12th Grade > Physics
ALTERNATING CURRENT MCQs
Total Questions : 30
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Answer: Option B. -> The glow increases
:
B
The glow increases because, when frequency vis increased, the capacitive reactance Xc=12πvCdecreases and hence the current through the bulb increases which will lead to increase in glow
:
B
The glow increases because, when frequency vis increased, the capacitive reactance Xc=12πvCdecreases and hence the current through the bulb increases which will lead to increase in glow
Answer: Option A. -> √3π Henry
:
A
cosϕ=12⇒ϕ=60∘,tan60∘=ωLR⇒L=√3πH
:
A
cosϕ=12⇒ϕ=60∘,tan60∘=ωLR⇒L=√3πH
Answer: Option A. -> 2A
:
A
Given XL=XC=5Ω this is the condition of resonance. Since VL=VCnet voltage across L and C combination will be zero.
Voltage drop across the resistance = 110v. Thus current flowing though Ammeter reading =110v55=2A.
:
A
Given XL=XC=5Ω this is the condition of resonance. Since VL=VCnet voltage across L and C combination will be zero.
Voltage drop across the resistance = 110v. Thus current flowing though Ammeter reading =110v55=2A.
Answer: Option D. -> 45 ohm
:
D
iL=9030=3A, ic=9090=1A
Net current through circuit i=ic−iL=2A
∴Z=Vi=902=45Ω
:
D
iL=9030=3A, ic=9090=1A
Net current through circuit i=ic−iL=2A
∴Z=Vi=902=45Ω
Answer: Option D. -> n> nr
:
D
The current will lag behind the voltage when reactance of inductance is more than the reactance of condenser. Thus, ωL>1wCor ω>1√LC or n>12π√LCor n>nrwhere nr= resonant frequency
:
D
The current will lag behind the voltage when reactance of inductance is more than the reactance of condenser. Thus, ωL>1wCor ω>1√LC or n>12π√LCor n>nrwhere nr= resonant frequency
Question 6. Following figure shows an ac generator connected to a "block box" through a pair of terminals. The box contains possible R, L, C or their combination, whose elements and arrangements are not known to us. Measurements outside the box reveals that e = 75 sin (sin ωt) volt, i = 1.5 sin (ωt + 45∘) amp then, the wrong statement is
Answer: Option B. -> There must be no inductor in the box
:
B
Since voltage is lagging behind the current, so there must be no inductor in the box.
:
B
Since voltage is lagging behind the current, so there must be no inductor in the box.
Answer: Option B. -> The voltage leads the current by Π2
:
B
At t = 0, phase of the voltage is zero, while phase of the current is −π2 i.e., voltage leads by π2
:
B
At t = 0, phase of the voltage is zero, while phase of the current is −π2 i.e., voltage leads by π2
Answer: Option B. -> √3Ω
:
B
V0=ioZ⇒200=100Z⇒Z=2Ω
Also Z2=R2+X2L⇒(2)2=(1)2+X2L⇒XL=√3Ω
:
B
V0=ioZ⇒200=100Z⇒Z=2Ω
Also Z2=R2+X2L⇒(2)2=(1)2+X2L⇒XL=√3Ω
Answer: Option A. -> −Π6rad
:
A
Phase difference relative to the current ϕ=(314t−π6)−(314t)=−π6
:
A
Phase difference relative to the current ϕ=(314t−π6)−(314t)=−π6
Answer: Option D. -> 12
:
D
iwL=irmssinϕ⇒√3=2sinϕ⇒sinϕ=√32
⇒ϕ=60∘ so p.f =cosϕ=cos60∘=12.
:
D
iwL=irmssinϕ⇒√3=2sinϕ⇒sinϕ=√32
⇒ϕ=60∘ so p.f =cosϕ=cos60∘=12.