Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Physics Class 12

Class 12 : Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

The document Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev is a part of the Class 12 Course Physics Class 12.
All you need of Class 12 at this link: Class 12

4. 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 VR is also along positive x-direction and VL 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 VR = i0R & V0 = i0XL

i = i0 sin wt 

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

VR(t) = i0 Rsin ωt

VL(t) = i0 XL sin (ωt + p/2)

hence we can write

V(t) = i0R sin ωt + i0 XL sin (ωt + p/2)

V0 = i0Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

where Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev is known as impedence (z) of the circuit.

now we can write

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

where tan β= Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

hence β = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Ex.6 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 = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev with Z = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

So when dc is applied, ω = 0, so z = R

and hence i = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev i.e., R = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 200 ?

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRevi.e., ω2L2 = Z2-R2

i.e., (2πfL)2 = 2002 - 1002 = 3 × 104 (as ω = 2πf)

So, L = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 0.55 H

Ex.7 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 = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 13 ohm

Current in the circuit i = E/Z = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 10 amp

Potential difference across resistance

VR = iR = 10 × 12 = 120 volt

Inductive reactance of coil XL = ωL = 2πfL

Therefore, XL = 2π × 50 × Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 5 ohm

Potential difference across inductance

VL = i × XL = 10 × 5 = 50 volt

5. 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 VR is also along positive x-direction but VC 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,

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

VR = I0 R sin ωt Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Potential difference across capacitor

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Potential at any instant t

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

V(t) = V0 sin (ωt + b)

tan α = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Ex.8 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

irmsSeries L-R, C-R and C-R Circuit Class 12 Notes | EduRev ...(i)

When frequency is changed to ω/3, the current is halved. Thus

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev ...(ii)

From equations (i) and (ii), we have

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Solving this equation, we get Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Hence, the ratio of reactance to resistance is Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Ex.9 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 = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 200 ? and the maximum current i = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev ; so when the lamp is put in series with a capacitance and run at 200 V ac, from V =iZ we have,

Z = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 400?

Now as in case of C-R circuit, Z = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev,

i.e., R2Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 160000

or, Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 16 × 104 - (200)2 = 12 × 104

So, Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev × 102

or C = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

i.e., C = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev = 9.2 μF

5. 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 Xc is capacitance reactance,

XL is Inductance reactance,

i = i0 sin wt current flowing through the circuit

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

VC(t) = i0 XC sin (ωt - π/2)

VL(t) = i0 XL sin (ωt + π/2)

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

= i0 XC sin wt cos π/2 - i0 XC cos wt sin π/2 + i0 XL sin wt cos π/2 + i0 XL cos wt sin π/2

= i0 cos w t(XL - XC)

V(t) = V0sin (wt + p/2)

V0 = i0Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Z = (X- XC)

cos Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev= 0

VCO = i0 XCSeries L-R, C-R and C-R Circuit Class 12 Notes | EduRev

6. SERIES L-C-R CIRCUIT

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 VR is along positive x-direction, VL along positive y-direction and VC 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 = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

L - R - C circuit

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Impedance phasor of above circuit

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

 

& Impedance triangle

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

here B is phase angle By triangle tan β= Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Power factor cosSeries L-R, C-R and C-R Circuit Class 12 Notes | EduRevSeries L-R, C-R and C-R Circuit Class 12 Notes | EduRev Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Let I be the current in the series circuit of any instant then

(1) Voltage V(t) = V0 sin (ωt + β) = i0 z sin (ωt + β)

here v0 = i0z & vrms = irmsz

(2) Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev here voltage VL across the inductance is ahead of current I in phase by π/2 rad

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

(3) VC(t) = VO C sin (ωt - π/2)

here voltage VC across the capacitance lags behind the current I in phase by π/2 rad

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

(4) VR(t) = i0 R sin ωt

here voltage VR across the resistor R has same phase as I

VO R = IR

Special Case :

(1) When XL > XC or VL > VC then emf is ahead of current by phase β which is given by

tan β = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev or cos f = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

The series LCR circuit is said to be inductive

(2) When XL < XC or VL < VC then current is ahead of emf by phase angle β which is given by

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev or Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

The series LCR circuit is said to be capacitive

(3) When XL = XC or VL = VC, 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

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev or Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev or IRmsSeries L-R, C-R and C-R Circuit Class 12 Notes | EduRev

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) ;

xL = 10 ?, XC = 6 ?, R = 3 ?

Calculate Z, i0, irms, vrms, VL O, VC O, VR O, b,

VL Rms, VC Rms, VRms, i(t), VL(t), Vc(t), and VR(t)

 

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRevXL > XC

Sol. V = 10 sin (wt + p/4) so V0 = 10 volt

VrmsSeries L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Therefore, Z = Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRevSeries L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRevSeries L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRevSeries L-R, C-R and C-R Circuit Class 12 Notes | EduRev

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

i(t) = 20 sin (ωt + π/4 - 53°)

VL(t) = 20 sin (ωt + π/4 - 53° + π/2)

= 20 sin (ωt + Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev - 53° - Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev) = 12 sin (ωt - Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev - 53°)

Series L-R, C-R and C-R Circuit Class 12 Notes | EduRev

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

VR = iR = 5 × 16 = 80 volt

Potential difference across inductance

VLi × (wL) = 5 × 24 = 120 volt

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