The document Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev is a part of the Class 12 Course Physics For JEE.

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**Series L-R Circuit**

- Now consider an ac circuit consisting of a resistor of resistance R and an inductor of inductance L in series with an ac source generator.
- Suppose in phasor diagram, current is taken along positive x-direction. The V
_{R}is also along positive x-direction and V_{L}along positive y-direction as we know that potential difference across a resistance in ac is in phase with current and it leads in phase by 90º with current across the inductor, and as we know V_{R}= i_{0}R & V_{0}= i_{0}X_{L}

i = i_{0}sin wt

V_{R}(t) = i_{0}Rsin ωt

V_{L}(t) = i_{0}X_{L}sin (ωt + p/2)

hence we can write

V(t) = i_{0}R sin ωt + i_{0}X_{L}sin (ωt + p/2)

V_{0}= i_{0}

where is known as impedence (z) of the circuit.

now we can write

where tan β=

hence β =

**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**

**Series**** C-R Circuit**

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 x-direction. Then V_{R} is also along positive x-direction but V_{C} is along negative y-direction 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 R-C 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 C-R 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**

**L.C. Circuit**

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 x-direction. Then V_{R} is along positive x-direction, V_{L} along positive y-direction and V_{C} along negative y-direction, 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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