# Inductor

## Trending Questions

**Q.**

In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC=5 V, βdc=200, RB=100 kΩ, RC=1 kΩ and VBE=1.0 V. The minimum base current and the input voltage at which the transistor will go to saturation, will be, respectively:

- 20 μA and 3.5 V
- 20 μA and 2.8 V
- 25 μA and 3.5 V
- 25 μA and 2.8 V

**Q.**The transfer characteristic curve of a transistor, having input and output resistances 100 Ω and 100 kΩ respectively, is shown in the figure. The voltage gain is

- 1×104
- 5×104
- 3×104
- 4×104

**Q.**A 40 mH inductor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly

- 2.05 A
- 16 A
- 25.1 A
- 1.7 A

**Q.**A pure inductor of 26 mH is connected to an ac source of 220 V. Given frequency of the source as 50 Hz, the rms current in the circuit is

[1 Mark]

- 7 A
- 14 A
- 28 A
- 742 A

**Q.**Match List-I with List-II :

List–I | List–II | ||

(a) | ωL>1ωC | (i) | Current is in same phase with emf. |

(b) | ωL=1ωC | (ii) | Current lags behind the applied emf. |

(c) | ωL<1ωC | (iii) | Maximum current occurs. |

(d) | Resonant frequency | (iv) | Current leads the emf. |

Choose the correct answer from the options given below :

- (a)–(ii);(b)–(i);(c)−(iv);(d)–(iii)
- (a)–(iii);(b)–(i);(c)−(iv);(d)–(ii)
- (a)–(ii);(b)–(i);(c)−(iii);(d)–(iv)
- (a)–(iv);(b)–(iii);(c)−(ii);(d)–(i)

**Q.**Four different circuit components are given in column I-and all the components are connected across an AC source of same angular frequency ω=200 rad/s.

The information of phase difference between the current and source voltage in such situation of column -I is given in column-II.

Match the circuit components in column-I with corresponding result in column-II.

Which of the following options is the correct match?

- (A)→q, r (B)→p, t (C)→p, r (D)→q, t
- (A)→p, r (B)→p, t (C)→p, r (D)→q, t
- (A)→q, r (B)→q, t (C)→q, r (D)→q, t
- (A)→p, r (B)→q, t (C)→p, r (D)→p, t

**Q.**An inductor (L=200 mH) is connected to an AC source of peak emf 210 V and frequency 50 Hz. The value of peak current and instantaneous voltage of the source when the current is at its peak value is respectively :

- 3.3 A, 210 V
- 6.6 A, 0 V
- 3.3 A, 0 V
- 6.6 A, 210 V

**Q.**A 1 Ω voltmeter has a range of 1 V. Find the additional resistance which has to be joined in series with this voltmeter, to increase the range of the voltmeter to 100 V.

- 10 Ω
- 199 Ω
- 99 Ω
- 100 Ω

**Q.**A 200 V, 100 Watt bulb is to be used across an A.C source of peak voltage 400 V and an angular frequency of 103 rad/s. The inductance of a choke coil to be used in series with the bulb so that the bulb won't blow off is

**Q.**In the following series LCR circuit, find the peak value of source current i0, voltage across capacitor VC and voltage across inductor VL at resonance, respectively.

- 5 A, 100 V, 100 V
- 2 A, 100 V, 200 V
- 10 A, 20 V, 30 V
- 5 A, 50 V, 50 V

**Q.**An ideal choke takes a current of 10 A when connected to an ac supply of 125 V and 50 Hz. A pure resistor under the same conditions takes a current of 12.5 A. If the two are connected to an AC supply of 100 V and 40 Hz, then the current in a series combination of the above resistor and inductor is

- 5√2 A
- 12.5 A
- 20 A
- 10 A

**Q.**The figure represents the voltage applied across a pure inductor. The diagram which correctly represents the variation of current i with time t is given by

**Q.**Given below are two statements.

Assertion - Resonance frequency will decrease in a series LCR circuit if a dielectric slab is inserted in between the plates of the capacitor.

Reason - By doing so, capacity of capacitor will increase.

Based on the above statements, choose the correct alternative.

- Both assertion and reason are true, and reason is a correct explanation of assertion.
- Both assertion and reason are true, but reason is not a correct explanation of assertion.
- Assertion is true, but reason is false.
- Assertion is false, but reason is true.

**Q.**A coil has negligible resistance and an inductive reactance of 20 Ω at 50 Hz. If an AC source of 200 V and 100 Hz frequency is connected across the coil, the rms current in the coil will be

- 2.0 A
- 5.0 A
- 7.0 A
- 10.0 A

**Q.**If the reading of voltmeter V shown in the figure at resonance is 200 V, then the quality factor of the circuit is,

- 2
- 4
- 1
- 3

**Q.**An AC circuit has R=100 Ω, C=2 μF and L=80 mH connected in series. The quality factor of the circuit is :

- 2
- 0.5
- 20
- 400

**Q.**A 100W/200V bulb and an inductor are connected in series to a 220V/50Hz supply. Find the power consumed by the bulb.

- 100W
- 92W
- 74W
- 84W

**Q.**An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V(rms), 50 Hz supply, the series inductor needed for it to work, is close to-

- 80 H
- 0.08 H
- 0.044 H
- 0.065 H

**Q.**An RLC series circuit has L=10 mH, R=3 Ω and C=1 μF connected in series to a source of E=15cosωt V. If the frequency of the voltage is 10% less than the resonant frequency, then

- Peak value of current is 0.704 A
- Impedance of the circuit is 21.32 Ω
- Power factor of the circuit is 0.141
- Resonant frequency of the circuit is 9×103 rad/s

**Q.**As the frequency of an AC circuit increases, the current first increases and then decreases. What combination of circuit elements is most likely to comprise the circuit?

- Inductor and capacitor
- Resistor and inductor
- Resistor, inductor and capacitor
- (a) and (c) above

**Q.**An inductor L=20 mH, a resistor R=100 Ω and a battery E=10 V are connected in series. After a long time the battery is short-circuited. Find the current in the circuit, 1 ms after short-circuiting.

[Use e−5=0.00673]

- 6.7×10−4A
- 2.5×10−4A
- 0.5×10−4A
- 2×10−3A

**Q.**An electric bulb, a capacitor, battery and a switch are all in series in a circuit. How does the intensity of light vary when the switch is turned on?

- Continues to increase gradually
- Gradually increases for some time and then becomes steady.
- Sharply rises initially and then gradually decreases
- Gradually increases for some then time and then gradually decreases

**Q.**A series LCR circuit containing a resistor of resistance 120 Ω has resonant frequency 4×105 rad/s. At resonance, the voltage across resistor and inductor are 60 V and 40 V respectively. Find the value of C.

- 12 μF
- 2 μF
- 19 μF
- 132 μF

**Q.**The angular frequency of alternating current in a LCR circuit is 100 rad/s. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.

- 0.8 H and 250 μF
- 0.8 H and 150 μF
- 1.33 H and 250 μF
- 1.33 H and 150 μF

**Q.**A series LCR circuit is designed to resonate at an angular frequency ω0=105 rad/s. The circuit draws 16 W power from 120 V source at resonance. The value of resistance R in the circuit is

**Q.**The inductive reactance of a coil is 1000Ω. If its self inductance and frequency both are increased two times then inductive reactance will be

- 1000 Ω
- 2000 Ω
- 4000 Ω
- 16000 Ω

**Q.**An inductive circuit contains resistance of 1 Ω and an inductance of 2.0 H. If an AC voltage of 120 V and frequency 60 Hz is applied to this circuit, the current would be nearly

- 0.8 A
- 0.48 A
- 0.16 A
- 0.32 A

**Q.**

An amplitude modulated voltage is expressed as

e=10(1+0.8cos2000πt)cos3×106πtvolt. The depth of modulation is :- 8
- 0.8
- 800
- None of these

**Q.**An LCR circuit contains resistance of 110 Ω and a supply of 220 V at 300 rad/s angular frequency. If only capacitance is removed from the circuit, current lags behind the voltage by 45∘. If on the other hand, only inductor is removed the current leads by 45∘ with the applied voltage. The rms current flowing in the circuit will be:

- 2.5 A
- 2 A
- 1 A
- 1.5 A

**Q.**A charged capacitor discharges through a resistance R with time constant τ. The two are now placed in series across an AC source of angular frequency ω=1τ . The impedance of the circuit will be:

- R√2
- √2 R
- R
- 2R