NCERT Class XII Chapter
Alternating Current
Questions With Solutions

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Question : 22
Total: 26
Answer the following questions.
(a) In any ac circuit, is the applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the series elements of the circuit? Is the same true for rms voltage?
(b) A capacitor is used in the primary circuit of an induction coil.
(c) An applied voltage signal consists of a superposition of a dc voltage and an ac voltage of high frequency. The circuit consists of an inductor and a capacitor in series. Show that the dc signal will appear across C and the ac signal across L.
(d) A choke coil in series with a lamp is connected to a dc line. The lamp is seen to shine brightly. Insertion of an iron core in the choke causes no change in the lamp’s brightness. Predict the corresponding observations if the connection is to an ac line.
(e) Why is choke coil needed in the use of fluorescent tubes with ac mains?
Why can we not use an ordinary resistor instead of the choke coil?
Solution:  

(a) It is true that applied instantaneous voltage is equal to algebraic sum of instantaneous potential drop across each circuit element is series.
E = VR+VL+VC
E0 sinωt +
E0
XL
sin (ωt - π/2) +
E0
Xc
sin (ωt + π/2)
But the rms voltage applied is equal to vector sum of potential drop across each element, as voltage drops are in different phases.

Ev = (VLVC)2+VR2
(b) At the time of broken circuit of the induction coil, the induced high voltage charges the capacitor. This avoid sparking in the circuit.
(c) Inductive reactance, XL = 2πfL
For a.c., XC ∝ f, For d.c., f = 0, XL = 0
Capacitive reactance, XC =
1
2πfC

For a.c., X ∝
1
f
, For d.c, f = 0 , XC = ∞
So, superimpose applied voltage will have all d.c. potential drop across XC and will have most of a.c potential drop across XL
(d) Inductor offer no hinderance to d.c. XL = 0, so insertion of iron core does not effect the d.c. current or brightness of lamp connected.
But it definitely effect a.c. current as insertion of iron core increases ‘L’,
L = µm nI
thus increases XL (2πf L). A.c. current in the circuit reduces Iv =
Ev
XL
and brightness of the bulb also reduces
(e) A fluorescent tube is connected directly across a 220 V source, it would draw large current which may damage the filaments of the tube. So a choke coil which behaves as L-R circuit reduces the current to appropriate value, and that also with a lesser power loss.
Iv =
Ev
R2+XL2
, P = EvIv cos ϕ
An ordinary resistor used to control the current would have maximum power wastage as heat
Iv =
Ev
R
, Pmax = EvIv
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