Question 1

The current i through a resistor of pure resistance 2 k&#937; varies with time t according to the equation i = 5 sin (2&#960; x 100t + 45&#186;) mA. The voltage across the resistor will thus be:&#10;A) v = 5 sin (2 &#960; % 100t + 45&#186;) V&#10;B) v = 5 sin (2 &#960; % 100t - 45&#186;) V&#10;C) v = 10 sin (2 &#960; % 100t + 45&#186;) V&#10;D) v = 10 sin (2 &#960; % 100t + 135&#186;) V

Accepted Answer

The voltage across a resistor is given by Ohm's Law as V = IR, where R is the resistance and I is the current flowing through the resistor. In this case, I = 5 sin (2π x 100t + 45º) mA and R = 2 kΩ = 2000 Ω. Therefore,

V = IR = (5 sin (2π x 100t + 45º) mA) x (2000 Ω) 
= 10 sin (2π x 100t + 45º) V

So the correct answer is option C.

Question 2

The voltage v across a 1 &#956;F pure capacitor varies with time t according to the equation v = 16 sin 2000t V. The current through the capacitor will thus be:&#10;A) i = 16 sin 2000t mA&#10;B) i = 16 sin (2000t + 90&#186;) mA&#10;C) i = 32 sin 2000t mA&#10;D) i = 32 sin (2000t + 90&#186;) mA

Accepted Answer

The current through a capacitor is given by $i = C \frac{dv}{dt}$, where $C$ is the capacitance and $v$ is the voltage across the capacitor. Differentiating $v = 16 \sin 2000t$ with respect to $t$ gives $i = 32000C \cos 2000t$. Since $C = 1 \mu F = 1 \times 10^{-6} F$, $i = 32 \cos 2000t$ mA. Since $\cos \theta = \sin(\theta + 90^\circ)$, the current can be written as $i = 32 \sin(2000t + 90^\circ)$ mA, which corresponds to option D.

Question 3

The voltage v across a 4 H pure inductor varies with time t according to the equation v = 18 sin (2π x
1000t - 30º) V. The current through the capacitor will thus be:

A) i = 18 sin (2π x 1000t - 30º) mA
B) i = 18 sin (2π x 1000t - 120º) mA
C) i = 0.72 sin (2π x 1000t - 30º) mA
D) i = 0.72 sin (2π x 1000t - 120º) mA

D

Accepted Answer

The current through an inductor is given by $i = \frac{V}{j\omega L}$, where $V$ is the voltage, $\omega$ is the angular frequency, and $L$ is the inductance. Here, $\omega = 2\pi \times 1000$, and $L = 4$ H. The phase shift for an inductor is -90º (current lags voltage by 90º in an inductor), so the phase in the given equation should be shifted by -90º, changing -30º to -120º. The magnitude of the current is calculated by dividing the voltage amplitude by the reactance, $X_L = \omega L = 2\pi \times 1000 \times 4$, resulting in a current amplitude of $0.72$ mA.

Question 4

Decide whether each of these statements is True (T) or False
An alternating current i varies with time t and its variation is described by the equation i = 2 sin 500t.
(i) The average current over a cycle is 2 A.
(ii) The root-mean-square current is 1.41 A.

A) (i) T (ii) T
B) (i) T (ii) F
C) (i) F (ii) T
D) (i) F (ii) F

Accepted Answer

The reactance of an inductor (XL = 2πfL) doubles with the doubling of frequency, making statement (i) true. However, the reactance of a capacitor (XC = 1/(2πfC)) is halved when the frequency is doubled, making statement (ii) false.

Question 5

A circuit has a 300 &#937; resistor in series with a 200 mH inductor. When a voltage v = 17 sin 2000t V is&#10;Applied to it, the impedance is:&#10;A) 400 &#937; at 53.1&#186; lagging the current&#10;B) 400 &#937; at 90&#186; lagging the current&#10;C) 500 &#937; at 53.1&#186; lagging the current&#10;D) 500 &#937; at 90&#186; lagging the current

Accepted Answer

The impedance (Z) of a circuit with a resistor (R) and an inductor (L) in series can be found using the following formula:

Z = sqrt(R^2 + (ωL)^2)

where ω is the angular frequency, given by 2πf, where f is the frequency in Hertz.

In this problem, R = 300 Ω, L = 200 mH = 0.2 H, and v(t) = 17 sin 2000t V. The angular frequency is therefore ω = 2πf = 2π(2000) = 12,566 rad/s.

Plugging in these values, we get:

Z = sqrt((300)^2 + (12,566 x 0.2)^2) = sqrt(90,000 + 3,163,264) = sqrt(3,253,264) ≈ 1,804 Ω

However, we need to also include the phase angle, which is given by:

φ = arctan(ωL/R)

Plugging in the values we get:

φ = arctan((12,566 x 0.2)/300) ≈ arctan(0.8377) ≈ 41.2º

Therefore, the impedance of the circuit is approximately 1,804 Ω at an angle of 41.2º lagging the current. However, this is not one of the answer choices.

To get the closest answer choice, we can round the impedance to the nearest 100 Ω and round the phase angle to the nearest 10º. This gives:

Z ≈ 1,800 Ω and φ ≈ 40º

Using the rules of significant figures, we can also round the angle to one significant figure, giving:

Z ≈ 1,800 Ω and φ ≈ 40º

Now we can look at the answer choices and see which one is closest:

A) 400 Ω at 53.1º lagging the current - This is too low for Z and too high for φ.
B) 400 Ω at 90º lagging the current - This is too low for Z and too high for φ.
C) 500 Ω at 53.1º lagging the current - This is the closest to our calculated values.
D) 500 Ω at 90º lagging the current - This is too high for Z and too high for φ.

Therefore, the best choice is C) 500 Ω at 53.1º lagging the current, which is the closest to our calculated values of Z ≈ 1,800 Ω and φ ≈ 40º.

Question 6

Decide whether each of these statements is True (T) or False
An alternating current i varies with time t and its variation is described by the equation i = sin 3141t.
(i) The frequency of the alternating current is 500 Hz.
(ii) The periodic time of the alternating current is 2 ms.

A) (i) T (ii) T
B) (i) T (ii) F
C) (i) F (ii) T
D) (i) F (ii) F

Accepted Answer

(i) At resonance, the reactance of the inductor and the capacitor cancel out, resulting in equal and oppositely directed voltage phasors for both components.
(ii) At resonance, the circuit impedance is purely resistive, and therefore equal to the value of the resistance.

Question 7

Decide whether each of these statements is True (T) or False
An alternating current i varies with time t and its variation is described by the equation i = sin 3141t.
(i) The frequency of the alternating current is 500 Hz.
(ii) The periodic time of the alternating current is 2 ms.

A) (i) T (ii) T
B) (i) T (ii) F
C) (i) F (ii) T
D) (i) F (ii) F

Accepted Answer

The angular frequency ω is 3141 rad/s. Frequency f = ω/(2π) = 500 Hz, so (i) is True. The period T = 1/f = 2 ms, so (ii) is True.

Question 8

Decide whether each of these statements is True (T) or False
An alternating current i varies with time t and its variation is described by the equation i = sin 3141t.
(i) The frequency of the alternating current is 500 Hz.
(ii) The periodic time of the alternating current is 2 ms.

A) (i) T (ii) T
B) (i) T (ii) F
C) (i) F (ii) T
D) (i) F (ii) F

Accepted Answer

(i) The average current over a cycle of an alternating current is 0 A, not 2 A. (ii) The root-mean-square (RMS) current of an AC current i = I_max sin(ωt) is I_max/√2. Given I_max = 2 A, the RMS current is 2/√2 = 1.41 A.

Deck 7: Alternating Current, and Series A.C Circuits