Chapter 13 - Resistors

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Chapter 13 - Resistors 13.1 - Resistors in Series Series - The entire current flows out of the power source with a certain potential, and through the first resistor. When it comes out of the first resistor, it has lost part of its potential energy, but not all of it. The current then passes though the second resistor, losing more of its potential, then through the third. The electrons come out of the third resistor in the ground state and flow back into the power supply to be re-energized. Example 1 - Two resistors of 10each are placed in series with a 6V battery. Determine the potential at a point between the resistors as well as the current flowing through them. Example 2 - Using Ohm’s Law, what is the equivalent resistance of the total circuit?

Transcript of Chapter 13 - Resistors

Page 1: Chapter 13 - Resistors

Chapter 13 - Resistors

13.1 - Resistors in Series

Series -

The entire current flows out of the power source with a certain potential, and throughthe first resistor. When it comes out of the first resistor, it has lost part of its potentialenergy, but not all of it. The current then passes though the second resistor, losing moreof its potential, then through the third. The electrons come out of the third resistor in theground state and flow back into the power supply to be re-energized.

Example 1 - Two resistors of 10⌦ each are placed in series with a 6V battery. Determinethe potential at a point between the resistors as well as the current flowing through them.

Example 2 - Using Ohm’s Law, what is the equivalent resistance of the total circuit?

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Example 3 - Repeat the previous example for a circuit with three 10⌦ resistors and a 6Vbattery.

When placed in series, the equivalent resistance of a circuit, Requivalent is simply,

(1)

the sum of the resistances.

Example 4 - Three resistors, 22⌦, 33⌦, and 47⌦ are placed in series with a 12V battery.What is the equivalent resistance of the circuit?

Example 5 - What is the potential at points A and B in the previous circuit? What is thepotential di↵erence across each resistor?

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R(equivalent) = R1 + R2 + … + Rn
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22 + 33 + 47 = 102Ohm’s
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R22 -12 - Va = I(22)
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R33 - Va - Vb = I(33)
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R47 -Vb - 0 = I(47)
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Total -V = IR 12 - 0 = I(102)I = 0.118A
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12 - Va = I(22)12 - Va = (0.118)(22)Va = 9.41V
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Vb - 0 = (0.118)(47)Vb = 5.53V
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Example 6 - A certain piece of lab equipment requires 6A and 12V to operate. Whenplaced on a 14V circuit, the current going through it is 7A. You wish to know when themachine is on, so you place a small 1⌦ lightbulb in series between the 14V power supplyand the lab equipment. Although the light is on, the equipment does not operate. Why?

13.2 - Resistors in Parallel

Parallel -

Example 7 - Two 10⌦ resistors are placed parallel to a 6V battery. Find the current flowingthrough the circuit.

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Equipment - V = IR 14 = 7(R)R = 2Ω
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Lamp -14 - Va = I(1) 28 - 2Va = I(2) = Va - 028 - 2Va = Va28 = 3VaVa = 9.3VNot enough potential.
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Lab Equipment - Va - 0 = I(2)
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When resistors are placed parallel to one another, their ends are connected so current is divided among the different pathways.
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The potential differences across each resistor placed parallel to one another are all the same.
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Rr -V = IR 6 - 0 = I(10)I = 0.6A
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R(L) - V = IR6 - 0 = I(10)I = 0.6A
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Total Current = 1.2 A = I
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Example 8 - What is the equivalent resistance of the circuit?

Example 9 - Three 10⌦ resistors are placed parallel to a 6V battery. Find the current flowingthrough the circuit.

Example 10 - What is the equivalent resistance of the circuit?

By placing resistors in parallel, electrons are given alternate pathways and therefor feel lessresistance.

When resistors are placed in parallel, the equivalent resistance is the inverse sum of theindividual resistances.

(2)

Example 11 - Determine the equivalent resistance through the following circuit, as well asthe current through the battery.

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V = IR 6 = 1.2(R) 6/1.2 = 5 Ω
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I = 3(0.6)I = 1.8 A
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V = IR6 = 1.8RR = 3.33 Ω
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1/Requivalent = 1/R(1) + 1/R(2) … + 1/Rn
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1/Re = 1/33 + 1/471/Re = 47/1551 + 33/15511/Re = 80/1551Re = 1551/80 = 19.4Ω
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V = IR 12 = I(19.4)I = 0.619 A
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Example 12 - Determine the unknown resistance that will give the circuit a total resistanceof 5K.

Example 13 - Determine the unknown resistance that will cause the circuit to have a totalcurrent of 0.5mA

13.3 Resistors in Complex Circuits

To simplify a circuit, find resistors between two nodes that are either in series or paral-lel, and replace them with a single resistor. Redraw the circuit and repeat until all resistorshave been combined into one. This will help you determine the current coming out of thepower source and flowing through the entire circuit.

Example 14 - Determine the equivalent resistance, the potential at points A and B, andthe potential di↵erence across each resistor.

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1/Re = 1/R1 + 1/R2 + 1/R31/5 = 1/ R1 + 1/33 + 1/221/5 = 726/726R + 22R/726R = 33R/726R1/5 = (726 + 22R + 33R)/726R726R = 5(726+55R)726R = 3630 = 275R451R = 3630 R = 8.05 Ω
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V = IR3 = (0.0005)RR = 6000 = 6KΩ1/6 = 1/22 + 1/33 + 1/R1/6 - 1/22 - 1/33 = 1/R726/4356 - 198/4356 - 132/4356 = 1/R396/4356 = 1/RR = 11K
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1/22 + 1/56 = 1/R22+56/(22)(56) = 1/RR = 15.69 Ω
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Re = R1+R2+R3Re = 33 + 15.8 + 10Re = 58.8 Ω
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33 -V = IR12 - Va = I(33)
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22, 56 -V=IRVa - Vb = I(15.8)
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10 -Vb - 0 = 10I
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Total -V=IR12 = I(58.8)I = 0.204A
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12 - Va = 0.204(33)
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Vb = 2.04V
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Va = 5.26 V
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V(33) = 12 - 5.26V(33) = 6.74
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V22 = V56 = 5.26 - 2.04
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V22 = V56 = 3.22V
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V10 = 2.04V
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Example 15 - Determine the equivalent resistance, the potential at points A and B, and thepotential di↵erence across each resistor.

Example 16 - Determine the equivalent resistance, the potential at points A and B, and thepotential di↵erence across each resistor.

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Va = 18VVb = 0VV22 = 18VV33 = 18VV56 = 18V
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1/22 + 1/33 + 1/56 = 1/R1848/40656 + 1232/40656 + 726/60656 = 1/R3806/40656 = 1/RR = 40656/3784 = 10.74R = 10.7Ω
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1/Re = 1/33 + 1/44 + 1/56Re = 14.1 Ω
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Va = 9VV33 = 9VVb = 0VV56 = 9V
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C
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22T =V=IR9-Vc = I(22)
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22B -V = IRVc - 0 = I(22)
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9-Vc = Vc - 09 = 2VcVc = 4.5V
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V22t = 9-4.5VV22t = 4.5V
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V22b = 4.5V
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Example 17 - Determine the equivalent resistance, the potential at points A, B, and C, andthe potential di↵erence across each resistor.

Example 18 - Determine the equivalent resistance, the potential at points A and B, and thepotential di↵erence across each resistor.

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1/Re = 1/33 + 1/22 1/Re = 1/10 + 1/56Re = 13.2Ω Re = 8.48Ω
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The goal is to make everything one resistor and find what that one is.
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Re = 21.68 Ω
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AB -BC -C⏚ -V=IRV= IRV=IRVa - Vb = IRVb - 0 = I(8.48)6 - Vb = I(13.2)6 - [(I)*(8.48)] = I(13.2)I = 0.277AVb = (0.277)*(8.48)Vb = 2.35Va = 6VVc = 0V
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V33 = V22 = 6 - 2.35V33 = V22 = 3.65VV10 = V56 = 2.35V
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1/Re = 1/33 + 1/88Re = 24 ΩVa = 12 VVb = 0V
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V33 = 12V
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12 - Vc = 22I
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Vc -VD = 56I12 - 9 = 3VV22 = 3V
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Vd - 0 = 10IVd = 10(0.136)Vd = 1.36V
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12 - 0 = 88II = 0.136A
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9 - 1.36 = V56V56 = 7.64V
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Example 19 - Determine the equivalent resistance, the potential at points A and B, and thepotential di↵erence across each resistor.

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Right -18 - Vc = I(22)Vc - Vo = I(33)Vd - 0 = I(33)18 - Vc = 0.277(22)Vc = 11.9V11.9-9.141 = I(10)Vd = 0.277(33)I = 0.277AVd = 9.1411/Re = 1/65 + 1/65 = ?/6518-0 = 65IRe = 32.5Ω18 - 11.9 = V22V10=2.8VV33 = 9.1VV22 = 6.1VVa = 18VVb = 0V
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Left Side -18-Ve = I(10)18-Ve = (0.277)(10)Ve = 15.2VV10 = 18 - 15.2V10 = 2.8V
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Ve - Vf = 33I
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15.2 - 6.1V33 = 9.1
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Vf - 0 = 22IVf = 22(0.277)Vf = 6.09VV22 = 6.1V
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Homework

13.1 Resistors in Series

1. Determine the equivalent resistance and current through the following circuit.

2. Determine the equivalent resistance as well as the potential across each resistor,and the current through the circuit.

3. Determine the equivalent resistance as well as the potential across each resistor,and the current through the circuit.

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4. Determine the resistance of R1 required to cause 0.67A

13.2 Resistors in Parallel

5. Determine the equivalent resistance, the current flowing through each resistor,and the current through the power supply.

6. Determine the equivalent resistance, the current flowing through each resistor,and the current through the power supply.

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0.067
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7. Determine the equivalent resistance, the current flowing through each resistor,and the current through the power supply.

8. What should the resistance be of R1 to make an equivalent resistance of 6.48K?What current flows through the circuit?

9. What should the resistance of R1 be to cause 0.287mA of current to flow throughthe circuit?

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13.3 Resistors in Complex Circuits

10. Determine the equivalent resistance, the potential at points A and B, and thepotential di↵erence across each resistor.

11. Determine the equivalent resistance, the potential at points A and B, and thepotential di↵erence across each resistor.

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12. Determine the equivalent resistance, the potential at points A, B, C, and D, andthe potential di↵erence across each resistor.

13. Determine the equivalent resistance, the potential at points A, B, C, and D, andthe potential di↵erence across each resistor.

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14. Determine the equivalent resistance, the potential at points A, B, and C, and thepotential di↵erence across each resistor.

15. Determine the equivalent resistance, the potential at points A, B, C, and D, andthe potential di↵erence across each resistor.

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16. Determine the equivalent resistance, the potential at points A and B, and thepotential di↵erence across each resistor.

17. Determine the equivalent resistance, the potential at points A and B, and thepotential di↵erence across each resistor.

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18. Determine the equivalent resistance, the potential at points A, B, and C, and thepotential di↵erence across each resistor.

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