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

2. 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)

3. 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: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω 20V R2R2 R1R1 R3R3

4. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =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.

5. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω 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 R 1 and R 2 are in parallel. We can combine them into one single resistor: This shortcut formula will work for any pair of parallel resistors.

6. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 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 R 1 and R 2 are in parallel. We can combine them into one single resistor: This shortcut formula will work for any pair of parallel resistors. The next step is to combine the remaining resistors, which are in series. The formula is simple – just add them together.

7. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 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 R 1 and R 2 are in parallel. We can combine them into one single resistor: This shortcut formula will work for any pair of parallel resistors. The next step is to combine the remaining resistors, which are in series. The formula is simple – just add them together. 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: 2 Amps

8. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps 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) R 1 =6Ω R 2 =12Ω R 3 =6Ω Battery

9. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps 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) R 1 =6Ω R 2 =12Ω R 3 =6Ω2 Amps Battery2 Amps20 volts Take a look at the original circuit. Notice that all the current has to go through R 3. So I 3 = 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.

10. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps 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) R 1 =6Ω R 2 =12Ω R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts Take a look at the original circuit. Notice that all the current has to go through R 3. So I 3 = 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:

11. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps 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) R 1 =6Ω R 2 =12Ω R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts 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!).

12. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps 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) R 1 =6Ω8 volts R 2 =12Ω8 volts R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts 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 2 nd diagram with the combined resistance of 4Ω), or notice that the total voltage is 20V, and R 3 uses 12V, so there is 8V left over for R 1 or R 2. Now we can use Ohm’s Law again for the individual resistors to find the current through each:

13. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps 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) R 1 =6Ω4/3 Amps8 volts R 2 =12Ω2/3 Amps8 volts R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts Total = 2 Amps 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 2 nd diagram with the combined resistance of 4Ω), or notice that the total voltage is 20V, and R 3 uses 12V, so there is 8V left over for R 1 or R 2. Now we can use Ohm’s Law again for the individual resistors to find the current through each:

14. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps 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) R 1 =6Ω4/3 Amps8 volts R 2 =12Ω2/3 Amps8 volts R 3 =6Ω2 Amps12 volts Battery2 Amps20 volts Finally, we can calculate the power for each circuit element. You have your choice of formulas:

15. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 20V 12Ω 6Ω6Ω 6Ω6Ω 20V 6Ω6Ω4Ω4Ω 10Ω Example: Find the current through, and power used by each resistor in this circuit. Resistances are given as follows: R 1 =6Ω, R 2 =12Ω, R 3 =6Ω Parallel – R eq =4Ω Series – R eq =10Ω 2 Amps 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) R 1 =6Ω4/3 Amps8 volts32/3 Watts R 2 =12Ω2/3 Amps8 volts16/3 Watts R 3 =6Ω2 Amps12 volts24 Watts Battery2 Amps20 volts40 Watts Finally, we can calculate the power for each circuit element. You have your choice of formulas: 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.