Resistance and Ohms Law More Practice Resistance and

Resistance and Ohm’s Law: More Practice

Resistance and Ohm’s Law: More Practice

Resistance and Ohm’s Law: More Practice

Resistance and Ohm’s Law: More Practice

Kirchoff’s Laws: Success Criteria I can state Kirchhoff's laws and Ohm's law and use them to explain, in quantitative terms, direct current, potential difference, and resistance in mixed circuit diagrams. n I can construct real and simulated mixed direct current circuits and analyse them in quantitative terms to test Ohm's and Kirchhoff's laws. n

Kirchoff’s Laws and Equivalent Resistance SPH 4 C

Kirchoff’s Current Law n At any junction point in an electrical circuit, the total current into the junction equals the total current out of the junction.

Kirchoff’s Current Law At any junction point in an electrical circuit, the total current into the junction equals the total current out of the junction. (“What goes in must come out. ”) n In the diagram at right, I 1 + I 2 = I 3

Kirchoff’s Voltage Law n In any complete path in an electrical circuit, the sum of the potential increases equals the sum of the potential drops.

Kirchoff’s Voltage Law In any complete path in an electrical circuit, the sum of the voltage increases equals the sum of the voltage drops. (“What goes up must come down. ”) n

The Laws for a Series Circuit The current is the same at all points in the circuit: IT = I 1 = I 2 =. . . The total voltage supplied to the circuit is equal to the sum of the voltage drops across the individual loads: VT = V 1 + V 2 +. . .

Equivalent Resistance in Series Given VT = V 1 + V 2 +. . .

Equivalent Resistance in Series Given VT = V 1 + V 2 +. . . From Ohm’s Law, V = IR I T R T = I 1 R 1 + I 2 R 2 +. . .

Equivalent Resistance in Series Given VT = V 1 + V 2 +. . . From Ohm’s Law, V = IR I T R T = I 1 R 1 + I 2 R 2 +. . . Since IT = I 1 = I 2 =. . . IRT = IR 1 + IR 2 +. . .

Equivalent Resistance in Series Given VT = V 1 + V 2 +. . . From Ohm’s Law, V = IR I T R T = I 1 R 1 + I 2 R 2 +. . . Since IT = I 1 = I 2 =. . . IRT = IR 1 + IR 2 +. . . Divide all terms by I and the equivalent resistance is the sum of the individual resistances: Req or RT = R 1 + R 2 +. . .

Series Resistance Example

Series Resistance Example Req = R 1 + R 2 + R 3

Series Resistance Example Req = R 1 + R 2 + R 3 Req = 17 W + 12 W + 11 W = 40 W

Series Resistance Example IT = VT/Req

Series Resistance Example IT = VT/Req IT = 60 V/40 W = 1. 5 A

Series Resistance Example IT = VT/Req IT = 60 V/40 W = 1. 5 A And I 1 = I 2 = I 3 = IT = 1. 5 A

Series Resistance Example V 1 = I 1 R 1 = (1. 5 A)(17 W) = 25. 5 V V 2 = I 2 R 2 = (1. 5 A)(12 W) = 18 V V 3 = I 3 R 3 = (1. 5 A)(11 W) = 16. 5 V

The Laws for a Parallel Circuit At a junction: IT = I 1 + I 2 +. . . But the total voltage across each of the branches is the same: VT = V 1 = V 2 =. . .

Equivalent Resistance in Parallel Given IT = I 1 + I 2 +. . . From Ohm’s Law, V = IR or I = V/R VT/RT = V 1/R 1 + V 2/R 2 +. . . Since VT = V 1 = V 2 =. . . V/RT = V/R 1 + V/R 2 +. . . Divide all terms by V and the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances: 1/Req or 1/RT = 1/R 1 + 1/R 2 +. . .

Equivalent Resistance in Parallel 1/Req = 1/R 1 + 1/R 2 + 1/R 3 1/Req = 1/(12 W) + 1/(12 W) Using a calculator. . . 1/Req = 0. 25 W-1 Req = 1 / (0. 25 W-1) Req = 4. 0 W

Equivalent Resistance in Parallel 1/Req = 1/R 1 + 1/R 2 + 1/R 3

Equivalent Resistance in Parallel 1/Req = 1/R 1 + 1/R 2 + 1/R 3 1/Req = 1/(5. 0 W) + 1/(7. 0 W) + 1/(12 W) 1/Req = 0. 42619 W-1 Req = 1 / (0. 42619 W-1) Req = 2. 3 W

Combination Circuits What do we do if a circuit has both series and parallel loads? Find the equivalent resistance of the loads in parallel and continue the analysis. E. g. :

More Practice n Equivalent Resistance Lab Activity