Class 12 Physics Alternating Current | Phasor diagram for Series LCR circuit |

**Phasor diagram for Series LCR circuit **

- Resistor, inductor and capacitor all are in series.
- Source voltage V = V
_{m}sinωt and current I = I_{m}sin (ωt + f). - V
_{L}= voltage across inductor, V_{C}=voltage across capacitor, V_{R }= voltage across resistor and V = source voltage. - Peak values: - V
_{L}= I_{m}X_{L}, V_{C}= I_{m}X_{C}, V_{R }= I_{m}R and V =V_{m}. - Resistor V
_{R}and I are in phase .Inductor I lag behind the V_{L}. - In Capacitor V
_{C}lag behind the I_{L}. - From the phasor diagram we can see that the V
_{L}and V_{C}are exactly in opposite direction with each other and are in same line. - The length of the phasors represents the peak values (I
_{m}X_{L}, I_{m}X_{C}and I_{m}R) . - From the circuit , V
_{R}+ V_{C}+ V_{L}= V. - Refer figure(2) from the phasors a right triangle is obtained whose hypotenuse =V .
- Using Pythagoras theorem, V
^{2}_{R}+ (V_{C}- V_{L})^{2}= V^{2}_{m} - V
^{2}_{m}= (I_{m}R)^{2}+ [I_{m}X_{C}- I_{m}X_{L}]^{2} - = I
^{2}_{m}R^{2}+ I^{2}_{m}(X_{C}- X_{L})^{2} - V
^{2}_{m}= I^{2}_{m}[R^{2}+ (X_{C}- X_{L})^{2}] - V
_{m }= I_{m}√( R^{2}+ (X_{C}- X_{L})^{2}) - Comparing the above equation with V=IR , then R = √( R
^{2}+ (X_{C}- X_{L})^{2}). - Therefore
**Z = √( R**^{2}+ (X_{C}- X_{L})^{2})

(a) Relation between the phasors V_{L}, V_{R}, V_{C}, and I,

(b) Relation between the phasors V_{L}, V_{R}, and (V_{L} + V_{C}) for the circuit.

(a) Phasor diagram of V and I.

(b) Graphs of v and i versus w t for a series LCR circuit where X_{C} > X_{L}.

** Problem:- **A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 W, L = 25.48 mH, and C = 796 μF.

Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.

** Answer:**-

(a) To find the impedance of the circuit, we first calculate X_{L} and X_{C}.

X_{L} = 2 π ν L

= 2 × 3.14 × 50 × 25.48 × 10^{–3} W = 8 W

X_{C} = (1/2 π ν C)

= (1/ 2 × 3.14 × 50 × 796 x 10^{-6})

= 4Ω

Therefore,

Z = √R^{2} + (X_{L} − X_{C} )^{2} =√(3)^{2} + (8 − 4)^{2}

= 5 Ω

(b) Phase difference, f = tan^{–1} ϕ (X_{C} – X_{L})/(R)

=tan^{-1} ((4-8)/(3))

=-53.1^{0}

Since ϕ is negative, the current in the circuit lags the voltage across the source.

(c) The power dissipated in the circuit is

P^{2} = I R

Now, I = (I_{M})/(√2)

=(1/√2 )(283/5)

=40 A

Therefore, P = (40A )^{2}× 3W = 4800

(d) Power factor = cos ϕ = cos53.1° = 0.6

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