Alternating Current(11th And 12th > Physics ) Questions and Answers
Explanation:-
Answer: Option B. -> Bulb will give more intense light:
B
When a bulb and a capacitor are connected in series to an ac source, then on increasing the frequency the current in the circuit is increased, because the impedance of the circuit is decreased. So the bulb will give more intense light.
Explanation:-
Answer: Option D. -> 2ω:
D
The instantaneous values of emf and current in inductive circuit are given by E=E0 sin ωt and i=i0 sin(ωt−π2)respectively.
So, Pinst=Ei=E0 sinωt× i0 sin(ωt−π2)
=E0i0 sin ωt(sin ωt cosπ2−cos ωt sinπ2)
=E0i0 sin ωt cosωt
=12E0i0 sin 2ωt (sin 2ωt=2sin ωt cos ωt)
Hence, angular frequency of instantaneous power is 2ω.
Explanation:-
Answer: Option A. -> The peak voltage of the source is 100 volts:
A
V=100× 2sin 100π t cos 100 π t=100 sin 200π t
⇒ V0=100 Volts and Frequency =100Hz
Explanation:-
Answer: Option C. -> 8.46V, 1.4 A:
C
Z=√(R)2+(XL−Xc)2
R=12Ω,XL=wL=2000× 5× 10−3=10Ω
Xc=1wC=12000× 50× 10−6=10Ω i.e.Z = 10Ω
Maximum current i0=V0Z=2412=2A
Hence irms=2√2=1.4A
and Vrms=6× 1.41=8.46 V
Explanation:-
Answer: Option C. -> 1.4 mH:
C
Capacitance of wire
C=0.014× 10−6× 200=2.8× 10−6F=2.8μF
For impedance of the circuit to be minimum XL=Xc⇒ 2π vL=12π vC
⇒ L=14π2v2C=14(3.14)2× (2.5× 103)2× 2.8× 10−6
=1.4× 10−3H = 1.4mH
Explanation:-
Answer: Option C. -> 1. (ii), 2. (iii), 3. (i):
C
1. rms values=x0√2
2. x0sin ωt cosωt=x02sin2ωt⇒ rms value = x02√2
3. x0sin ωt+x0cosωt⇒rms value = √(x0√2)2+(x0√2)2
=√x20=x0