Alternating Current Test

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Alternating Current Test - 70

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Question 1
1 / -0

The self-inductance of the motor of an electric fan is 10H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of : (Take $${\pi}^2$$ = 10)

A
$$4\mu F$$

B
$$8\mu F$$

C
$$1 \mu F$$

D
$$2\mu F$$

Solution

$$L=10H, f=50 Hz, w=50\times 2\pi$$

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

$$C=\dfrac{1}{Lw^2}$$

$$C=\dfrac{1}{10\times \pi^2\times 10^4}$$

$$=10^{-6}f$$

$$=1\mu F$$
Question 2
1 / -0

A circuit containing resistance $$R_{1}$$, Inductance $$L_{1}$$ and capacitance $$C_{1}$$ connected in series resonates at the same frequency $$n$$ as a second combination of $$R_{2}$$, $$L_{2}$$ and$$C_{2}$$. If the two are connected in series, then the circuit will resonate at:

A
$$n$$

B
$$2n$$

C
$$\sqrt{\dfrac{L_{2}C_{2}}{L_{1}C_{1}}}$$

D
$$\sqrt{\dfrac{L_{1}C_{1}}{L_{2}C_{2}}}$$

Solution

Condition of resonance $$\omega = \dfrac {1} { \sqrt {LC}}$$

So $$\omega_{1} = \dfrac {1} { \sqrt {L_{1}C_{1}}}$$

$$\omega_{2} = \dfrac {1} { \sqrt {L_{2}C_{2}}}$$

Since both frequency are same $$L_{1}C_{1} = L_{2}C_{2}$$

Equivalent capacitance of $$C_{1} \ and \ C_{2}$$ in series = $$ \dfrac {C_{1} C_{2}} { C_{1}+c_{2}}$$

Equivalent inductance of $$L_{1}$$ and $$L_{2} = L_{1} + L_{2}$$

Resonance for combination = $$ \sqrt{\dfrac { (C_{1} +C_{2})} { C_{1}C_{2}(L_{1}+L_{2})}}$$

Using $$L_{1}C_{1} = L_{2}C_{2}$$, resonance of combination = $$\dfrac{1}{\sqrt{L_{1}C_{1}}}$$

So the frequency of resonance remains same = $$n$$
Question 3
1 / -0

A 120V, 60Hz a.c. power is connected 800$$\Omega $$ non-inductive resistance and unknown capacitance in series. The voltage drop across the resistance is found to be 102V, then voltage drop across capacitor is

A
8V

B
102V

C
63V

D
55V

Solution

For RC circuits,

$$V^2 = V_R^2 +V_C^2$$

$$120^2 = 102^2 + V_C^2$$

$$V_C^2 = 63.214 \approx\ 63 V$$
Question 4
1 / -0

An alternating e.m.f. of $$E= 100 \sin(100 \pi$$t ) is connected to a choke of negligible resistance and current oscillations of amplitude 1 A are produced. The inductance of the choke should be

A
$$100 H$$

B
$$\dfrac{1}{\pi } \ H$$

C
$$1 H$$

D
$$\pi H$$

Solution

Angular frequency $$\omega=100\pi$$

Reactance $$X_L=100\pi\times L$$

$$I_o=1 \ A$$

$$E_o=100 \ V$$

$$\therefore \dfrac{E_o}{X_L}= \dfrac{100}{100\pi\times L}=1$$

$$L=\dfrac{1}{\pi}H$$
Question 5
1 / -0

A coil has an inductance of 0.7H and is joined in series with a resistance of 220 $$\Omega $$ . When an alternating e.m.f. of 220V at 50 c.p.s. is applied to it, then the wattless component of the current in the circuit is

A
5 ampere

B
0.5 ampere

C
0.7 ampere

D
7 ampere

Solution

Impedance of the circuit is given by:

$$Z = \sqrt {R^2 + (2 \pi f L)^2}$$

$$Z = \sqrt {220^2 + (2 \pi \times 50 \times 0.7)^2}$$

$$Z \approx 220\sqrt 2\ \Omega$$

Power factor is given by:

$$\cos \phi = \dfrac{R}{Z} = \dfrac{\pi}{4}$$

Total current in the circuit flowing is:

$$I = \dfrac{V}{Z} = \dfrac{220}{220\sqrt 2} = \dfrac{1}{\sqrt 2} A$$

Wattless component of current in the circuit is:

$$I_a = I \sin (\phi)$$

$$I_a = \dfrac{1}{\sqrt 2} \times \dfrac{1}{\sqrt 2} = 0.5 A$$
Question 6
1 / -0

In the given circuit the readings of the voltmeter $$V_1$$ and the ammeter $$A$$ are:

A
220 V, 2.2 A

B
110 V; 1.1A

C
220 V, 1.1 A

D
110 V; 2.2 A

Solution

$$w=100\pi\Omega$$

we observe that $$w^2=\dfrac{1}{LC}$$

i.e. this frequency is resonance frequency
all the voltage drop across resistor.

$$\therefore v_1=220V$$

and $$i=\dfrac{V_1}{R}$$

$$=\dfrac{220}{100}$$

$$=2.2A$$
Question 7
1 / -0

Assertion: A resistance is connected to an ac source. Now a capacitor is included in the series circuit. The average power absorbed by the resistance will remain same.
Reason: By including a capacitor or an inductor in the circuit average power across resistor does not change.

A
A and R both are true and R is correct explanation of A

B
A and R both are true but R is not the correct explanation of A

C
A is true R is false

D
A is false and R is true

Solution

After connecting the capacitor Irms will charge because impedance is changed.
$$\therefore$$ A is false
Question 8
1 / -0

A certain choke coil of negligible resistance draws a current of 8A, when connected to a supply of 100 volts, at 50Hz. A certain non-inductive resistance, under the same conditions carries a current of 10 A. If the two are transferred to a supply system working at 150 V, at 40 Hz, the total current they will take if joined in series is

A
1.06 A

B
10.6 A

C
0.106 A

D
100.6 A

Solution

In choke coil, $$I=8A, Erms=100V$$

$$\omega=100\pi \ rad/s$$

$$\dfrac{100}{100\pi\times L}=8=L=\dfrac{1}{8\pi}H$$

In resistance, $$I=10 \ amp$$

$$\therefore R=\dfrac{100}{10}=10 \ \Omega$$

New source V=150 V, f= 40 Hz, $$\omega'=40\times 2\pi$$

Connected in series, $$z=\sqrt{10^2+(\dfrac{80\pi}{8\pi})}$$

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

$$Irms=\dfrac{150}{10\sqrt{2}}$$

$$=10.61 \ A$$
Question 9
1 / -0

A capacitor has a resistance of $$1200 M\Omega$$ and capacitance of $$22 \mu F$$. When connected to an AC supply of frequency 80Hz, the alternating voltage supply required to drive a current of 10 virtual amperes is

A
$$904\sqrt{2} V$$

B
904 V

C
$$904/\sqrt{2} V$$

D
452 V

Solution

$$X_c=1200\times 10^6\Omega$$

$$C=22\mu F$$

$$w=160\pi\ rad/s$$

$$\dfrac{V_o}{X_c}=10A$$

$$V_o=\dfrac{10}{160\pi(22\times 10^{-6})}=904.29V$$
Question 10
1 / -0

The potential difference across a $$2 H$$ inductor as a function of time is shown in figure. At time $$t=0$$, current is zero. Current at $$t=2$$ second is:

A
$$1 A$$

B
$$3 A$$

C
$$4 A$$

D
$$5 A$$

Solution

$$t=0, i=0$$

$$t=2, i=?$$

$$L=2H$$

$$V=L\dfrac{dI}{dt}$$$$V=2\dfrac{dI}{dt}$$

$$\int_{0}^{2}Vdt=2\int_{0}^{I}dI$$

area of shaded portion

$$\dfrac{1}{2}\times 2\times 10=2\times I$$

$$I=5A$$

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 70

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Alternating Current Test - 70

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SHARING IS CARING

Question 1

$$4\mu F$$

$$8\mu F$$

$$1 \mu F$$

$$2\mu F$$

Solution

Question 2

$$n$$

$$2n$$

$$\sqrt{\dfrac{L_{2}C_{2}}{L_{1}C_{1}}}$$

$$\sqrt{\dfrac{L_{1}C_{1}}{L_{2}C_{2}}}$$

Solution

Question 3

8V

102V

63V

55V

Solution

Question 4

$$100 H$$

$$\dfrac{1}{\pi } \ H$$

$$1 H$$

$$\pi H$$

Solution

Question 5

5 ampere

0.5 ampere

0.7 ampere

7 ampere

Solution

Question 6

220 V, 2.2 A

110 V; 1.1A

220 V, 1.1 A

110 V; 2.2 A

Solution

Question 7

A and R both are true and R is correct explanation of A

A and R both are true but R is not the correct explanation of A

A is true R is false

A is false and R is true

Solution

Question 8

1.06 A

10.6 A

0.106 A

100.6 A

Solution

Question 9

$$904\sqrt{2} V$$

904 V

$$904/\sqrt{2} V$$

452 V

Solution

Question 10

$$1 A$$

$$3 A$$

$$4 A$$

$$5 A$$

Solution

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