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Physics 6B

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:. Units are Coulombs/second, or Amperes (A).

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Physics 6B

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  1. Physics 6B Electric Current And DC Circuit Examples Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  2. Electric Current is the RATE at which charge flows (usually through a wire). We can define it with a formula: Units are Coulombs/second, or Amperes (A) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  3. 20V R1 R3 R2 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 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  4. 20V 6Ω 6Ω 12Ω 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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  5. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 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. Notice that R1 and R2 are in parallel. We can combine them into one single resistor: Parallel – Req=4Ω This shortcut formula will work for any pair of parallel resistors. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  6. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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. Notice that R1 and R2 are in parallel. We can combine them into one single resistor: Parallel – Req=4Ω This shortcut formula will work for any pair of parallel resistors. Series – Req=10Ω The next step is to combine the remaining resistors, which are in series. The formula is simple – just add them together. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  7. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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. Notice that R1 and R2 are in parallel. We can combine them into one single resistor: Parallel – Req=4Ω This shortcut formula will work for any pair of parallel resistors. Series – Req=10Ω The next step is to combine the remaining resistors, which are in series. The formula is simple – just add them together. 2 Amps 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: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  8. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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 Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Parallel – Req=4Ω Series – Req=10Ω 2 Amps Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  9. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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 Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Parallel – Req=4Ω 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 2 Amps Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  10. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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 Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Parallel – Req=4Ω 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: Series – Req=10Ω 2 Amps Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  11. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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 Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Parallel – Req=4Ω 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!). Series – Req=10Ω 2 Amps Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  12. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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 Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Parallel – Req=4Ω 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: Series – Req=10Ω 2 Amps Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  13. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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 Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Parallel – Req=4Ω 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: Series – Req=10Ω 2 Amps Total = 2 Amps Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  14. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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 Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Parallel – Req=4Ω Finally, we can calculate the power for each circuit element. You have your choice of formulas: Series – Req=10Ω 2 Amps Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  15. 20V 6Ω 6Ω 12Ω 20V 4Ω 6Ω 20V 10Ω 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 Now that we have the total current, we have to find the individual currents. It might help to make a table like this: Parallel – Req=4Ω Finally, we can calculate the power for each circuit element. You have your choice of formulas: Series – Req=10Ω 2 Amps I suggest using the simplest one. Plus it’s easy to remember because you probably live there… 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

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