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NCERT Class XII Chapter
Alternating Current
Questions With Solutions

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Question : 20
Total: 26
A series LCR circuit with L = 0.12 H, C = 480 nF, R = 23 Ω is connected to a 230 V variable frequency supply.
(a) What is the source frequency for which current amplitude is maximum. Obtain this maximum value.
(b) What is the source frequency for which average power absorbed by the circuit is maximum. Obtain the value of this maximum power.
(c) For which frequencies of the source is the power transferred to the circuit half the power at resonant frequency? What is the current amplitude at these frequencies?
(d) What is the Q-factor of the given circuit?
Solution:  

Soln. (a) At resonant frequency, the current amplitude is maximum.
f =
1
2π√LC
 =
1
2π√0.12×480×10−9
 = 663 Hz
Iv =
Ev
R
 , I0 = Iv√2 =
Ev√2
R
 =
230√2
23
 = 14.14 A
(b) Maximum power loss at resonant frequency, P = EvIv cos ϕ
P = Ev
Ev
R
 cos 0° =
Ev2
R
 =
(230)2
23
 = 2300 W
(c) Let at an angular frequency, the source power is half the power at resonant frequency.
P = EvIv cos ϕ
1
2
[
Ev2
R
] =
EvEv
Z
R
Z
Z2 = 2R2,R2 + (XL–XC)2 = 2R2
XL–XC = R
ωL -
1
ωC
 = R or ω2−
1
LC
 =
R
L
 ω
where resonant angular frequency, ωr =
1
LC
 =
1
0.12×480×10−9
so , ω2−ωr2 = ±
R
L
 ω
two quadratic equations can be formed
ω2−
R
l
 ω - ωr2 = 0 and ω2+
R
L
 ω - ωr2 = 0
On solving, we get, ω1 =
R
2L
+[ωr2+R2∕4L2]1∕2 = ωr+Δω and
ω2 = -
R
2L
+[ω2+
R2
4L2
]1∕2 = ωr - Δω

Now, ω1–ω2 = R/L
[ωr + Δω] – [ωr – Δω] = R/L or Δω = R/L
Δω =
R
2L
 bandwidth of angular frequency
So, band width of frequency
Δf =
Δω
2π
 =
R
4πL
 =
23
4×3.14×0.12
 = 15.26 Hz
Hence the two frequencies for half power
fZ = f2 – Δf and f1 = fr + Δf
fZ = 663 – 15.26 = 647.74 Hz and f1 = 663 + 15.26 = 678.26 Hz
At these frequencies the current amplitude is I =
I0
√2
 = 10 A
(d) Q factor, Q =
1
R
√
L
C
 ⇒ Q =
1
23
√
0.12
480×10−9
 = 21.7
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