| Trigonometry Demystified, 2nd edition |
| Stan Gibilisco |
| Explanations for Quiz Answers in Chapter 10 |
| 1. In Fig. 10-12, the current vector lies along the 0º axis, while the voltage vector appears to go out at an angle of about 45º. Therefore, the current lags the voltage in phase by approximately 45º. Current lags voltage in an AC circuit if and only if that circuit contains both resistance and inductive reactance. The correct choice is C. If choice A were correct, the current would lag the voltage by a full 90º. If choice B were right, the current would lead (rather than lag) the voltage by a full 90º. If choice D were right, the current would lead (rather than lag) the voltage by some angle larger than 0º but smaller than 90º. |
| 2. Based on the appearance of Fig. 10-12, the current appears to lag the voltage by about 45º, which is π/4 rad. The correct choice is A. |
| 3. In an AC circuit, the current and voltage follow each other along in phase coincidence if and only if no inductive or capacitive reactance exists. The circuit must comprise pure resistance. (We can consider perfect conductivity, such as we'd find in a short length of copper wire, as a pure resistance of 0 ohms.) The answer is A. |
| 4. Figure 10-13 shows an amplitude-versus-time graph of the voltage and current waves in an AC circuit where the current leads the voltage by 17.0º. (Remember, as we move toward the left in a graph of this sort, we go earlier in time; as we move toward the right, we go later in time.) Because the current leads the voltage by some angle greater than 0º but smaller than 90º, we know that the circuit contains some resistance and some capacitive reactance. The correct choice is D. |
| 5. In the situation of Fig. 10-13, we have a certain amount of capacitive reactance
(call it XC), whose absolute value we've been given as 100 ohms.
Therefore, we know that |XC| = 100 We have a certain resistance R, which we want to quantify. The graph explicitly tells us that the current leads the voltage by 17.0º. If we call the RC phase angle φRC then φRC = 17.0º When we input these two known values into the formula for phase angle in terms of capacitive reactance and resistance, we get 17.0º = Arctan (100 / R) We can take the tangent of both sides of the foregoing equation to get tan 17.0º = tan [Arctan (100 / R)] which, with the help of a scientific calculator set to work in degrees (not radians!), simplifies to 0.30573 = 100 / R and ultimately solves to R = 327 ohms rounded off to three significant figures, the maximum accuracy that we can "legally" claim. The correct choice is B. |
| 6. The period equals the reciprocal of the frequency. We've been told that a certain wave has a frequency of 500 kHz. When we convert that to hertz and write it in scientific notation, we get 5.00 x 105 Hz. A calculator gives us the reciprocal as 2.00 x 10-6 s. The correct choice is B. |
| 7. For the wave described above, 90º of phase corresponds to exactly 1/4 of
the period. When we calculate, we get a time span of
2.00 x 10-6 / 4.00 The answer is C. |
| 8. We've been told that for a certain AC wave, 0.001000 s represents π/4 rad of
phase, the equivalent of 1/8 of a cycle (because a full wave cycle contains 2π rad of
phase, and π/4 equals 1/8 of 2π). Therefore, a full cycle requires 8 x 0.001000 s, or
0.008000 s, to occur. That's the period T of the wave. When we take the reciprocal
of the period, we get a frequency of f = 1 / T We recall the fact that the angular frequency of a wave equals 2π times the conventional frequency. We can therefore state the angular frequency ω of this particular wave as ω = 2π f The correct choice is A. |
| 9. When we increase the frequency of the AC that we apply to a resistance-capacitance
(RC) circuit, we decrease the capacitive reactance XC
(negatively). In other words, we cause the capacitive reactance to get closer to zero
while remaining negative. Capacitive reactance varies inversely according to the
negative of the frequency. That action decreases the absolute value of XC
(positively). However, the resistance R remains constant; it doesn't depend on the
AC frequency. We remember that the formula for phase angle φRC in
terms of XC and R is φRC = Arctan (|XC| / R) In this situation, we start out with a phase angle of 30º, so 30º = Arctan (|XC| / R) As we raise the frequency, we decrease |XC| in this formula, which causes the ratio |XC| / R to grow smaller (positively). That change reduces the value of the Arctangent, which in turn reduces the phase angle. We can conclude that if we raise the AC frequency in this circuit, the current will lead the voltage by less than 30º. The correct choice is C. |
| 10. When we double the frequency of the AC that we apply to a resistance-inductance (RL)
circuit, we double the inductive reactance XL, because inductive
reactance varies directly with the frequency. But the resistance R remains
constant as the AC frequency increases. We recall that the formula for phase angle φRL
in terms of XL and R is φRL = Arctan (XL / R) In this situation, we start out with a phase angle of 45º, so 45º = Arctan (XL / R) Let's calculate the ratio XL / R. (We can figure out the ratio, even though we don't know the exact inductive reactance or the exact resistance). Taking the tangent of both sides in the above formula, we get tan 45º = tan [Arctan (XL / R)] A calculator tells us that tan 45º = 1.0, so we have XL / R = 1.0 If we double the frequency, thereby doubling XL, we get XL / R = 2.0 Going back to the formula for phase angle, we can calculate φRL = Arctan (XL / R) rounded off to two significant figures. The answer is B. |