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Alternating
Current and LCR circuit
1. Consider alternating
currents set up in an LCR (inductance-capacitance-resistance)
circuit by a time varying emf, E = E0 sin wt,
where w
is the angular frequency. The corresponding alternating current is
given by I = I0 sin (wt - f),
where
f is the phase difference between the emf and the current, and I0
is the peak value of the current in
the circuit.
2. For such a circuit the
average power dissipated is given by:
Pav = 1/2 E0 I0 cosf,
or
where Erms =
E0/ and
Irms - I0/
are the root mean square (rms) values of the emf and the current
respectively. Cosf
is the power factor. Hence,
a. RMS value of emf,  ,
(E0 is
the peak value or amplitude of the sinusoidal time varying voltage).
b. RMS value of current,  ,
(I0 is
the peak value of the current).
c. Power factor = cosf,
(f is
the phase difference between the emf and the current).
3. The overall ‘resistance’
of the LCR circuit is called Impedance, which is defined as
 .
a. XL
is called the Inductive Reactance. XL = wL
= 2p¦L,
where
w
and
¦
are the angular frequency, and frequency, respectively of the emf,
and L is the value of the inductance in Henry.
b. XC is called the Capacitive Reactance.  ,
where C is the value of the capacitance in Farad.
4. The peak value of the
current is related to the peak value of the emf via:
5. The voltage across L is simply VL = IXL.
VL always leads the current I by p/2.
6. Similarly, the voltage across C is VC = IXC.
VC always lags behind the current I by p/2.
7. The phase difference
between the emf (which is made up of VR, VL,
and VC)
and the current is given by
 .
- If XL > XC,
f
is positive. This means that the emf leads the current by f.
- If XL < XC,
f
is
negative. This implies that the emf lags behind the current by
f.
8. Note that the value of the
current I is maximum (= I0)
when the impedance Z
is minimum, or when XL = XC.
This condition can be written as
w
= 1/ |
w0
= 1/
is the natural frequency of oscillation of the LCR circuit and the
condition w
= w0
is the condition for resonance of the circuit.
Examples:
A. The resonance frequency
of an oscillator, whose inductance and capacitance values are 
H and 25
mF
respectively, is
[TNPCEE 1997]
1. 3140 Hz
2. 314 Hz
3. 10-3
Hz
4. 103 Hz
We know that for resonance  .
Hence,
B. A fuse having a current
rating of 6 A indicates that the current through the fuse wire can
have a peak value of
[TNPCEE
1997]
1. 7.07 A
2. 8.48 A
3. 0.707 A
4. 84.8 A
C. In an AC circuit with
capacitance only, the current
[TNPCEE
1997]
1. leads the voltage by 
2. leads the voltage by p
3. lags behind the voltage by
4. Is in phase with the voltage
The current leads the voltage by .
D. When 100
V AC of frequency 50 Hz is applied to a coil, a current of 0.5 A
flows through it. What is the impedance of the coil?
[TNPCEE
2000]
1. 50 W
2. 200 W
3. 2500 W
4. 4 W
Impedance,  .
Note that the
inductance of the coil can be determined by using
E. In an LCR
circuit in series with a 220 V, 50 Hz AC, it is observed that L =
0.2 H, C = 
F and R = 35 W.
Then in the circuit
[TNPCEE 1999]
1. current leads voltage by 60°
2. voltage leads current by 60°
3. current leads voltage by 30° 4. current and voltage are in
phase
In an LCR circuit, the phase difference F
between the voltage and current is given by  .
Now, 
and  .
Hence, tan F
= 0.
In other words, the current and the voltage are in phase with each
other.
F. A capacitor, an inductor
and a 30 W
resistor are connected in series with a 220 V, 50 Hz AC. If the
reactance of the circuit is
-40 W,
what is the RMS value of the current in the circuit?
[TNPCEE
1998]
1. 4.40 A
2. 50 A
3. 3.33 A
4. 5.50 A
Impedance, 
Hence, 
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