Series L-R, C-R & C-R Circuit | Physics for JEE Main & Advanced PDF Download

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 = i₀R & V₀ = i₀X_L
  i = i₀ sin wt
   
  V_R(t) = i₀ Rsin ωt
  V_L(t) = i₀ X_L sin (ωt + p/2)
  hence we can write
  V(t) = i₀R sin ωt + i₀ X_L sin (ωt + p/2)
  V₀ = i₀
  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., ω²L² = Z²-R²

i.e., (2πfL)² = 200² - 100² = 3 × 10⁴ (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 = I₀ R sin ωt

Potential difference across capacitor

Potential at any instant t

V(t) = V₀ 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., R² +  = 160000

or,  = 16 × 10⁴ - (200)² = 12 × 10⁴

So,  =  × 10²

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 = i₀ sin wt current flowing through the circuit

V_C(t) = i₀ X_C sin (ωt - π/2)

V_L(t) = i₀ X_L sin (ωt + π/2)

= i₀ X_C sin wt cos π/2 - i₀ X_C cos wt sin π/2 + i₀ X_L sin wt cos π/2 + i₀ X_L cos wt sin π/2

= i₀ cos w t(X_L - X_C)

V(t) = V₀sin (wt + p/2)

V₀ = i₀Z

Z = (X_L - X_C)

cos = 0

V_CO = i₀ X_C ;

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 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) = V₀ sin (ωt + β) = i₀ z sin (ωt + β)

here v₀ = i₀z & v_rms = i_rmsz

(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) = i₀ 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, i₀, 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 V₀ = 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