Chapter 24 Solutions - Problems 1, 3, 4, 7, 11
  1. Figure P24.3 displays a voltage v(t) versus time t curve. Its formula is given by Eq. (24.1):
            v(t) = v0 sin (2πft)

    1. The maximum voltage can be read off from this curve:
      Vmax = v0 = 25 volts
    2. The rms voltage is Vmax = = 17.7 volts (see Sec 24.2)
    3. The period T = 1 / f can be read from the graph: T= 0.1 s

    4. The frequency f = 1 / T = 10 Hz.

  1. The current is described by i = 2 cos (40 t) = i0 cos (2πft)
    Comparing the last two expressions used immediately we see:

    1. peak current i0= 2 A

    2. effective current

    3. rms current = effective current = 1.41 A

    4. 40 = 2πf, thus f 6.37 Hz

  2. From the results of Problem 1, we can write:

                            v(t)= 25 sin (20 π t)
  1. The current i(t)=5 sin (20 t) flows through a resistor of R = 15 Ω. The power lost is given by:
                    
    or         

  1. This problem refers to the situation in Fig. 24.5. The capcitance C = 2µF, the resistance is R = 5 x 106Ω, and the DC voltage of the battery is V = 12 V.

    1. The initial current i(0) = V / R = 2.4 x 10-6 A

    2. The time constant τ = RC = 10 s

    3. The final charge q (t = ∞ ) = CV = 2.4 x 10-5C

    4. q(t = RC) = 0.63 CV = 1.51 x 10-5C

    5. i(t = RC) = 0.37 i(0) = 8.9 x 10-7A