1. 1 DC ELECTRICAL CIRCUITS OHMS LAW

2. 2 DC ELECTRICAL CIRCUITS Ohms law is the most important and basic law of electricity and electronics. It defines the relationship between the three fundamental electrical quantities; current, voltage and resistance. Ohms law is the most important and basic law of electricity and electronics. It defines the relationship between the three fundamental electrical quantities; current, voltage and resistance. THIS IS A DIAGRAM THAT EXPRESSES THE OHMS LAW RELATIONSHIP

3. 3 DC ELECTRICAL CIRCUITS The three formulas for ohms law as seen in the diagram are: E = I x R I = E / R R = E / I The three formulas for ohms law as seen in the diagram are: E = I x R I = E / R R = E / I

4. 4 DC ELECTRICAL CIRCUITS By covering each of the letters you want to solve for you can see the formula for finding that value, in the diagram below the first circle has the E covered so the formula to find voltage (E) is I x R.

5. 5 DC ELECTRICAL CIRCUITS To solve for current (I) you cover the I and the formula is E divided by R (E/R).

6. 6 DC ELECTRICAL CIRCUITS To solve for resistance (R) you cover up the R and the formula is E divided by I or (E/R).

7. 7 DC ELECTRICAL CIRCUITS In the series circuit below we have the voltage and resistance of the circuit but no current, we can solve for current using ohms law; I = E/R 20/10 = 2, the amperage of this circuit is 2Amps. In the series circuit below we have the voltage and resistance of the circuit but no current, we can solve for current using ohms law; I = E/R 20/10 = 2, the amperage of this circuit is 2Amps.

8. 8 DC ELECTRICAL CIRCUITS The value missing in this circuit is voltage, so using the ohms law formula E = I x R, 40 x 2 = 80 volts. The value missing in this circuit is voltage, so using the ohms law formula E = I x R, 40 x 2 = 80 volts.

9. 9 DC ELECTRICAL CIRCUITS We need to solve for resistance in this series circuit; R = E/I or 24/.05 = 480 ohms. Notice the direction of current flow, from the (-) terminal of the battery. We need to solve for resistance in this series circuit; R = E/I or 24/.05 = 480 ohms. Notice the direction of current flow, from the (-) terminal of the battery.

10. 10 DC ELECTRICAL CIRCUITS Using ohms law we can find the current for this circuit however we have to find the total resistance first, Rt + R1+R2+R3 or 95 ohms. I = E/R, 25/95 =.263 amps or 263 mA (milliamps)

11. 11 DC ELECTRICAL CIRCUITS We can apply ohms law to parallel circuits too. There are some circuit rules you must know when working with parallel and series parallel circuits; CURRENT STAYS THE SAME IN A SERIES CIRCUIT, BUT DIVIDES IN PARALLEL. We can apply ohms law to parallel circuits too. There are some circuit rules you must know when working with parallel and series parallel circuits; CURRENT STAYS THE SAME IN A SERIES CIRCUIT, BUT DIVIDES IN PARALLEL.

12. 12 DC ELECTRICAL CIRCUITS The circuit that we solved current for shows that in a series circuit the current will remain the same through each resistor. IN A SERIES CIRCUIT CURRENT STAYS THE SAME THROUGH EACH RESISTOR.

13. 13 DC ELECTRICAL CIRCUITS Voltage does the opposite of current in series and parallel circuits. VOLTAGE DROPS IN SERIES AND STAYS THE SAME IN A PARALLEL CIRCUIT. Voltage does the opposite of current in series and parallel circuits. VOLTAGE DROPS IN SERIES AND STAYS THE SAME IN A PARALLEL CIRCUIT. CURRENT STAYS THE SAME IN A SERIES CIRCUIT, BUT DIVIDES IN PARALLEL.

14. 14 DC ELECTRICAL CIRCUITS In this parallel circuit the 120 volts would appear across each resistive branch, 120V @ R1, R2 and R3. The current would divide at each branch. In this parallel circuit the 120 volts would appear across each resistive branch, 120V @ R1, R2 and R3. The current would divide at each branch.

15. 15 DC ELECTRICAL CIRCUITS Using the same series circuit that we’ve been working with we can solve for the voltage drops around the circuit. E = R1 x I1 (10 x.263) = 2.63V E = R2 x I2 (25 x.263) = 6.57V E = R3 x I3 (60 x.263) = 15.78V ADD ALL OF THE VOLTAGE DROPS; 2.63+ 6.57+ 15.78= 24.98 or 25Volts

16. 16 DC ELECTRICAL CIRCUITS In a series parallel circuit some of the rules for series circuits will apply and some of the rules for parallel circuits will apply. An important point when analyzing and solving for this type of circuit you must subtract your voltage drops to determine the remaining voltage for the rest of the circuit. In a series parallel circuit some of the rules for series circuits will apply and some of the rules for parallel circuits will apply. An important point when analyzing and solving for this type of circuit you must subtract your voltage drops to determine the remaining voltage for the rest of the circuit.

17. 17 DC ELECTRICAL CIRCUITS When solving a series parallel circuit start at the opposite end of the power source and solve towards the power source. Redraw the circuit as you reduce your parallel loads. When solving a series parallel circuit start at the opposite end of the power source and solve towards the power source. Redraw the circuit as you reduce your parallel loads.

18. 18 DC ELECTRICAL CIRCUITS The first thing we solve for in series parallel circuits will typically be Rt and then It, with that information we can then solve for individual values around the circuit.

19. 19 DC ELECTRICAL CIRCUITS Rt = R2 x R3 120 x 60 = 7200 divided by 120 + 60 (180) R2 + R3 7200/180= 40 Ω plus R1 (30 Ω ) = 70 Ω. The total current It = E/R or 70/70 = 1 Amp. The voltage drop @ R1 is E=IxR, I stays the same in series. Rt = R2 x R3 120 x 60 = 7200 divided by 120 + 60 (180) R2 + R3 7200/180= 40 Ω plus R1 (30 Ω ) = 70 Ω. The total current It = E/R or 70/70 = 1 Amp. The voltage drop @ R1 is E=IxR, I stays the same in series.

20. 20 DC ELECTRICAL CIRCUITS 1 x 30 = 30 Volts, this means that 30 volts is dropped at the R1 resistor, that leaves only 40 volts for the rest of the circuit, because voltage stays the same in parallel the voltage across R2 and R3 is 40 Volts.

21. 21 DC ELECTRICAL CIRCUITS We can find the amperage around the circuit using ohms law, I= E/R, at R1 the amperage is 1A because current stays the same in series, at R2 its 40V/120 Ω =.333Amps, at R3 its 40V/60Ω =.666Amps,.333+.666 =.999 or rounded up 1amp (our total amperage available for the circuit.