Alternating Current Test

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Alternating Current Test - 47

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Question 1
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An inductor (L = 20 H), a resistor (R = 100 $$\Omega$$) and a battery (E = 10 V) are connected in series. After a long time, the circuit is short-circuited and then the battery is disconnected. Find the current in the circuit at 1 ms after short circuiting.

A
$$4.5 \times 10^5 A$$

B
$$3.2 \times 10^{-5} A$$

C
$$9.9\times 10^{-2} A$$

D
$$6.7 \times 10^{-4} A$$

Solution

Given data :
$$L = 20 H $$
$$R = 100 \Omega$$
$$E = 10 V$$
Time after short circuiting the circuit, $$t = 1 ms.$$
The initial current, $$I'$$ in the circuit is,
$$I' = \dfrac{E}{R}$$

$$I' = \dfrac{10}{100}$$

$$I' = 0.1 A$$ ........(1)

The time constant, $$\tau$$ is,
$$\tau = \dfrac{L}{R}$$

$$\tau = \dfrac{20}{100}$$

$$\tau = 0.2 s$$ ........(2)

Now, the current at $$t = 1\ ms = 0.001\ s$$, is:
$$I = I' e^{-\dfrac{t}{\tau}}$$

Using equations (1) and (2) here, we get:
$$I = (0.1)(e^{-\dfrac{0.001}{0.2}})$$
$$I = (0.1)(e^{-0.005})$$
$$I = (0.1)(0.995)$$
$$I = 0.0995 A$$
$$I = 9.95 \times 10^{-2} A$$
Question 2
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In a series RLC circuit that is operating above the resonant frequency, the current

A
lags the applied voltage

B
leads the applied voltage

C
is in phase with the applied voltage

D
is zero

Solution

Capacitive reactance is given by $$X_C = \dfrac{1}{wC}$$
Inductive reactance is given by $$X_L = wL$$
At resonance, $$X_L = X_C$$ $$\implies \ wL = \dfrac{1}{wC}$$
But a frequency higher than resonance frequency, $$X_L > X_C$$
So the circuit behaves as a inductive circuit at a frequency higher that resonant frequency and the current lags behind the voltage in an inductive circuit.
Thus option A is correct.
Question 3
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If a capacitance C is connected in series with an inductor of inductance L, then the angular frequency will be

A
$$\sqrt{\dfrac{1}{LC}}$$

B
$$\sqrt{\dfrac{L}{C}}$$

C
LC

D
$$\sqrt{LC}$$

Solution

The peak current of an $$RLC$$ circuit is $$\frac{1}{\sqrt{R^{2}+(\frac{1}{{\omega C}}-\omega L)^{2}}}$$ , where $$\omega$$ is angular frequency
It varies maximum when
We can get the angular frequency by equating $$\frac{1}{\omega C}-\omega L=0$$
$$\Rightarrow \dfrac{1}{\omega C}=\omega L$$
$$\Rightarrow \omega^2=\dfrac{1}{LC}$$
$$\omega=\dfrac{1}{\sqrt{LC}}.$$
Question 4
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A resistor of $$3 k \Omega$$, a $$0.05 \mu F$$ capacitor and a $$120 mH$$ coil are in series across a $$5 kHz, 20 V$$ ac source.What is the impedance, expressed in polar form?

A
$$636\Omega$$

B
$$3,769\Omega$$

C
$$433\Omega$$

D
$$4,337\Omega$$

Solution

Given,

Resistance, $$R = 3 \, k\Omega = 3000 \Omega$$

Capacitance, $$C = 0.05 \, \mu F = 0.05 \times 10^{-6} \, F$$

Inductance, $$L = 120 \, mH = 120 \times 10^{-3} \, H$$
$$f = 5 \, kHz = 5000 \, Hz$$
It is asked to find the impedance in polar form

We have ,

Impedence, $$Z = \sqrt{R^2 + (X_L – X_C)^2}$$

Also

Inductive Reactance, $$X_L = L \omega , X_C = \dfrac{1}{C \omega}$$

We have $$\omega = 2 \pi F = 2 \times 3.14 \times 5000 = 31416$$

$$X_L = L \omega = 120 \times 10^{-3} \times 31416 = 3769.9 $$

Capacitive Reactance, $$X_C = \dfrac{1}{C \omega} = \dfrac{1}{31416 \times 0.05 \times 10^{-6}} = 636.61 $$

$$Z = \sqrt{3000^2 + (3769.9 – 636.61)^2} = \sqrt{9000,000 + 9819090.9} = 4337.2 \approx 4337 \Omega $$
Question 5
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A $$12 \Omega$$ resistor, a $$40 \mu F$$ capacitor, and an $$8 mH$$ coil are in series across an ac source. The resonant frequency is

A
$$28.1 Hz$$

B
$$281 Hz$$

C
$$2,810 Hz$$

D
$$10 Hz$$

Solution

$$\omega=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{8\times 10^{-3}\times 40\times 10^{-6}}}$$

$$\implies \omega=\dfrac{10^4}{\sqrt{32}}$$

$$\omega=2\pi\nu$$ where $$\nu$$=frequency

frequency$$=\nu=\dfrac{\omega}{2\pi}=\dfrac{10^4}{2\times 3.143\times \sqrt{32}}$$

$$\implies \nu=281Hz$$

Answer-(B)
Question 6
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Which one of the following graphs in following figure represents variation of reactance $$'X_c'$$ of a capacitor with frequency 'f' of an ac supply?

A

B

C

D

Solution

Capacitor reactance is given by: $$X_c = \dfrac{1}{2\pi f C}, C$$ is the capacitance.
$$X_c\ \&\ f$$ are inversely proportional.
Question 7
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If the value of C in a series RLC circuit is decreased, the resonant frequency

A
is not affected

B
increases

C
is reduced to zero

D
decreases

Solution

Resonant frequency in the series RLC circuit $$\nu_r = \dfrac{1}{2\pi\sqrt{LC}}$$
$$\implies \ \nu_r\propto \dfrac{1}{\sqrt{C}}$$
Thus resonant frequency of the circuit increases if the value of $$C$$ decreases.
Question 8
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For a certain load, the true power is $$150 W$$ and the reactive power is $$125 W$$. The apparent power is

A
$$19.52 W$$

B
$$195.2 W$$

C
$$275 W$$

D
$$25 W$$

Solution

True power $$P = 150 W$$
Reactive power $$P_r = 125 W$$
Thus apparent power $$P_a = \sqrt{P^2 + P_r^2}$$
$$\therefore$$ $$P_a = \sqrt{(150)^2+ (125)^2}$$
$$\implies$$ $$P_a = 195.2 W$$
Question 9
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A $$90 \Omega$$ resistor, a coil with $$30 \Omega$$ of reactance, and a capacitor with $$50 \Omega$$ of reactance are in series across a $$12 V$$ ac source. The current through the resistor is

A
$$9 mA$$

B
$$90 mA$$

C
$$13 mA$$

D
$$130 mA$$

Solution

Resistance=$$R=90\Omega$$

Reactance of coil$$=X_L=30\Omega$$

Reactance of capacitor$$=X_C=50\Omega$$

Impedance of circuit$$=X=\sqrt{R^2+(X_C-X_L)^2}$$

$$\implies X=\sqrt{90^2+(50-30)^2}=\sqrt{8100+400}$$

Current$$=I=\dfrac{E}{X}=\dfrac{12}{\sqrt{8500}}$$

$$\implies I=0.13A=130mA$$

Answer-(D)
Question 10
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In a series RC circuit, when the frequency and the resistance are halved, the impedance

A
doubles

B
is halved

C
is reduced to one-fourth

D
cannot be determined without values

Solution

Capacitive reactance is given by $$X_C = \dfrac{1}{2\pi f C}$$
where $$f$$ is the frequency.
Impedance of the RC circuit $$Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + \dfrac{1}{4\pi^2 f^2 C^2}}$$
If frequency and resistance are halved, we cannot determined unless the values are given.

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 47

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Alternating Current Test - 47

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SHARING IS CARING

Question 1

$$4.5 \times 10^5 A$$

$$3.2 \times 10^{-5} A$$

$$9.9\times 10^{-2} A$$

$$6.7 \times 10^{-4} A$$

Solution

Question 2

lags the applied voltage

leads the applied voltage

is in phase with the applied voltage

is zero

Solution

Question 3

$$\sqrt{\dfrac{1}{LC}}$$

$$\sqrt{\dfrac{L}{C}}$$

LC

$$\sqrt{LC}$$

Solution

Question 4

$$636\Omega$$

$$3,769\Omega$$

$$433\Omega$$

$$4,337\Omega$$

Solution

Question 5

$$28.1 Hz$$

$$281 Hz$$

$$2,810 Hz$$

$$10 Hz$$

Solution

Question 6

Solution

Question 7

is not affected

increases

is reduced to zero

decreases

Solution

Question 8

$$19.52 W$$

$$195.2 W$$

$$275 W$$

$$25 W$$

Solution

Question 9

$$9 mA$$

$$90 mA$$

$$13 mA$$

$$130 mA$$

Solution

Question 10

doubles

is halved

is reduced to one-fourth

cannot be determined without values

Solution

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