2V and 4V depending on the LED you are using. • This is important because you need to take this into account when calculating current through the circuit. Current required = 15mA R=? V drop = ? 9v 2v drop LED In Circuit Example Continued (Putting it all together) Example: You have an LED with a voltage drop of 2 volts and the source voltage is 9 volts. If you want a constant current of 15mA, what resistor do you need? Here is where Ohm’s Law is needed! Current required = 15mA R=? V drop = ? 9v Voltage/Current = Resistance (The problem is the RESISTOR determines the CURRENT in the circuit, therefore you need to MINUS the voltage drop of the LED from the Source Voltage to get the correct current flowing through the resistor) (Source Voltage - LED Voltage Drop ) / Current = OHMs 9v - 2v = 7v / .015A = 466 Ω Closest resistor is 470ohms 2v drop Positive Wire Lab #1 - Setup Negative Wire Meter #1 Voltage Make sure you are on the DC voltage setting or the multimeter will not give you the correct reading. Meter #2 Current Make sure the meter dial is on the 200mA range and that the red lead is in the mA hole and the black lead is in the COM hole Watt’s Law Watt’s Law • When electricity is being used by a load (led, motor, toaster, etc), the electrical energy is being converted into another form (light, heat, motion). • Power is the amount of electrical energy being converted by a load. The unit of measure is the Watt. P V I P = Watts, V = Voltage, I = Current Calculating Power; P=VxI Watt’s Law Continued Examples You have a voltage of 10v and a current of 2A, what amount of power is being consumed? P=VxI P = 10V x 2A P = 20W P V I You have light bulb that is 60W in your house. How much current is the light P = Watts, V = Voltage, I = Current bulb drawing? I=P/V I = 60W / 120V I = .5A P=VxI I =P/V V=P/I Limiting Current using Resistors Resistors are an important part of electronics. They are used to control the amount of current being used in a circuit and specifically by components. Resistors come in different sizes (1/8, ¼, ½, 1, 5, 10watt….etc) and values (1 Ohm – 20M Ohm). It is EXTREMELY IMPORTANT to match up a resistors wattage to the amount of current that will flow through it in a circuit. Watt’s Law and Resistors • Watt’s Law is needed when working with resistors. Once the current that will flow through a resistor is calculated using Ohm’s Law, you need to figure out the size of the resistor (wattage) so it can handle the current and NOT fail. • Example – You have a voltage of 10v and a resistor of 100Ω. What is the current? What wattage resistor is needed to handle that current? V=IxR I = 10V / 100 I = .1A P=VxI P = 10v x .1A P = 1 Watt Therefore, the current is .1A and you would require a 1W resistor in the circuit. P V I 50V 100Ω 47Ω Resistor @ 9V Example P V I You have a 47Ω resistor connected to a 9V battery, what happens? V=IxR P=VxI I = 9V / 47Ω I = .1915A or 191.5mA P = 9v x .1915A P = 1.7 Watt What about a 100Ω resistor? V=IxR I = 9V / 100Ω I = .09A or 90mA P=VxI P = 9v x .09A P = .81 Watts 9V 47Ω