Table of contents  
Series LR Circuit  
Series CR Circuit  
L.C. Circuit  
Series LCR Circuit 
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Example 1. When 100 volt dc is applied across a coil, a current of 1 amp flows through it; when 100 V ac of 50 Hz is applied to the same coil, only 0.5 amp flows. Calculate the resistance of inductance of the coil.
Sol. In case of a coil, i.e., L  R circuit.
i = with Z = =
So when dc is applied, ω = 0, so z = R
and hence i = i.e., R = = = 200 ?
i.e., ω^{2}L^{2} = Z^{2}R^{2}
i.e., (2πfL)^{2} = 200^{2}  100^{2} = 3 × 10^{4} (as ω = 2πf)
So, L = = = 0.55 H
Example 2. A 12 ohm resistance and an inductance of 0.05/p henry with negligible resistance are connected in series. Across the end of this circuit is connected a 130 volt alternating voltage of frequency 50 cycles/second. Calculate the alternating current in the circuit and potential difference across the resistance and that across the inductance.
Sol. The impedance of the circuit is given by
Z = =
= = = 13 ohm
Current in the circuit i = E/Z = = 10 amp
Potential difference across resistance
V_{R} = iR = 10 × 12 = 120 volt
Inductive reactance of coil X_{L} = ωL = 2πfL
Therefore, X_{L} = 2π × 50 × = 5 ohm
Potential difference across inductance
V_{L} = i × X_{L} = 10 × 5 = 50 volt
Now consider an ac circuit consisting of a resistor of resistance R and an capacitor of capacitance C in series with an ac source generator.
Suppose in phasor diagram current is taken along positive xdirection. Then V_{R} is also along positive xdirection but V_{C} is along negative ydirection as potential difference across a capacitor in ac lags in phase by 90º with the current in the circuit. So we can write,
V_{R} = I_{0} R sin ωt
Potential difference across capacitor
Potential at any instant t
V(t) = V_{0} sin (ωt + b)
tan α =
Example 3. An A.C. source of angular frequency w is fed across a resistor R and a capacitor C in series. The current registered is i. If now the frequency of the source is changed to ω/3 (but maintaining the same voltage), the current in the circuit is found to be halved. Calculate the ratio of reactance to resistance at the original frequency.
Sol. At angular frequency w, the current in RC circuit is given by
i_{rms} = ...(i)
When frequency is changed to ω/3, the current is halved. Thus
= ...(ii)
From equations (i) and (ii), we have
=
Solving this equation, we get
Hence, the ratio of reactance to resistance is
Example 4. A 50 W, 100 V lamp is to be connected to an ac mains of 200 V, 50 Hz. What capacitance is essential to be put in series with the lamp?
Sol. As resistance of the lamp R = = 200 ? and the maximum current i = = = ; so when the lamp is put in series with a capacitance and run at 200 V ac, from V =iZ we have,
Z = = = 400?
Now as in case of CR circuit, Z = ,
i.e., R^{2} + = 160000
or, = 16 × 10^{4}  (200)^{2} = 12 × 10^{4}
So, = × 10^{2}
or C =
i.e., C = = 9.2 μF
As shown in figure a capacitor and inductance are connected in series method and alternating voltage is applied across the circuit.
Let X_{c} is capacitance reactance,
X_{L} is Inductance reactance,
i = i_{0} sin wt current flowing through the circuit
V_{C}(t) = i_{0} X_{C} sin (ωt  π/2)
V_{L}(t) = i_{0} X_{L} sin (ωt + π/2)
= i_{0} X_{C} sin wt cos π/2  i_{0} X_{C} cos wt sin π/2 + i_{0} X_{L} sin wt cos π/2 + i_{0} X_{L} cos wt sin π/2
= i_{0} cos w t(X_{L}  X_{C})
V(t) = V_{0}sin (wt + p/2)
V_{0} = i_{0}Z
Z = (X_{L } X_{C})
cos = 0
V_{CO} = i_{0} X_{C} ;
Now consider an ac circuit consisting of a resistor of resistance R, a capacitor of capacitance C and an inductor of inductance L are in series with an ac source generator.
Suppose in a phasor diagram current is taken along positive xdirection. Then V_{R} is along positive xdirection, V_{L} along positive ydirection and V_{C} along negative ydirection, as potential difference across an inductor leads the current by 90° in phase while that across a capacitor, lags by 90°.
V =
L  R  C circuit
Impedance phasor of above circuit
& Impedance triangle
here B is phase angle By triangle tan β=
Power factor cos=
Let I be the current in the series circuit of any instant then
(1) Voltage V(t) = V_{0} sin (ωt + β) = i_{0} z sin (ωt + β)
here v_{0} = i_{0}z & v_{rms} = i_{rms}z
(2) here voltage V_{L} across the inductance is ahead of current I in phase by π/2 rad
(3) V_{C}(t) = V_{O C} sin (ωt  π/2)
here voltage V_{C} across the capacitance lags behind the current I in phase by π/2 rad
(4) V_{R}(t) = i_{0} R sin ωt
here voltage V_{R} across the resistor R has same phase as I
V_{O R} = I_{O }R
Special Case :
(1) When X_{L} > X_{C} or V_{L} > V_{C} then emf is ahead of current by phase β which is given by
tan β = or cos f =
The series LCR circuit is said to be inductive
(2) When X_{L} < X_{C} or V_{L} < V_{C} then current is ahead of emf by phase angle β which is given by
or
The series LCR circuit is said to be capacitive
(3) When X_{L} = X_{C} or V_{L} = V_{C}, b = 0, the emf and current will be in the same phase. The series LCR circuit is said to be purely resistive. It may also be noted that
or or I_{Rms} =
Susceptance : The reciprocal of the reactane of an a.c. circuit is called its susceptance.
Admittance : The reciprocal of the impedance of an a.c. circuit is called its admittance.
Ex.10 Figure shows a series LCR circuit connected to a variable voltage source V = 10 sin (ωt + p/4) ;
x_{L} = 10 ?, X_{C} = 6 ?, R = 3 ?
Calculate Z, i_{0}, i_{rms}, v_{rms}, V_{L O}, V_{C O}, V_{R O},_{ }b,
V_{L Rms}, V_{C Rms, }V_{Rms}, i(t), V_{L}(t), V_{c(t), }and V_{R(t)}
X_{L} > X_{C}
Sol. V = 10 sin (wt + p/4) so V_{0} = 10 volt
V_{rms} =
Therefore, Z =
;
;
;
i(t) = 20 sin (ωt + π/4  53°)
V_{L}(t) = 20 sin (ωt + π/4  53° + π/2)
= 20 sin (ωt +  53°  ) = 12 sin (ωt   53°)
Ex.11 A resistor of resistance R, an inductor of inductance L and a capacitor of capacitance C all are connected in series with an a.c. supply. The resistance of R is 16 ohm and for a given frequency the inductive reactance of L is 24 ohm and capacitive reactance of C is 12 ohm. If the current in the circuit is 5 amp., find
(a) the potential difference across R, L and C (b) the impedance of the circuit
(c) the voltage of a.c. supply (d) phase angle
Sol. (a) Potential difference across resistance
V_{R} = iR = 5 × 16 = 80 volt
Potential difference across inductance
V_{L} = i × (wL) = 5 × 24 = 120 volt
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