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1. Welcome to
Classes
BYJU’S
Alternating Current
What you already know
What you will learn
S6: Power in AC circuits
1 . Pure AC circuits
2 . RC and L R AC circuits
3 . Im pe dance
1 . Total work done in a
cy cle
2 . Av e rage po we r de liv e re d
in pu re re sistiv e and
pu re re activ e circu its
3 . Powe r in RC and RL
circu its

2. Impedance
The relation b/w peak current and peak voltages can be written as 𝑖0 =
E0
𝑍
𝑍 = 𝑅 𝑍 = 𝜔𝐿
𝑍 is called impedance.
Impedance is defined as the opposition any circuit presents when voltage is applied to it.
SI unit is Ohm (Ω)
𝑍 = 1/𝜔𝐶

3. RC combination in AC circuits
𝑍 = 𝑅2 + 1/𝜔𝐶 2
tan 𝜙 =
1
𝜔𝐶𝑅
𝑖0 =
E0
𝑍
=
E0
𝑅2 + 1/𝜔𝐶 2
Steady state current (𝑖) in the circuit,
𝑖 =
E0
𝑍
sin(𝜔𝑡 + 𝜙)
E0 sin 𝜔𝑡
𝐶
𝑅
The current leads the emf by 𝜙.
𝑉𝑅
𝜙
𝑉𝐶 E0
𝑖0
𝜙
𝑋𝐶
𝑅
𝑍

4. LR combination in AC circuits
tan 𝜙 =
𝜔𝐿
𝑅
Steady state current (𝑖) in the circuit
𝑖 =
E0
𝑍
sin(𝜔𝑡 − 𝜙)
E0 sin 𝜔𝑡
𝐿
𝑅
𝑉𝑅
𝜙
𝑉𝐿
E0
𝑖0
𝑍 = 𝑅2 + 𝜔𝐿 2
𝑖0 =
E0
𝑍
=
E0
𝑅2 + 𝜔𝐿 2
The current lags the emf by 𝜙.
𝜙
𝑋𝐿
𝑅
𝑍

5. A series R-C circuit is connected to an alternating voltage source. Consider two situations
(a) When capacitor is air filled.
(b) When capacitor is mica filled.
Current through resistor is 𝐼 and voltage across capacitor is 𝑉 then-
a b c
𝑉
𝑎 = 𝑉𝑏 𝑉
𝑎 < 𝑉𝑏 𝑉
𝑎 > 𝑉𝑏 𝐼𝑎 > 𝐼𝑏
d

6. 𝑉
𝑎 > 𝑉𝑏
𝑋𝐶 ∝
1
𝐶
When capacitor is filled with mica, its capacitance increases.
As 𝐶 increases, 𝑋𝐶 decreases
As 𝑋𝐶 decreases, voltage across capacitor decreases (𝑋𝐶 ∝ 𝑉).
𝐶 𝐶′
𝐶′ > 𝐶
𝑋𝐶 =
1
𝜔𝐶
a
𝑉
𝑎 = 𝑉𝑏 𝑉
𝑎 < 𝑉𝑏 𝑉
𝑎 > 𝑉𝑏 𝐼𝑎 > 𝐼𝑏
d
c
b

7. An A.C. voltage is applied to a resistance 𝑅 and an inductor 𝐿 in series. If 𝑅 and the
inductive reactance are both equal to 3 Ω, the phase difference between the applied
voltage and the current in the circuit is
E0 sin 𝜔𝑡
𝑋𝐿 = 3 Ω
𝑅 = 3 Ω
a b c d
Zero 𝜋/6 𝜋/4 𝜋/2

8. tan 𝜙 =
𝑋𝐿
𝑅
tan 𝜙 =
𝑋𝐿
𝑅
= 1
𝜙 = 45° or 𝜋/4
E0 sin 𝜔𝑡
𝑋 = 3 Ω
𝑅 = 3 Ω
𝜙
𝑋𝐿
𝑅
𝑍
a b d
Zero 𝜋/6 𝜋/4 𝜋/2
c

9. In an A.C. circuit, an alternating voltage, 𝜀 = 200 2 sin(100𝑡) volt is connected to
capacitor of capacity 1 𝜇𝐹. The r.m.s. value of current in the circuit is
𝜀 = 200 2 sin(100𝑡)
1 𝜇𝐹
a b c d
10 𝑚𝐴 100 𝑚𝐴 200 𝑚𝐴 20 𝑚𝐴

10. Alternating voltage, ε = 200 2 sin(100𝑡)
comparing with ε = 𝜀0 sin 𝜔𝑡
𝜔 = 100 𝑟𝑎𝑑/𝑠 𝜀0 = 200 2 𝑣𝑜𝑙𝑡
𝑋𝐶 =
1
𝜔𝐶
=
1
100 × 10−6
Ω = 104 Ω
𝐼0 =
𝜀0
𝑋𝐶
=
200 2
104
𝐴 = 2 2 × 10−2
𝐴 𝐼𝑟𝑚𝑠 =
𝐼0
2
=
2 2×10−2
2
= 2 × 10−2𝐴 = 20 𝑚𝐴
𝜀 = 200 2 sin(100𝑡)
1 𝜇𝐹
𝜀 = 𝜀0 sin 𝜔𝑡
a b c
10 𝑚𝐴 100 𝑚𝐴 200 𝑚𝐴 20 𝑚𝐴
c

11. E = E0 sin 𝜔𝑡
𝑖 = 𝑖0 sin(𝜔𝑡 + 𝜙)
𝑑𝑊 = E0𝑖0 sin 𝜔𝑡 sin(𝜔𝑡 + 𝜙) 𝑑𝑡
= E0𝑖0 sin2
𝜔𝑡 cos 𝜙 + sin 𝜔𝑡 cos 𝜔𝑡 sin 𝜙 𝑑𝑡
sin(𝜔𝑡 + 𝜙)
Total work done in a cycle is,
𝑊 = E0𝑖0 cos 𝜙 ‫׬‬
0
𝑇
sin2 𝜔𝑡 𝑑𝑡 + E0𝑖0 sin 𝜙 ‫׬‬
0
𝑇
sin 𝜔𝑡 cos 𝜔𝑡 𝑑𝑡
= E0𝑖0 cos 𝜙 ‫׬‬
0
𝑇
(1 − cos 2𝜔𝑡) 𝑑𝑡 + E0𝑖0 sin 𝜙 ‫׬‬
0
𝑇 1
2
sin 2𝜔𝑡 𝑑𝑡
𝑊 =
1
2
E0𝑖0 𝑇 cos 𝜙
The work done by source in time interval 𝑑𝑡 is,
𝑑𝑊 = E 𝑖 𝑑𝑡 𝐿 𝐶 𝑅
=
E0𝑖0 cos 𝜙
2
‫׬‬
0
𝑇
𝑑𝑡 − ‫׬‬
0
𝑇
cos 2𝜔𝑡 𝑑𝑡 + E0𝑖0 sin 𝜙 ‫׬‬
0
𝑇 1
2
sin 2𝜔𝑡 𝑑𝑡

12. Total work done in a cycle is,
𝑊 =
1
2
E0𝑖0 𝑇 cos 𝜙
Average Power delivered,
𝑃𝑎𝑣𝑔 =
𝑊
𝑇
=
E0
2
𝑖0
2
cos 𝜙
𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙
𝑃𝑎𝑣𝑔 =
1
2
E0 𝑖0 cos 𝜙
E = E0 sin 𝜔𝑡
𝑖 = 𝑖0 sin(𝜔𝑡 + 𝜙)
𝐿 𝐶 𝑅

13. E = E0 sin 𝜔𝑡
𝑖 = 𝑖0 sin(𝜔𝑡 + 𝜙)
𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙
Power Factor
For purely resistive circuit,
𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 0°
𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠
Power drawn is maximum in a purely resistive circuit
𝑅
𝜀 = 𝜀0 sin 𝜔𝑡
𝑡
𝑖, 𝜀
0 𝜋 2𝜋
𝜋
2
3𝜋
2
𝜀0
𝑖0

14. 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙
Power Factor
For purely reactive circuit -
𝑡
𝑖, 𝜀
0 𝜋 2𝜋
𝜋
2
3𝜋
2
𝜀 = 𝜀0 sin 𝜔𝑡
𝑖 = 𝑖0 sin 𝜔𝑡 −
𝜋
2
𝜀0
𝑖0
𝑖0
𝑡
𝑖, 𝜀
0 𝜋 2𝜋
𝜋
2
3𝜋
2
𝜀 = 𝜀0 sin 𝜔𝑡
𝜀0
𝑖0
𝑖0 𝑖 = 𝑖0 sin 𝜔𝑡 +
𝜋
2
𝜙 = 𝜋/2 𝜙 = −𝜋/2
𝜙 = 𝜋/2 or 𝜙 = −𝜋/2
𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos ±
𝜋
2
𝑃𝑎𝑣𝑔 = 0
No power is absorbed for a full cycle in purely
inductive or purely capacitive circuits

15. |
𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙
E0 sin 𝜔𝑡
𝐶
𝑅
cos 𝜙 =
𝑅
𝑍
cos 𝜙 =
𝑅
𝑅2 + 1/𝜔𝐶 2
Average Power in RC circuits is,
𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠𝑖𝑟𝑚𝑠
𝑅
𝑍
𝜙
𝑋𝐶
𝑅
𝑍

16. 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙
cos 𝜙 =
𝑅
𝑍
cos 𝜙 =
𝑅
𝑅2 + 𝜔𝐿 2
Average Power in LR circuits is,
𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠𝑖𝑟𝑚𝑠
𝑅
𝑍
E0 sin 𝜔𝑡
𝐿
𝑅
E0 sin 𝜔𝑡
𝜙
𝑋𝐿
𝑅
𝑍
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