ch03_ism - Simple Resistive Circuits 3 Assessment Problems...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
3 Simple Resistive Circuits Assessment Problems AP 3.1 Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 6 Ω resistor and the 10 Ω resistor in series: 6 Ω + 10 Ω = 16 Ω Now combine this 16 Ω resistor in parallel with the 64 Ω resistor: 16 Ω 64 Ω = (16)(64) 16 + 64 = 1024 80 = 12 . 8 Ω This equivalent 12 . 8 Ω resistor is in series with the 7 . 2 Ω resistor: 12 . 8 Ω + 7 . 2 Ω = 20 Ω Finally, this equivalent 20 Ω resistor is in parallel with the 30 Ω resistor: 20 Ω 30 Ω = (20)(30) 20 + 30 = 600 50 = 12 Ω Thus, the simplified circuit is as shown: 3–1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
3–2 CHAPTER 3. Simple Resistive Circuits [a] With the simplified circuit we can use Ohm’s law to find the voltage across both the current source and the 12 Ω equivalent resistor: v = (12 Ω)(5 A ) = 60 V [b] Now that we know the value of the voltage drop across the current source, we can use the formula p = vi to find the power associated with the source: p = (60 V )(5 A ) = 300 W Thus, the source delivers 300 W of power to the circuit. [c] We now can return to the original circuit, shown in the first figure. In this circuit, v = 60 V, as calculated in part (a). This is also the voltage drop across the 30 Ω resistor, so we can use Ohm’s law to calculate the current through this resistor: i A = 60 V 30 Ω = 2 A Now write a KCL equation at the upper left node to find the current i B : 5 A + i A + i B = 0 so i B = 5 A i A = 5 A 2 A = 3 A Next, write a KVL equation around the outer loop of the circuit, using Ohm’s law to express the voltage drop across the resistors in terms of the current through the resistors: v + 7 . 2 i B + 6 i C + 10 i C = 0 So 16 i C = v 7 . 2 i B = 60 V (7 . 2)(3) = 38 . 4 V Thus i C = 38 . 4 16 = 2 . 4 A Now that we have the current through the 10 Ω resistor we can use the formula p = Ri 2 to find the power: p 10 Ω = (10)(2 . 4) 2 = 57 . 6 W AP 3.2 [a] We can use voltage division to calculate the voltage v o across the 75 k resistor: v o (no load) = 75 , 000 75 , 000 + 25 , 000 (200 V ) = 150 V
Image of page 2
Problems 3–3 [b] When we have a load resistance of 150 k then the voltage v o is across the parallel combination of the 75 k resistor and the 150 k resistor. First, calculate the equivalent resistance of the parallel combination: 75 k Ω 150 k Ω = (75 , 000)(150 , 000) 75 , 000 + 150 , 000 = 50 , 000 Ω = 50 k Now use voltage division to find v o across this equivalent resistance: v o = 50 , 000 50 , 000 + 25 , 000 (200 V ) = 133 . 3 V [c] If the load terminals are short-circuited, the 75 k resistor is effectively removed from the circuit, leaving only the voltage source and the 25 k resistor. We can calculate the current in the resistor using Ohm’s law: i = 200 V 25 k = 8 mA Now we can use the formula p = Ri 2 to find the power dissipated in the 25 k resistor: p 25 k = (25 , 000)(0 . 008) 2 = 1 . 6 W [d] The power dissipated in the 75 k resistor will be maximum at no load since v o is maximum. In part (a) we determined that the no-load voltage is 150 V, so be can use the formula p = v 2 /R to calculate the power: p 75 k (max) = (150) 2 75 , 000 = 0 . 3 W AP 3.3 [a]
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern