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Download AP Physics 2 Electrical Circuits 2015-16
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Drawing a Circuit • When illustrating an electrical circuit for everyone to understand, you must draw a standardized picture called a schematic. • A schematic is simply a picture of an electrical circuit that uses standardized symbols to represent the different parts of an electrical circuit. Basic Schematic Symbols Battery/ Voltage source Resistor Switch Fuse Series Circuit • Because a series circuit has only one loop, the amount of current that flows through one resistor must be the same amount of current that flows for all the other resistors as well. .5 A .5 A .5 A .5 A Series Circuit • Because no one resistor is connected directly to both terminals of the power source. No one resistor receives all of the voltage, the voltage is divided out amount each of the resistors 50 V 30 V 20 V 100 V Kirchhoff’s Voltage Rule • We know from the conservation from energy that all the energy that the power source provides must equal the total work that each resistor does. • So then we can say the the energy that each charge has must to be used by all the resistors. • Since voltage is the energy of one charge, we can say that the voltage of the power supply must be used up by all the resistors. • Kirchhoff’s Voltage law – Vpower supply = Sum of all Voltage drops – Vin = VR1 + VR2 + VR3 + VR4 …. Simplifying a Series Circuit • When trying to analyze a series circuit, normally the first step is to reduce the circuit of several resistors to an equivalent circuit of only 1 resistor and 1 power source • For any simplified circuits we will have a/an – Equivalent Voltage (Veq) - How much voltage is truly being supplied to the circuit – Equivalent Current (Ieq) - How much current the power source has entering the circuit – Equivalent resistance (Req) - How much resistance the circuit actually has – Equivalent power (Peq) - The total power of the circuit Original Circuit Equivalent Circuit Lets start with a series circuit with 3 resistors R2 R3 R1 Vin We can simply a series circuit by using 3 basic principles Kirrcoff’s Voltage law: VEq = VR1 + VR2 + VR3 Because there is 1 pathway for current in a series circuit the current for all resistors is the same. IEq = IR1 = IR2 = IR3 But we will still need to find the equivalent resistance. For that we will need to use Ohm’s Law: V=IR Lets start with a series circuit with 3 resistors Because: And since: We can say: R2 R3 R1 Vin Veq = VR1 + VR2 + VR3 V = IR IeqReq = IR1R1 + IR2R2 + IR3R3 Because All I’s are the same we call each one I We can say: IReq = IR1 + IR2 + IR3 The current is a common multiple on both sides Now we have an equation for the equivalent Resistance: Req = R1 + R2 + R3 Equations for a Series Circuit • Equivalent Voltage – Veq = VR1 + VR2 + VR3…. • Equivalent Current – IEq = IR1 = IR2 = IR3…. • Equivalent Resistance – Req = R1 + R2 + R3… • Equivalent power – PowEq = PowR1 +PowR2 + PowR3 … Sample problem For the circuit shown what is the equivalent resistance? How much current leaves the battery? For a series Req = R1 + R2 + R3 Req = 10 W + 10 W + 10 W Req = 30 W Using Ohms Law: 300 V = I(30 W) I= 10 Amps 10 W 10 W 10 W 300 V Solve a Series Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. 15 W + 12 V - 4W 5W Solve a Series Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. 15 W + 12 V - 4W 3A 5W Solve a Series Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. 3A 15 W + 12 V - 4W 3A 5W 3A 3A Solve a Series Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. + 45 V 3A 15 W 12 V + 4W 3A + 15 V 5W 3A 3A - Solve a Series Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. + 45 V 3A 15 W + 12 V - 4W 3A + 15 V 5W 3A 3A 45 W - V battery = 12V + 45V + 15V = 72 V Parallel Circuit • A parallel circuit is a circuit were each resistor has its own loop, being connected directly to the power source itself. Because of this each resistor receives the full voltage from the power supply. + 20 V - + 20 V - + 20 V - + 20 V - Parallel Circuit • Because each resistor has its own connection to both terminals of the power source each resistor operates independently of all the others, thus each can have their own current, and in one is off the others can still be on. 5A 5A 6A 1A 5A 4A 6A 10 A 10 A 10 A Kirchhoff’s Current Rule • We know from the conservation of charges that the total charge of a system can not change. • Kirchhoff’s Current rule states when a group of charges enter an intersection of wire, the total number of charges that leave the intersection must equal, the total number of charges that entered the intersection.. Iin I2 I1 Iin = I1 + I2 Lets simplify a parallel circuit with 3 resistors We can simply a parallel circuit by using 3 basic principles Kirchhoff’s Current law: IEq = IR1 + IR2 + IR3 Because each resistors is connected directly to the power source. VEq = VR1 = VR2 = VR3 But we still need to find the equivalent resistance. For that we will still need to use Ohm’s Law: V=IR Finding Req of a parallel circuit with 3 resistors Because: And since: Ieq = IR1 + IR2 + IR3 V = IR or even better I = V/R We have: (Veq/Req) = (V1/R1) + (V2/R2) + (V3/Re3) Because All V’s are the same we call each one V We have: (V/Req) = (V/Req) + (V/Req) + (V/Req) (V/Req) The Voltage is a common multiple on both sides Now we have an equation for the equivalent Resistance: (1/Req)= (1/R1) + (1/R2)+ (1/R3) Equations for a Parallel Circuit • Equivalent Voltage – Veq = VR1 = VR2 = VR3…. • Equivalent Current – IEq + IR1 + IR2 + IR3…. • Equivalent Resistance – 1/Req = 1/R1 + 1/R2 + 1/R3… • Equivalent power – PowEq = PowR1 +PowR2 + PowR3 … Two Circuits, but What are They Good for? • Series Circuit is great for control – Switches are connect in series to turn things on and off – Fuses are connected in series as to turn the circuit off – “Dimmers”/Rheostats are used in series to regulate the current flow and voltage of another object • Parallel is great when you want things to run independently – Power strips – Power outlets of a house – The Different electronics of a car Sample problem For the circuit shown what is the equivalent resistance? How much current leaves the battery? 10 W For a parallel 1/Req = 1/R1 + 1/R2 + 1/R3 1/Req = 1/10 W + 1/10 W + 1/10 W 1/Req = 3/10 W Req = 10/3 W = 3.33 W Using Ohms Law: 300 V = I(3.33W) I= 90.09 Amps 10 W 10 W 300 V Using the PVIR Box The box is just an easy way to organize the information provided Because V = IR, and P = IV we can use The box to solve any given series or parallel circuit Solve a Parralell Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. 2A 10 W 5W 20W Solve a Parralell Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. - 20 V + 2A 10 W 5W 20W Solve a Parralell Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. - 20 V + 2A 10 W 4A 5W 1A 20W Solve a Parralell Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. - 20 V + 2A 10 W 4A 5W 1A 20W 7A Solve a Parralell Circuit Find the voltage of the battery, the current leaving the battery, and the power of the 5 ohm resistor. - 20 V + 2A 10 W 4A 5W 1A 20W 7A P = IV P = (4A)(20V) P = 80 Watts