An L-C-R series circuit with 100 Ω resistance is conne

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 Multiple Choice QuestionsMultiple Choice Questions

201.

A step down transformer operated on a 2.5 kV line. It supplies a load with 20 A. The ratio of the primary winding to the secondary is 10 : 1. If the transformer is 90% efficient, the voltage in the secondary is

  • 225 V

  • 550 V

  • 250 V

  • 200 V


202.

When a given coil of ohmic resistance 100 Ω and inductive reactance 1003 Ω is connected to a 200 V DC source, the current flows in the circuit is

  • 1 A

  • 1.5 A

  • 2 A

  • 1003 Ω


203.

The ratio of inductive reactance to capacitive reactance in an AC circuit will be

  • ω LC

  • ω2 L

  • zero

  • ω2 LC


204.

In a step up transformer, the turns ratio of field is primary and secondary is 1 : 4. A  leclanche cell of emf 1.5 V is connected across the primary. The voltage developed across the secondary will be

  • 3 V

  • 6 V

  • 1 V

  • zero


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205.

When an AC source of emf e = E0 sin (100t) is connected across a circuit, the phase difference between the emf e and the current i in the circuit is observed to be π/4, as shown in the diagram. If the circuit consists possibly only of R-C or R-L or L-C is series, find the relationship between the two elements.

     

  • R = 1 kΩ , C = 10 µF

  • R = 1 kΩ , C = 1 µF

  • R = 1 kΩ , L = 10 H

  • R = 1 kΩ , L = 1 H


206.

For the circuit shown in figure, the voltage across 4 Ω resistance is

     

  • 1 V

  • 2 V

  • 4 V

  • 5 V


207.

A LED has a voltage drop of 2V across it and passes a current of 10 mA. When it operates with a 6V battery through a limiting resistor R. The value of R is

  • 40 kΩ

  • 200 Ω

  • 4 kΩ

  • 400 Ω


208.

A series R-L-C circuit driven with Erms = 100V and frequency fd = 50 Hz is shown below.

What new capacitance should be connected to maximize the average power, if other parameters of the circuit are not changed ?

  • 41.6 µF

  • 33.5 µF

  • 39.8 µF

  • 31.3 µF


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209.

An L-C-R series circuit with 100 Ω resistance is connected to an AC source of 200 V and angular frequency 300 rad s-1 . When only the capacitance is removed, the current lags behind the voltage by 60°. When only the inductance is removed, the current leads the voltage by 60°. The average power dissipated in this L-C-R circuit is

  • 0

  • 400 W

  • 200 W

  • 800 W


B.

400 W

When capacitance is removed, then          tan ϕ = XLRor   tan 60° = XLR          XL = 3 R      .... (i)When inductance is removed, then          tan ϕ = XCRor   tan 60° =  XCR         XC = 2 R     .... (ii)From Eqs. (i) and (ii), we havewe see that,      XC =XLSo, the L-C-R circuit is in resonance.Hence,                 Z= R      Irms = VrmsZ = 200100 = 2 ANow, average power dissipated      <P> = VrmsIrms cos ϕAt resonance, current and voltage are in phasei.e.         ϕ = 0   <P> = (200)(2)(1) = 400 W


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210.

Given,

R1 = 10 Ω, C1 = 2µF, R2 = 20 and C2 =  4F

The time constants (in µs) for the circuits I, II, III are respectively.

  • 18, 18/9, 4

  • 4, 8/9, 18

  • 18, 4, 8/9

  • 8/9, 18, 4


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