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ALTERNATING CURRENT Quiz-20

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. The average power dissipated in a pure inductor of inductance L when an ac current is passing through it, is (Inductance of the coil L and current I)
  •  1/2 LI2
  •  1/4 LI2
  •  2 Li2
  •  Zero
Solution
It is a fact.

Q2. The time taken by AC of 50 Hz in reaching from zero to the maximum value is
  •  50×10-3s
  •  5×10-3s
  •  1×10-3s
  •  2×10-3s
Solution
The time taken by AC in reaching from zero to maximum value is t=T/4=1/(4 f) 
 = 1/(50×4)=5×10-3s

Q3. A 100 V, AC source of frequency 500 Hz is connected to an L-C-R circuit with L=8.1 mH, C=12.5 μF, R=10 Ω all connected in series as shown in figure. What is the quality factor of circuit?

  •  2.02
  •  2.5434
  •  20.54
  •  200.54
Solution
We know that Q - factor of series resonant circuit is given as Q=(ωrL)/R 
Here, L=8.1 mH,C=12.5 μF,R=10Ω,f=500 Hz 
∴ Q=(ωrL)/R=2πfL/R =(2×π×500×8.1×10-3)/10
=8.1π/10
=2.5434

Q4. Two coils A and B have coefficient of mutual inductance M = 2H. The magnetic flux passing through coil A changes by 4 Wb in 10 s due to change in current in B. Then
  •  Change in current in B in this time interval is 0.5 A
  •  Change in current in B in this time interval is 8 A
  •  The change in current in B in this time interval is 2 A
  •  A change in current of 1 A in coil A will produce a change in flux passing through B by 4 Wb
Solution
Here, M=2H,dϕ=4 Wb,dt=10 s 
As ϕ =M i 
dϕ=M di 
Or di=dϕ/M=4/2=2 A 
Also, dϕ=M (di)=2(1) 
 = 2 Wb

Q5. In an LCRcircuit R=100 ohm. When capacitance C is removed, the current lags behind the voltage by π/3. When inductance L is removed, the current leads the voltage by π/3. The impedance of the circuit is
  •  50 ohm
  •  100 ohm
  •  200 ohm
  •  400 ohm
Solution
When C is removed circuit becomes RL circuit 
hence tan⁡ π/3=XL/R…(i) 
When L is removed circuit becomes RC circuit hence tan⁡ π/3=XC/R …(ii) 
From equation (i) and (ii) we obtain XL=XC. This is the condition of resonance and in resonance Z=R=100Ω

Q6. The instantaneous values of current and emf in an ac circuit are I=1/√2 sin⁡ 314 t amp and E=√2 sin⁡ (314 t-π/6) V respectively. The phase difference between E and I will be
  •  -π/6 rad
  •  -π/3 rad
  • π/6 rad
  •  π/3 rad
Solution
Phase difference relative to the current 
ϕ=(314t-π/6)-(314 t)=-π/6

Q7. In the given circuit diagram the current through the battery and the charge on the capacitor respectively in steady state are

  •  1 A and 3 μC
  •  17 A and 0 μC
  •  6/7 A and 12/7 μC
  •  11 A and 3 μC
Solution
In steady state current through the branch having capacitor is zero. 
∴ 1/R=1/1+1/2+1/3 
1/R=(6+3+2)/6 
R=6/11 
As V=iR 
∴ 6=i×6/11 
Current through the battery i=11A 
Charge on the capacitor q=CV ⇒ q=0.5×10-6×6 
q=3μC

Q8. In a L – R circuit, the value of L is (0.4/π)H and the value of Ris 30 Ω. If in the circuit, an alternating emf of 200 V at 50 cycle/s is connected, the impedance of the circuit and current will be
  •  11.4 Ω,17.5 A
  •  30.7 Ω,6.5 A
  •  40.4 Ω,5 A
  •  50 Ω,4 A
Solution
Z^2=R^2+(2πfL)2 
=(30)^2+(2π×50×0.4/π)2 
=(900+1600)=2500  
or Z=50 Ω 
Also, I=V/Z=200/50=4 A

Q9. The time constant of the given circuit is

  •  3RC/5
  •  6RC/5
  •  5RC/6
  •  None of these
Solution
Time constant of R – C circuit is τ=RC 
Here effective resistance of the circuit 
 = (2R×3R)/(2R+3R)=6R/5 
∴ τ=6R/5×C=6RC/5

Q10. A bulb and a capacitor are in series with an ac source. On increasing frequency how will glow of the bulb change
  •  The glow decreases
  •  The glow increases
  •  The glow remain the same
  • The bulb quenches
Solution
This is because, when frequency v is increased, the capacitive reactance XC=1/2πvC decreases and hence the current through the bulb increases

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