## Example 2.1

How many cycles are there?## Solution

The answer is 2 cycles.The saw tooth wave at C goes negative first but that doesn't change the number of cycles or the definition of a cycle.

T = 1 / 60 = 0.0167 = 1.67 x 10^{-2} (1.67e-2)

Worked without using scientific notation.

f = 1 / .004 seconds

f = 250 cycles

I'm not going to humor you very long about learning scientific notation. Eventually you're going to have to go back to the end of the DC chapter and learn it. You can do it now or do it later; but eventually you will do it.

Worked using scientific notation.

f = 1 / 4 x 10^{-3} = 250 cycles = 2.50 x 10^{2} = (2.50e2)

Delta V = 150 - 120 = 30 volts.

Delta t = 16 milliseconds = 16 x 10

It's alright to break the rules once in a while. I won't tell if you won't.

C = 40 x 10

Delta V = 10 volts,

Delta t = 8 milliseconds = 8 x 10

I = 100 mA = 0.1 A.

C = 8 x 10^{-5}= (8e-5) farads.

C = 80 microfarads.

X

X

X_{C} = 1/(2 pi f C)

X_{C} = 1/(2 pi 60 x 40 x 10^{-6})

X_{C} = 6.63 x 10^{1} = 66.3 ohms.

Now we use the AC version of ohm's law to calculate the current.

I = V / X_{C}

I = 120 volts / 66.3 ohms = 1.81 amps.

X

X

(a)

X

X

Ar = R / Square root of (R^{2} + Xc^{2})

Ar = 1.59 x 10^{4} / Square root of ([1.59 x 10^{4}]^{2} + [1.59 x 10^{6}]^{2})

Ar = 1.59 x 10^{4} / Square root of (2.53 x 10^{8} + 2.53 x 10^{12})

Ar = 1.59 x 10^{4} / Square root of (0.000253 x 10^{12} + 2.53 x 10^{12})

When 0.000253 is added to 2.53 it gives 2.53253 and then when it is rounded off to 3 digits it gives 2.53. So we have,

Ar = 1.59 x 10^{4} / Square root of (2.53 x 10^{12})

Ar = 1.59 x 10^{4} / 1.59 x 10^{6}

Ar = 1.00 x 10^{-2}

(b)

X

Ar = R / Square root of (R^{2} + Xc^{2})

Ar = 1.59 x 10^{4} / Square root of ([1.59 x 10^{4}]^{2} + [1.59 x 10^{5}]^{2})

Ar = 1.59 x 10^{4} / Square root of (2.53 x 10^{8} + 2.53 x 10^{10})

Ar = 1.59 x 10^{4} / Square root of (0.0253 x 10^{10} + 2.53 x 10^{10})

When 0.0253 is added to 2.53 it gives 2.5553 and then when it is rounded off to 3 digits it gives 2.56. So we have,

Ar = 1.59 x 10^{4} / Square root of (2.56 x 10^{10})

Ar = 1.59 x 10^{4} / 1.60 x 10^{5}

Ar = 0.994 x 10^{-1} = 9.94 x 10^{-2} = (9.94e-2)

(c)

X

Ar = R / Square root of (R^{2} + Xc^{2})

Ar = 1.59 x 10^{4} / Square root of ([1.59 x 10^{4}]^{2} + [1.59 x 10^{4}]^{2})

Ar = 1.59 x 10^{4} / Square root of (2.53 x 10^{8} + 2.53 x 10^{8})

When 2.53 is added to 2.53 it gives 5.06. So we have,

Ar = 1.59 x 10^{4} / Square root of (5.06 x 10^{8})

Ar = 1.59 x 10^{4} / 2.25 x 10^{4}

Ar = 0.707 = 7.07 x 10^{-1}

X

Ar = R / Square root of (R^{2} + Xc^{2})

Ar = 1.59 x 10^{4} / Square root of ([1.59 x 10^{4}]^{2} + [1.59 x 10^{3}]^{2})

Ar = 1.59 x 10^{4} / Square root of (2.53 x 10^{8} + 2.53 x 10^{6})

Ar = 1.59 x 10^{4} / Square root of (2.53 x 10^{8} + 0.0253 x 10^{8})

When 2.53 is added to 0.0253 it gives 2.5553. When it is rounded off it gives 2.56. So we have,

Ar = 1.59 x 10^{4} / Square root of (2.56 x 10^{8})

Ar = 1.59 x 10^{4} / 1.60 x 10^{4}

Ar = 0.994 = 9.94 x 10^{-1}

(a)

X

X

Ar = X_{C} / Square root of (R^{2} + X_{C}^{2})

Ar = 1.59 x 10^{5} / Square root of ([1.59 x 10^{4}]^{2} + [1.59 x 10^{5}]^{2})

Ar = 1.59 x 10^{5} / Square root of (2.53 x 10^{8} + 2.53 x 10^{10})

Ar = 1.59 x 10^{5} / Square root of (0.0253 x 10^{10} + 2.53 x 10^{10})

Ar = 1.59 x 10^{5} / Square root of (2.56 x 10^{10})

Ar = 1.59 x 10^{5} / 1.60 x 10^{5}

Ar = 9.94 x 10^{-1} = (9.94e-1)

X

Ar = 0.707

(c)

X_{C} = 1.59 x 10^{3}

Ar = 9.94 x 10^{-2}

(d)

X_{C} = 1.59 x 10^{2}

Ar = 1.00 x 10^{-2} = 1.00e-2

X

f = 1000 cycles,

Delta f = 10 cycles.

Q = 100 / 1 = 100

Q = f / Delta f

Q = 1000 / 10 = 100

Q = 455 kc / 15 kc = 30.3

X

X

Q = X_{L} / R.

R = X_{L} / Q.

R = 1.43 x 10^{3} / 30.3 = 47.2 ohms.

L = 1 / ([2 pi f]

L = 1 / (2 pi 530 x 10

L = 2.47 x 10

(a)

(c) 6.02 dB

(d) 20 dB

(e) 30.0 dB

(f) 40 dB

(g) -6.02 dB

(h) -20 dB

(i) -40 dB