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ALTERNATING CURRENT Quiz-19

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. Alternating current cannot be measured by DC ammeter because
  •  AC cannot pass through DC ammeter
  •  AC changes direction
  •  Average value of current for complete cycle is zero
  •  DC ammeter will get damaged
Solution
The full cycle of alternating current consists of two half cycles. For one – half, current is positive and for second half, current is negative. Therefore, for an AC cycle, the net value of current average out to zero. While the DC ammeter, read the average value. Hence, the alternating current cannot DC measured by DC ammeter.

Q2. A voltage of peak value 283 V and varying frequency is applied to a series L – C – R combination in which R=3 Ω, L=25 mH and C=400μF. The frequency (in Hz)of the source at which maximum power is dissipated in the above, is
  •  51.5
  •  50.7
  •  51.1
  •  50.3
Solution
A series resonance circuit admits maximum current, as P=i2 R So, power dissipated is maximum at resonance. So, frequency of the source at which maximum power is dissipated in the circuit is 



Q3. A choke coil has
  •  High inductance and low resistance
  •  Low inductance and high resistance
  •  High inductance and high resistance
  •  Low inductance and low resistance
Solution
It is a fact.

Q4. The number of turns in the primary coil of a transformer is 200 and the number of turns in secondary coil is 10. If 240 V AC is applied to the primary, the output from secondary will be
  •  48 V
  •  24 V
  •  12 V
  •  6 V
Solution




Q5. If an 8 Ω resistance and 6 Ω reactance are present in an ac series circuit then the impedance of the circuit will be
  •  20 ohm
  •  5 ohm
  •  10 ohm
  •  14√2 ohm
Solution
 Impedance Z=√(R2+X2)=√(82+62)=10Ω

Q6. In an ac circuit with voltage V and current I, the power dissipated is
  •  VI
  •  1/2 VI
  •  1/√2 VI
  •  Depends on the phases between V and I
Solution
∵P=Vicos⁡ϕ,∴P∝cos⁡ϕ

Q7. In an AC circuit, the instantaneous values of emf and current are e=200 sin⁡ 314 t volt and I=sin ⁡(314t+π/3)amp. The average power consumed in watt is
  •  200
  •  100
  •  50
  •  25
Solution
Vrms=200/√2,Irms=1/√2 
∴ P=Vrms Irms cos⁡ϕ = 200/√2 1/√2 cos⁡ π/3 = 50W

Q8. In a region of uniform magnetic induction B=10-2 tesla, a circular coil of radius 30 cm and resistance π^2 ohm is rotated about an axis which is perpendicular to the direction of B and which forms a diameter of the coil. If the rotates at 200 rpm the amplitude of the alternating current induced in the coil is
  •  2 mA
  •  30 mA
  •  6 mA
  •  200 mA
Solution




Q9. A rectangular loop with a sliding connector of length l=1.0 m is situated in a uniform magnetic field B = 2T. Perpendicular to the plane of loop. Resistance of connector is r=2Ω. Two resistance of 6 Ω and 3 Ω are connected as shown in figure. The external force required to keep the connector moving with a constant velocity v=2 ms-1 is


  •  2 N
  •  1 N
  •  4 N
  •  6 N
Solution
Motion emf induced in the connector 
 e=Blv=2(1)(2)=4 V 
This acts as a cell of emf 4 V and internal resistance 2 Ω.6Ω and 3 Ω resistors are in parallel. 
∴ 1/RP =1/6+1/3=(1+2)/6=3/6=1/2 
 RP=2 Ω 

∴ Current through the connector (i) 
 =E/(RP+r)=4/(2+2)=1 A. 
Magnetic force on the connector 
=Bil=(1)(1)= 2 N 
Therefore, to keep the connector moving with a constant velocity, a force of 2 N has to be applied to the right side.

Q10. An AC voltage source has an output of ΔV=(200V) sin⁡ 2πft. This source is connected to a 100 Ω resistor. RMS current in the resistance is
  •  1.41 A
  •  2.41 A
  •  3.41 A
  • 0.71 A
Solution
I0=V0/R=200/100=2 A
Irms=I0/√2=1.414 A

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