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Transcript
Physics 6B Electric Current And DC Circuit Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Electric Current is the RATE at which charge flows (usually through a wire). We can define it with a formula: I Q t Units are Coulombs/second, or Amperes (A) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V R1 R3 R2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V 6Ω Before we can calculate the individual currents, we need to know how much current is supplied by the battery. So we need to find the equivalent resistance for the circuit. 6Ω 12Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω Before we can calculate the individual currents, we need to know how much current is supplied by the battery. So we need to find the equivalent resistance for the circuit. 20V 6Ω 6Ω Parallel – Req=4Ω 12Ω Notice that R1 and R2 are in parallel. We can combine them into one single resistor: 1 1 1 R R Req 1 2 Req R1 R2 R1 R2 Req 6 12 4 6 12 This shortcut formula will work for any pair of parallel resistors. 20V 4Ω 6Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω Before we can calculate the individual currents, we need to know how much current is supplied by the battery. So we need to find the equivalent resistance for the circuit. 20V 6Ω 6Ω Parallel – Req=4Ω 12Ω Notice that R1 and R2 are in parallel. We can combine them into one single resistor: 1 1 1 R R Req 1 2 Req R1 R2 R1 R2 Req This shortcut formula will work for any pair of parallel resistors. 6 12 4 6 12 20V Series – Req=10Ω 4Ω 6Ω The next step is to combine the remaining resistors, which are in series. The formula is simple – just add them together. 20V 10Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω Before we can calculate the individual currents, we need to know how much current is supplied by the battery. So we need to find the equivalent resistance for the circuit. 20V 6Ω 6Ω Parallel – Req=4Ω 12Ω Notice that R1 and R2 are in parallel. We can combine them into one single resistor: 1 1 1 R R Req 1 2 Req R1 R2 R1 R2 Req This shortcut formula will work for any pair of parallel resistors. 6 12 4 6 12 20V Series – Req=10Ω The next step is to combine the remaining resistors, which are in series. The formula is simple – just add them together. 4Ω 6Ω 2 Amps 20V Now that we finally have our circuit simplified down to a single resistor we can use Ohm’s Law to compute the current supplied by the battery: V 20V I 2Amps R 10 10Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V 6Ω Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I) Voltage (V) Power (P) 6Ω Parallel – Req=4Ω 12Ω R1=6Ω R2=12Ω R3=6Ω 20V Battery Series – Req=10Ω 4Ω 6Ω 2 Amps 20V 10Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V 6Ω Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I) Voltage (V) Power (P) 6Ω Parallel – Req=4Ω 12Ω R1=6Ω R2=12Ω R3=6Ω 2 Amps Battery 2 Amps 20V 20 volts Take a look at the original circuit. Notice that all the current has to go through R3. So I3 = 2 Amps. We can also fill in the information for the battery – we know its voltage and we already found the total current, which has to come from the battery. Series – Req=10Ω 4Ω 6Ω 2 Amps 20V 10Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V 6Ω Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I) Voltage (V) R3=6Ω 2 Amps 12 volts Battery 2 Amps 20 volts Power (P) 6Ω Parallel – Req=4Ω 12Ω R1=6Ω R2=12Ω Take a look at the original circuit. Notice that all the current has to go through R3. So I3 = 2 Amps. We can also fill in the information for the battery – we know its voltage and we already found the total current, which has to come from the battery. Now that we have the Current for resistor #3, we can use Ohm’s Law to find the voltage drop: V I R 2A 6 12V 20V Series – Req=10Ω 4Ω 6Ω 2 Amps 20V 10Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V 6Ω Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I) Voltage (V) R3=6Ω 2 Amps 12 volts Battery 2 Amps 20 volts Power (P) 6Ω Parallel – Req=4Ω 12Ω R1=6Ω R2=12Ω The current for R1 and R2 will be easier to find if we calculate the voltage drops first (they have to be the same voltage because they are in parallel – make sure you understand why!). 20V Series – Req=10Ω 4Ω 6Ω 2 Amps 20V 10Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V 6Ω Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I) Voltage (V) R1=6Ω 8 volts R2=12Ω 8 volts R3=6Ω 2 Amps 12 volts Battery 2 Amps 20 volts Power (P) The current for R1 and R2 will be easier to find if we calculate the voltage drops first (they have to be the same voltage because they are in parallel – make sure you understand why!). Either use Ohm’s law (use the 2nd diagram with the combined resistance of 4Ω), or notice that the total voltage is 20V, and R3 uses 12V, so there is 8V left over for R1 or R2. Now we can use Ohm’s Law again for the individual resistors to find the current through each: 6Ω Parallel – Req=4Ω 12Ω 20V Series – Req=10Ω 4Ω 6Ω 2 Amps 20V 10Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V 6Ω Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I) Voltage (V) R1=6Ω 4/3 Amps 8 volts R2=12Ω 2/3 Amps 8 volts R3=6Ω 2 Amps 12 volts Battery 2 Amps 20 volts Power (P) The current for R1 and R2 will be easier to find if we calculate the voltage drops first (they have to be the same voltage because they are in parallel – make sure you understand why!). Either use Ohm’s law (use the 2nd diagram with the combined resistance of 4Ω), or notice that the total voltage is 20V, and R3 uses 12V, so there is 8V left over for R1 or R2. Now we can use Ohm’s Law again for the individual resistors to find the current through each: 8V 4 Amps 6 3 8V 2 I2 Amps 12 3 6Ω Parallel – Req=4Ω 12Ω 20V Series – Req=10Ω 4Ω 6Ω 2 Amps 20V 10Ω I1 Total = 2 Amps Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V 6Ω Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I) Voltage (V) R1=6Ω 4/3 Amps 8 volts R2=12Ω 2/3 Amps 8 volts R3=6Ω 2 Amps 12 volts Battery 2 Amps 20 volts Power (P) Finally, we can calculate the power for each circuit element. You have your choice of formulas: V2 P I V I R R 6Ω Parallel – Req=4Ω 12Ω 20V Series – Req=10Ω 2 4Ω 6Ω 2 Amps 20V 10Ω Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R1=6Ω, R2=12Ω, R3=6Ω 20V 6Ω Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Current (I) Voltage (V) Power (P) R1=6Ω 4/3 Amps 8 volts 32/3 Watts R2=12Ω 2/3 Amps 8 volts 16/3 Watts R3=6Ω 2 Amps 12 volts 24 Watts Battery 2 Amps 20 volts 40 Watts Finally, we can calculate the power for each circuit element. You have your choice of formulas: V2 P I V I R R 12Ω 20V Series – Req=10Ω 2 I suggest using the simplest one. Plus it’s easy to remember because you probably live there… 6Ω Parallel – Req=4Ω 4Ω 6Ω 2 Amps 20V 10Ω As a final check you can add the powers to make sure they come out to the total power supplied by the battery. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB