1. Tips are provided for taking online classes effectively, including being prepared, treating it like a real course, practicing time management, and eliminating distractions. 2. Inside the class, students are advised to open their camera, mute audio when not speaking, and ask permission before adding notes or sharing. 3. Discipline is important for online students.

2. TIPS FOR TAKING ONLINE CLASS
FOR YOURSELF
1: Be ready before the starting time
2: Treat an online course like a real course
3: Hold yourself accountable
4: Practice time management
5: Create a regular study space and stay organized
6: Eliminate distractions
7: Figure out how you learn best
8: Actively participate
9: Leverage your network
INSIDE THE CLASS
1: Open your camera
2: Mute your audio
3: take your pen, copy and other material sources
3: Don’t add your note on the screen
4: Do ask, write and share after the permission. It
will be provided at the end of the class
5: Be more disciplined.
2

3. UDAY KHANAL
Department of Physics
CCRC
3

4. The essence of SCIENCE: ask an impertinent question, and
you are on the way to a pertinent answer.

29. OLD Questions
Discussion
AC and Chemical effect of current

30. 𝑧 = 𝑅2 + 𝑋𝐶
2
𝑖𝑟𝑚𝑠 =
𝑖𝑜
2
E = Eo sin ωt
𝑓 =
1
2𝜋 𝐿𝐶
Pav = Irms Erms cosf
C I V I L
Power factor, 𝑐𝑜𝑠𝜙 =
𝑅
𝑅2
+(ω𝐿 −
1
𝜔𝐶
)2
𝑍 = 𝑅2 + 𝑋𝑐 − 𝑋𝐿 2
𝑋𝑐 =
1
𝜔𝐶
=
1
2𝜋𝑓𝐶
𝑋𝐿 = 𝜔𝐿 = 2𝜋𝑓𝐿 P = IV = I2R
𝑇𝑎𝑛φ =
𝑋𝐿 − 𝑋𝐶
𝑅
𝐼 =
𝐸
𝑍
𝑉𝑐 = 𝐼𝑋𝐶
𝑇𝑎𝑛φ =
𝑋𝐿
𝑅
𝑇𝑎𝑛φ =
𝑋𝐶
𝑅
𝑍 = (𝑅 + 𝑟)2 + 𝑋𝑐 − 𝑋𝐿 2
𝑧 = 𝑅2 + 𝑋𝐿
2
Erms=
𝐸𝑜
2
𝑉𝐿 = 𝐼𝑋𝐿 I = Io sin (ωt+φ)
LIST OF FPRMULAS
m =
𝐴𝐼𝑡
𝐹𝑉

32. Short questions

33. 1.
The emf of an ac source is given by the expression, E = 300 sin 314 t volts.
Write the values of peak voltage and frequency of source. [2074 'A]
Comparing with E = Eo sin ωt,
we get
Eo= 300 and ω = 314 Hence, the peak value of
Eo = 300 V
f = 50 Hz Ans.

34. 2.
Define rms value of ac. How is it related with the peak value of ac?
[2074 'B', supp. 2070, 2nd Exam 2068, 1st Exam 2067, 2064]
It is the steady current which when passed through a given resistance for a given time
produces the same amount of heat as produced by the a.c for the same resistor for the
same time.
𝑖𝑟𝑚𝑠 =
𝑖𝑜
2
Peak Value

35. 3.
Why is choke coil preferable to resistor ? [Supp. 2073, 2071 ‘D’] OR
Choke coil is preferred to a resistor in an alternating current circuit
Why? [Set 'B'2069, 1st Exam 2068,2062, 2057 2055, 2053] OR
Explain why choke coil is used in an a.c. [Supp. 2062]
A choke coil decreases the current without wasting
electrical energy in the form of heat.

36. 4.
Alternating current passes through a capacitor where as direct current does not.
Explain this fact on the basis of capacitive reactance. [2073 ‘D’, Supp.2063]
Xc =
1
2𝜋𝑓𝐶
where f is frequency and C is capacitance
We see that.
(I)for ac current xc ~ small
(ii) for d.c. current f = 0 , xc ~ ∞

37. 5.
How does the resonance frequency of an L.C.R. series circuit change if the plates
of the capacitor are brought closer together ?
𝑐 =
𝜖𝑜𝐴
𝑑 d
+
+
+
+
+
-
-
-
-
-
𝑓 =
1
2𝜋 𝐿𝐶

38. 6.
What is wattles current? [2072'D', 2071 ‘C’ 2067 2nd Exam, 2058]
It doesn't consume any power in a circuit.
In case of wattles current phase difference between
current and e.m.f. should be p/2 in an circuit.
We have power in A.C. is given by
Pav = Irms Erms cosf
In case of inductor f = p/2
Pav = Irms Erms cos p/2 = 0
Hence no power is consumed by an inductor.

39. 7.
What are the advantages of A.C. over D.C. ? [2072'E’]
• Transmission and distribution cheaper
• Easily convertible to DC
• AC machine are stronger and easy to use
• Voltage can be stepped up or down and can reach
distant places

40. Reactance --------resistance offered by an inductor or a
capacitor. The alternating e.m.f. and the current differ
in phase by p/2 when a.c. flows through an inductor or
a capacitor.
Impedance --------effective resistance of a.c. --------------i.e.
the effective resistance offered by LR circuit or CR
circuit or LCR circuit.
𝑍 = 𝑅2 + 𝑋𝑐 − 𝑋𝐿 2
8.
Distinguish between reactance and impedance for an a.c. circuit?
[2054]

41. 9.
Which is more dangerous in use a.c. or d.c.? Why? [2070 set C]
Ans: The a.c. is more dangerous than d.c. of the same voltage.
For a.c. Erms = 220 V means peak value is E0 = 2 Erms = 2 x 220 = 311 V
whereas for 220 V d.c. peak value is (220 V), the same.

42. 10
What do you mean by power factor? On what factors does it
depend?
[2069 set A]
Pav = Irms Erms cosf
Power factor, 𝑐𝑜𝑠𝜙 =
𝑅
𝑅2
+(ω𝐿 −
1
𝜔𝐶
)2
Virtual
Real
Ans: The cosine of the phase angle between alternating
e.m.f. and current in an a.c. circuit is called power
factor.

43. 12.
For a capacitor in an a.c. circuit, explain why there is a phase difference
between current and voltage. [Supp. 2071]
If 𝐼 = 𝐼𝑜𝑠𝑖𝑛𝜔𝑡
Then, 𝐰𝐞 𝐠𝐞𝐭 𝐸 = 𝐸𝑜𝑠𝑖𝑛(𝜔𝑡 −
𝜋
2
)
When Capacitor is connected in the a.c.
Both alternating
C I V I L

44. 13
Fluorescent lights often use an inductor, to limit the current through the tubes. Why is
it better to use an inductor rather than a resistor for this purpose? [2070
'D’]
Inductor consumes
no power

45. 14.
At high frequencies, capacitor becomes a short-circuit and an inductor
becomes an open circuit. Explain. [2069 Set A]
𝑋𝑐 =
1
𝜔𝐶
=
1
2𝜋𝑓𝐶
𝑋𝐿 = 𝜔𝐿 = 2𝜋𝑓𝐿
𝑅 ≈ 0
𝑅 ≈ ∞

46. 15.
Sketch the symbols of "a capacitor", "an inductor" , "emf of a cell"
and "a resistor". [Set 'B' 20691

47. 14
Long-distance, electric power, transmission lines always operate at very high
voltage, some time as much as 750 K. V. What are the advantages of such high
voltages? [2nd Exam 2068]
P = IV = I2R
At Long distance R is significant

48. Numericals

49. 1.
A coil of inductance O.1H and negligible resistance is in series with a resistance 40Ω.
A supply voltage of 50V (rms) is connected to them. If the voltage across L is equal
to that across R, calculate the voltage across the inductor and frequency of the
supply. [2074]
C I V I L
VR
E
VL
Φ
𝐸2 = 𝑉2
𝑅 + 𝑉2
𝐿
𝑉𝑅 = 𝑉𝐿
50V
O.1H
40Ω.
𝑽𝑳 = 𝟑𝟓. 𝟑𝟓V
𝒇 = 𝟔𝟑. 𝟕𝑯𝒛
Phasor Diagram

50. 2.
An iron cored coil of inductance 3H and 50Ω resistance is placed in series with a
resistor of 550Ω and a 100V , 50Hz ac supply is connected across the arrangements.
Find the current flowing in the coil and voltage across the coil. [Supp 2073]
3H and 50Ω
550Ω
100V , 50Hz
𝑍 = (𝑅 + 𝑟)2 + 𝑋𝐿2
𝐼 =
𝐸
𝑍
𝑉𝐿 = 𝐼𝑋𝐿 = 𝐼 𝑟2 + 𝑋𝐿2
𝑰 = 𝟎. 𝟎𝟖𝟗𝟓
𝑽𝑳 = 𝟖𝟒. 𝟒𝟕𝑽

51. 3.
L-C-R alternating current series circuit of L = 1H, C = 1𝝁F and R =
100Ω are connected in series With a source of frequency 50 Hz. What
is the phase shift between current and voltage ? [2073 ‘C’]
1H
1𝝁F
100Ω
50 Hz
φ
VR
VL
Vc
𝑉𝐿 − 𝑉𝐶
𝑇𝑎𝑛φ =
𝑋𝐿 − 𝑋𝐶
𝑅
φ = −90𝑜
C I V I L

52. 4.
A coil having inductance and resistance is connected to an oscillator
giving a fixed sinusoidal output voltage of 5V rms. With the oscillator set
at a frequency of 50Hz, the rms current in the coil is 1A and at a
frequency of 100Hz, the rms current is 0.625,A. Determine the
inductance of the coil. [2072'c']
z1=
𝐸
𝐼1
z2=
𝐸
𝐼2
𝑍 = 𝑅2 + 𝑋𝐿2 𝑍2 = 𝑅2 + 𝑋𝐿2
𝑍1
2 = 𝑅2 + 𝑋1𝐿
2 𝑍2
2 = 𝑅2 + 𝑋2𝐿
2
∴ 𝐿 = 1.14 × 10−2H
𝑍1
2 − 𝑍2
2 = 𝑋1𝐿
2 − 𝑋2𝐿
2

53. 5.
A circuit consists of a capacitor of 2𝝁F and a resistor of 1000Ω an
alternating emf of 12V (rms) and frequency 50Hz is applied. Find the
current flowing, the voltage across capacitor and the phase angle between
the applied emf and current. [2071 ‘C’]
𝑧 = 𝑅2 + 𝑋𝐶
2
= 𝟏𝟖𝟕𝟗. 𝟔𝟑𝛀
2𝝁F
1000Ω
12V
50Hz
𝐼 =
𝐸
𝑍
= 𝟔. 𝟑𝟖 × 𝟏𝟎𝟏𝟎−𝟑𝑨
𝑉𝑐 = 𝐼𝑋𝐶
𝑇𝑎𝑛φ =
𝑋𝐶
𝑅
𝝓 = 𝟓𝟕. 𝟖𝟓
𝑶
= 𝟏𝟎. 𝟏𝟔𝐕

54. 6.
An iron cored coil of 2 H and 50 Ω resistance placed in series with a resistor of
450Ω and 200v, 50Hz a.c. supply is connected across the arrangement, find a)
the current flowing in the coil, b) its phase angle relative to the voltage supply,
c) the voltage across the coil. [Supp. 2070]
𝑧 = 𝑅2 + 𝑋𝐿
2
= 𝟖𝟎𝟐. 𝟗𝛀
𝐼 =
𝐸
𝑍
= 𝟎. 𝟐𝟓𝑨
𝑉𝐿 = 𝐼𝑍𝐿
𝑇𝑎𝑛φ =
𝑋𝐿
𝑅
𝝓 = 𝟓𝟏. 𝟓
𝑶
= 𝟏𝟓𝟕. 𝟓𝐕
50Ω
2H
450Ω
200V, 50Hz
𝑟2 + 𝑋𝐿
2

55. 7.
A 50V a.c. supply is connected to a resistor having resistance 50Ω in series
with a solenoid whose inductance is 0.25H. The potential difference
between the ends of the resistor is 25V. Find the resistance of the wire of
solenoid. Take frequency of the ac source 50Hz. [2070 ‘C’]
𝐈 =
𝐄
𝐙
=
𝐄
𝐑+𝐫 𝟐
+ 𝛚𝐋 𝟐
50V
50Ω 0.25H
50Hz
25V
IR = 25
r = 11.89 Ω

56. 8.
Alternating voltage in an ac circuit is represented by
E = 100 𝟐 sin (100 𝝅𝒕) volts. Find its root mean
square value and the frequency. [Set 'B' 2069]
𝑓 =
𝜔
2𝜋
= 50Hz
Erms=
𝐸𝑜
2
= 100V
E = Eo sin ωt

57. 9.
A 100 V, 50Hz AC source is connected to an LCR circuit containing L=8.1 mH, C =
12.5 𝝁𝑭 𝒂𝒏𝒅 𝑹 = 𝟏𝟎 𝜴 all connected in series. Find the potential difference across
the resistor. [1st Exam 2068]
8.1mH
1𝟐. 𝟓𝝁F
10Ω
100V, 50 Hz
𝑍 = 𝑅2 + 𝑋𝑐 − 𝑋𝐿 2
= 252.3Ω
𝐼 =
𝐸
𝑍
= 0.396 A
V = 𝐼𝑅 = 3.96𝑉

58. 10.
A circuit consists of an inductor of 200𝝁𝑯and resistance of 10 Ω in series with
a variable capacitor and a 0.10 V (r.m.s.), 1.0 MHz supply. Calculate (i) the
capacitance to give resonance (ii) the quality factor of the circuit at resonance.
[2nd Exam 2068]
𝑋𝐿 = 𝑋𝐶
𝐶 = 1.266 × 10−10 𝐹
𝑄 =
1
𝑅
𝐿
𝑅
= 1256.89

59. Chemical Effect
of Current

60. Short Questions

61. 1.
The conductivity of an electrolyte is very low as
compared to a metal at room temperature, Why?
Hints:
Ion density < electron density
Drift velocity(vd ) of ions is smaller
Resistance offered in electrolyte is more.

62. 2.
Why does voltammeter measure current more
accurately than an ammeter?
Hints
In a voltmeter, I =
𝑚
𝑧𝑡
, m can be measured upto third
decimal place and z is known upto five decimal places, so
value of I can be calculated with a very small error as
compared to that by an ammeter.

63. 3.
What is meant by Faraday constant ?
The quantity of charge required to liberate one gram
equivalent of a substance during electrolysis.
Or
Charge required to liberate one mole of a monovalent
element during electrolysis.
Its value is 96500C mol-1

64. 4.
Define one Faraday.
It is defined as 96500 C of charge required to deposit or
liberate one gram equivalent of a substance.

65. Numericals

66. 1.
Assuming Faraday constant to be 96500 C/mole and
relative atomic mass of copper 63, calculate the mass of
copper liberated by 0.5A current in 10 minutes. [2056]
Solution
For copper, m =
𝐴𝐼𝑡
𝐹𝑉
=
63 ×0.5×600
96500×2
Valency
𝒎 = 𝟗. 𝟖 × 𝟏𝟎−𝟓
Kg

67. 2.
In a copper plating system, an electrolysis current of 3
A is used. How many atoms of Cu++ are deposited in
1.5 hr? ( e = - 1.6 × 10-19 C)
Solution
Charge passed in 1.5 hr , Q = It = 3 × 5400 C
Charge carried by each Cu++ atom =2 × 1.6 × 10-19 C
No. of atoms deposited =
3×5400
2×1.6×10−19 = 5.06 × 1022

68. MCQ
Level 1
https://forms.office.com/Pages/ResponsePage.aspx?id=qZTHiefnX0yp8QngrubZNE32HdK_qMpMuj
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69. DID YOU ENJOYE THE CLASS?
Leave your valuable suggestions so that I will be
better for you all in the next class. Your
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NO?
Yes?

71. 71
“Thousands and thousands are dying due to
coronavirus and yet when this thing is over the
humanity will declare a victory! What victory?
Fools! There is no victory for the dead people!
And when it comes to the living people,
everyone should have a deep sadness in their
soul, not a brag of victory!”

72. STAY SAFE!
72

73. 73