ch23 - CHAPTER 23 ALTERNATING CURRENT CIRCUITS ANSWERS TO...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 23 ALTERNATING CURRENT CIRCUITS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (d) According to 2 rms / P V R = (Equation 20.15c), the average power is proportional to the square of the rms voltage. Tripling the voltage causes the power to increase by a factor of 3 2 = 9. 2. I rms = 1.9 A 3. (b) The current I rms through a capacitor depends inversely on the capacitive reactance X C , as expressed by the relation I rms = V rms / X C (Equation 23.1). The capacitive reactance becomes infinitely large as the frequency goes to zero (see Equation 23.2), so the current goes to zero. 4. (e) According to ( 29 C 1/ 2 X f C π = (Equation 23.2) and L 2 X f L π = (Equation 23.4), doubling the frequency f causes X C to decrease by a factor of 2 and X L to increase by a factor of 2. 5. I rms = 1.3 A 6. (a) The component of the phasor along the vertical axis is V 0 sin 2 π f t (see the drawing that accompanies this problem), which is the instantaneous value of the voltage. 7. (b) The instantaneous value of the voltage is the component of the phasor that lies along the vertical axis (see Sections 23.1 and 23.2). This vertical component is greatest in B and least in A, so the ranking is (largest to smallest) B, C, A. 8. (d) In a resistor the voltage and current are in phase. This means that the two phasors are colinear. 9. (c) Power is dissipated by the resistor, as discussed in Section 20.5. On the other hand, the average power dissipated by a capacitor is zero (see Section 23.1). 10. I rms = 2.00 A
Image of page 1

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

View Full Document Right Arrow Icon
217 ALTERNATING CURRENT CIRCUITS 11. (a) When the rms voltage across the inductor is greater than that across the capacitor, the voltage across the RCL combination leads the current (see Section 23.3). 12. (d) Since I rms = V rms / Z (Equation 23.6), the current is a maximum when the impedance Z is a minimum. The impedance is ( 29 2 2 L C Z R X X = + - (Equation 23.7), and it has a minimum value when X C = X L = 50 . 13. (c) The inductor has a very small reactance at low frequencies and behaves as if it were replaced by a wire with no resistance. Therefore, the circuit behaves as two resistors, R 1 and R 2 , connected in parallel. The inductor has a very large reactance at high frequencies and behaves as if it were cut out of the circuit, leaving a gap in the connecting wires. The circuit behaves as a single resistance R 2 connected across the generator. The situation at low frequency gives rise to the largest possible current, because the effective resistance of the parallel combination is smaller than the resistance R 2 . 14. (a) The capacitor has a very small reactance at high frequencies and behaves as if it were replaced by a wire with no resistance. Therefore, the circuit behaves as two resistors, R 1 and R 2 , connected in parallel. The capacitor has a very large reactance at low frequencies and behaves as if it were cut out of the circuit, leaving a gap in the connecting wires. Therefore, the circuit behaves as a single resistor R 1 connected across the generator. The situation at high frequencies gives rise to the largest possible current, because the effective resistance of the parallel combination is smaller than the resistance R 1 .
Image of page 2
Image of page 3
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