Alternating Current Test

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Alternating Current Test - 69

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Question 1
1 / -0

What is the phase angle in a series $$RLC$$ circuit at resonance?

A
$$180^{o}$$

B
$$90^{o}$$

C
$$0$$

D
$$-90^{o}$$

E
None of those answers is necessarily correct.

Solution

At resonance the inductive reactance and capacitive reactance add to zero: $$\phi=\tan^{-1}[(X_L-X_c)/R]=0$$.
Question 2
1 / -0

What is the rms voltage?

A
$$\sqrt{2}\Delta V_{max}$$

B
$$\Delta V_{max}$$

C
$$\Delta V_{max}/\sqrt{2}$$

D
$$\Delta V_{max}/2$$

E
$$\Delta V_{max}/4$$

Solution

The average of the squared voltage is $$([\Delta V]^2)_{avg}=\dfrac{(\Delta V_{max})^2}{2}$$. Then its square root is $$\Delta V_{max}=\dfrac{\Delta V_{max}}{\sqrt{2}}$$
Question 3
1 / -0

A 100 volt A.C. source of frequency 500 hertz is connected to a L-C-R circuit with L $$=$$ 8.1 millihenry, C $$=$$ 12.5 microfarad and R $$=$$ 10 ohm, all connected in series. The potential difference across the resistance will be:

A
10V

B
100V

C
50V

D
500V

Solution

Impedance of series RLC circuit is given by:

$$Z = \sqrt {R^2 + \left( 2 \pi f L - \dfrac{1}{2 \pi f C} \right)^2}$$

$$Z = \sqrt {10^2 + \left(2\pi \times 500 \times 8.1 \times10^{-3} - \dfrac{1}{2 \pi \times 500 \times 12.5 \times 10^{-6} } \right)^2}$$

$$Z \approx 10\ \Omega$$

Current flowing through the circuit is:

$$I = \dfrac{V}{Z} = 10A$$

Potential difference across the resistor, by ohm's Law is:

$$V_R = IR = 10 \times 10 = 100\ V$$
Question 4
1 / -0

An AC source of variable frequency is applied across a series L-C-R circuit. At a frequency double the resonance frequency, the impedance is $$\sqrt{10}$$ times the minimum impedance. The inductive reactance is

A
$$R$$

B
$$2R$$

C
$$3R$$

D
$$4R$$

Solution

Let $$\omega_0$$ be the resonance frequency
minimum impedance is at $$\omega_0$$

$$Z_{min}=X_R=R$$

At $$\omega = 2\omega_0$$,

$$Z=\sqrt{10}Z_{min}$$

$$\sqrt{{R}^2+(\omega L-\dfrac{1}{\omega C})^2}=\sqrt{10}\times R$$

$${R}^2+(2\omega_0 L-\dfrac{1}{2\omega_0 \times C})^2=10\times R^2$$

because at $$\omega_0 $$ , $$\omega_0 \times L=\dfrac{1}{\omega_0 C}$$ ,

$${R}^2+(2\omega_0 L-\dfrac{\omega_0 L}{2})^2=10\times R^2$$

$$(\dfrac{3}{2}\omega_ 0 L)^2=9\times R^2$$

$$\dfrac{3}{2}\times \omega_ 0 L=3\times R $$

$$X_L = \omega L = 2\omega_0 L=4 R$$ (at $$\omega=2\omega _0$$)
Question 5
1 / -0

At a frequency $$\omega_{0} $$ the reactance of a certain capacitor equals that of a certain inductor. If frequency is changed to 2 $$\omega_{0} $$ . The ratio of reactance of the inductor to that of the capacitor is

A
4 : 1

B
$$\sqrt{2}:1$$

C
$$1:2\sqrt{2}$$

D
1 : 2

Solution

at $$w_o, \dfrac{w_oL}{\dfrac{1}{w_oC}}=1$$

i.e. $$w_o^2LC=1$$

now at $$2w_o, \dfrac{2w_oL}{\dfrac{1}{2w_oC}}=\dfrac{X_L^1}{X_C^1}$$

$$4w_o^2LC=\dfrac{X_L^1}{X_C^1}$$

$$\dfrac{4}{1}=\dfrac{X_L^1}{X_C^1}$$
Question 6
1 / -0

An ideal inductor takes a current of $$10 A$$ when connected to a $$125$$ $$V$$, $$50$$ $$Hz$$ AC supply. A pure resistor across the same source takes $$12.5$$ $$A$$. If the two are connected in series across a $$100\sqrt{2}V$$, $$ 40$$ $$Hz$$ supply, the current through the circuit will be

A
$$10 A$$

B
$$12.5 A$$

C
$$20 A$$

D
$$25 A$$

Solution

Inductor
$$X_L=\dfrac{125}{10}=12.5\Omega$$

Resistor
$$X_R=\dfrac{125}{12.5}=10\Omega$$

$${ X }_{ L }'=12.5(\dfrac { { \omega }_{ 2 } }{ { \omega }_{ 1 } } )\Omega =12.5(\dfrac { 40 }{ 50 } )\Omega =10\quad \Omega$$

both are connected in series

$$z=\sqrt{10^2+(10)^2}$$

$$=10\sqrt { 2 } \Omega $$

$$Vrms=100\sqrt{2} \ V$$

$$Irms=\dfrac { 100\sqrt { 2 } }{ 10\sqrt { 2 } } =10A$$
Question 7
1 / -0

When a series combination of $$L$$ and $$R$$ are connected with a $$10V$$, $$50HZ$$ A.C. source, a current 1A flows in the circuit. The voltage leads the current by a phase angle of $$\pi/3$$ radian. Then the resistance is

A
5$$\Omega $$

B
10$$\Omega $$

C
15$$\Omega $$

D
20$$\Omega $$

Solution

$$Vrms=10V, Irms=1A$$

i.e. $$tan\dfrac{\pi}{3}=\dfrac{x_L}{x_R}$$

$$\sqrt{3}=\dfrac{x_L}{R}$$

also $$\dfrac{10}{z}=1$$

$$10^2=R^2\times X_L^2$$

$$10^2=R^2+3^2$$

$$100=4R^2$$

$$R=5\Omega$$
Question 8
1 / -0

In an A.C circuit, the resistance $$ R = 0.2\, \Omega$$. At a certain instant, $$V_A - V_B = 0.5\, V$$, $$ I = 0.5\, A$$, and the current is increasing at the rate of $$\dfrac{\Delta I}{\Delta t} = 8\,A/s$$. The inductance of the coil is:

A
$$0.05\,H$$

B
$$0.1\,H$$

C
$$0.2\,H$$

D
none of these

Solution

Given $$I=0.5\ A$$

From Ohm's Law,
$$V_R=(0.5)\times (.2)$$
$$=0.1\ V$$

From Kirchoff's Voltage Law,
$$V_L=V_a-V_b-V_R = 0.5-0.1$$
$$=0.4\ V$$

For an inductor,
$$V_L=L\dfrac{di}{dt}$$
$$0.4=L\times8$$
$$L=0.05H$$
Question 9
1 / -0

An AC source of angular frequency $$\omega $$ is fed across a resistor R and a capacitor C in series. The current registered is I. If now the frequency of source is changed to $$\omega /3$$ (but maintaining the same voltage), the current in the circuit is found to be halved. The ratio of reactance to resistance at the original frequency $$\omega $$ is

A
$$\sqrt{\dfrac{3}{5}}$$

B
$$\sqrt{\dfrac{5}{3}}$$

C
$$\dfrac{3}{5}$$

D
$$\dfrac{5}{3}$$

Solution

$$I=\dfrac{V}{\sqrt{R^2+(\dfrac{1}{wc})^2}}$$, for w

for $$\dfrac{w}{3}, \dfrac{I}{2}=\dfrac{V}{\sqrt{R^2+(\dfrac{3}{wc})^2}}$$$$\dfrac{2}{\sqrt{R^2+\dfrac{9}{w^2C^2}}}=\dfrac{1}{\sqrt{R^2+\frac{1}{w^2C^2}}}$$$$4R^2+\dfrac{4}{w^2c^2}=R^2+\dfrac{9}{w^2c^2}$$

$$3R^2=\dfrac{5}{w^2c^2}$$

$$3R^2=5(x_c)^2$$ [at w]

$$\sqrt{\dfrac{3}{5}}=\dfrac{x_c}{R}$$
Question 10
1 / -0

In the given circuit, $$R$$ is a pure resistor, $$L$$ is a pure inductor, S is a 100V, 50Hz AC source and A is an AC ammeter. With either $$K_1$$ or $$K_2$$ alone closed, the ammeter reading is $$I$$. If the source is changed to 100 V, 100 Hz, the ammeter reading with $$K_1$$ alone closed and with $$K_2$$ alone closed will be respectively

A
$$I , \dfrac{I}{ 2}$$

B
$$I , 2I$$

C
$$2I , I$$

D
$$2I , \dfrac{I }{ 2}$$

Solution

$$I=\dfrac{100}{R}=\dfrac{100}{2\pi(50)(L)}$$

now at $$100 Hz$$

new I when $$K_1$$ is close only

$$I_1=\dfrac{100}{R}=I$$

and new I when only $$K_2$$ is closed

$$I_2=\dfrac{100}{2\pi\times (100)\times L}=\dfrac{I}{2}$$

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 69

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Alternating Current Test - 69

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SHARING IS CARING

Question 1

$$180^{o}$$

$$90^{o}$$

$$0$$

$$-90^{o}$$

None of those answers is necessarily correct.

Solution

Question 2

$$\sqrt{2}\Delta V_{max}$$

$$\Delta V_{max}$$

$$\Delta V_{max}/\sqrt{2}$$

$$\Delta V_{max}/2$$

$$\Delta V_{max}/4$$

Solution

Question 3

10V

100V

50V

500V

Solution

Question 4

$$R$$

$$2R$$

$$3R$$

$$4R$$

Solution

Question 5

4 : 1

$$\sqrt{2}:1$$

$$1:2\sqrt{2}$$

1 : 2

Solution

Question 6

$$10 A$$

$$12.5 A$$

$$20 A$$

$$25 A$$

Solution

Question 7

5$$\Omega $$

10$$\Omega $$

15$$\Omega $$

20$$\Omega $$

Solution

Question 8

$$0.05\,H$$

$$0.1\,H$$

$$0.2\,H$$

none of these

Solution

Question 9

$$\sqrt{\dfrac{3}{5}}$$

$$\sqrt{\dfrac{5}{3}}$$

$$\dfrac{3}{5}$$

$$\dfrac{5}{3}$$

Solution

Question 10

$$I , \dfrac{I}{ 2}$$

$$I , 2I$$

$$2I , I$$

$$2I , \dfrac{I }{ 2}$$

Solution

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