Class 12 Physics Alternating Currents Notes

Class 12 Physics Alternating Currents Notes
Chapter 19 – Alternating Currents | Nepal eNotes
PHYSICS • CHAPTER 19

Alternating Currents

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PDF Page 1

Alternating Current

The current whose value and direction changes with time is called alternating current.

The instantaneous value of alternating current is:

I = I0 sinωt

The instantaneous value of alternating emf is:

E = E0 sinωt

Graph of alternating current

I t T

Advantages AC over DC

  • AC is cheaper than DC.
  • AC can be transmitted to longer distance.
  • AC can be converted AC.
  • High voltage AC can be converted to low voltage and vice-versa.
The third point above is preserved exactly as written in the source page.
PDF Page 2

AC through resistor

R E = E₀ sinωt

Consider a resistor of resistance ‘R’ is connected in series with an ac source as shown in figure.

The alternating emf is given by:

E = E0 sinωt    …(1)

Since,

I = E/R
= E0 sinωt / R
I = I0 sinωt    …(2)

From eqn (1) and (2), it is clear that E and I are in same phase.

Phasor diagram of R-circuit

I E
PDF Page 3

Fig: wave diagram of R-circuit

AC through Inductor

L

Consider a inductor of resistance ‘L’ is connected in series with an ac source as shown in figure.

The alternating emf is given by:

E = E0 sinωt    …(1)

The emf induced in just inductor is:

Eb = −L dI/dt    …(2)

Applying Kirchhoff’s law,

E + Eb = 0
PDF Page 4
E0 sinωt − L dI/dt = 0
E0 sinωt = L dI/dt
(E0/L) sinωt dt = dI
(E0/L) ∫ sinωt dt = ∫ dI
(E0/L)(−cosωt/ω) = I
(E0/ωL) sin(ωt − π/2) = I
(E0/XL) sin(ωt − π/2) = I

where,

XL = ωL    is inductive reactance
I = I0 sin(ωt − π/2)    …(3)

From eqn (1) and (3), it is clear that I lags behind E by π/2.

I E π/2
PDF Page 5

AC through Capacitor

Consider a capacitor of capacitance ‘C’ is connected in series with an ac source as shown in figure.

Alternating emf is given by:

E = E0 sinωt    …(1)

Since,

q = CE = CE0 sinωt
Then,
I = dq/dt = d(CE0 sinωt)/dt
I = CE0 d(sinωt)/dt
C
PDF Page 6
I = CE0 cosωt · ω
I = ωCE0 sin(ωt + π/2)
I = E0 / (1/ωC) · sin(ωt + π/2)
I = (E0/XC) sin(ωt + π/2)

where,

XC = 1/ωC    is capacitive reactance
I = I0 sin(ωt + π/2)    …(3)

From eqn (1) and (3), it is clear that I leads E by π/2.

E I π/2
PDF Page 7
The final two pages of the supplied “Chapter 19 – Alternating Currents” PDF switch to spring/SHM notes. They are preserved below because they are physically present in the source file.

From Hooke’s law:

F = −ky
F = ma
ma = −ky
a = −(k/m)y
a ∝ −y
F1 ∝ e
F1 = −ke
F2 ∝ (e + y)
F2 = −k(e + y)

As F1 = −ke    …(1)

The spring is displaced by y.

F2 = −k(e + y)
= −ke
PDF Page 8
m m e y

Here,

F1 ∝ e
F1 = −ke
F2 ∝ (e + y)
F2 = −k(e + y)

Suppose a load of mass (m) is attached to free end of spring & extension e is produced.

The restoring force (F1) = −ke    …(1)

The spring is displaced by y.

Restoring force (F2) = −k(e + y)
F2 = −ke − ky    …(2)
The page ends after this step; the continuation of the derivation is not present in the supplied PDF.
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