To study series and parallel R-L-C circuit.

Experiment No.: 17

Aim of The Experiment: –

To study series and parallel R-L-C circuit.

Objective:

To be written by student.

Apparatus Required: –

Sl. No.Equipment NameSpecificationMakeQuantity
1Single-phase auto transformer5KVA, (0-270)V1 Nos.
2Voltmeter(0-250)V, MI2 Nos.
3Voltmeter(0-150)V, MI2 Nos.
4Ammeter(0-5)A, MI1 Nos.
5Rheostat(0-80)Ω, 5A1 Nos.
6Inductor1 Nos.
7Capacitor1 Nos.
8Connecting WiresPVC Insulated CopperAs per required
Table 17.1

Circuit Diagram: –

Fig. 17.1
Fig. 17.2

Theory: –

In R-L-C series ad parallel AC circuit a resistor of resistance R ohm, an Inductor of inductance L henry and a Capacitor of capacitance C farad are connected across single phase ac supply of rms V volts as shown in above fig.

As the circuit shown is in series the current across each element R,L and C will remain the same, but the voltage drops among these elements will differ depending upon their respective capacity.

As the circuit shown is in parallel the current across each element R,L and C will differ depending upon their respective capacity, but the voltage drops among these elements will remain the same.

Let V = rms value of applied voltage and I = rms value of supply current flows in the circuit.

As shown in phasor diagram:

\[\color\red{V_R}\;\color\green{ =}\; \color\red{I\;R\;}\color\green{ where}\color\red{\; V_R}\color\green{\; is\;in\;phase\;with}\color\red{\;I,}\] \[\color\red{V_L}\;\color\green{ =}\;\color\red{ I\;X_L\;}\color\green{where,}\;\color\red{V_L}\;\color\green{leads\;}\color\red{I}\;\color\green{by\;}\color\magenta{90^0,}\] \[\color\red{V_c}\;\color\green{ =}\;\color\red{ I\;X_c}\;\color\green{where\;}\color\red{V_c\;}\color\green{lags}\;\color\red{I}\;\color\green{by\;}\color\magenta{90^0,}\]

It is better to say that the supply current I is in phase with VR , but lags with VL by 900 and leads with VC by 900.

From phasor diagram of R-L-C series circuit an impedance triangle can be drawn by dividing each side of phasor diagram by the same factor I, we get a triangle whose sides represent R, ( XL – XC ), Z.

The impedance triangle is a right angle triangle.

Fig. 17.4
Fig. 17.5

When |VL| > |VC|, the resultant of phasor VR and (VL – VC) will give the voltage across the impedance and should be equal to supply voltage.

\[\color\red{V_Z}\;\color\green{=}\;\color\red{\sqrt{V_R^2\;+\;(V_L-V_C)^2}}\] \[\color\green{=}\;\color\red{I\;Z}\] \[\color\green{=}\;\color\red{\sqrt{(IR)^2\;+\;(IX_L-IX_C)^2}}\] \[\color\green{=}\;\color\red{I\sqrt{R^2\;+\;(X_L-X_C)^2}}\]
\[\color\green{in\;parallel\;circuit\;}\color\red{ I_Z}\;\color\green{ =}\color\red{\; \sqrt{{I_R^2\;+\;(I_L-I_C)^2\ }}}\] \[\color\green{=}\;\color\red{\ \sqrt{\left\{\dfrac{V}{R}\right\}^2\;+\;\left\{\dfrac{V}{X_L}\;-\;\dfrac{V}{X_C}\right\}^2\ }}\] \[\color\green{=}\;\color\red{\dfrac{V}{Z}}\] \[\color\green{where,}\] \[\color\red{\; I_R\;=\;\dfrac{V}{R}}\] \[\color\red{\; I_L\;=\;\dfrac{V}{X_L}}\] \[\color\red{\;I_C=\dfrac{V}{X_C}}\]

Thus,

\[\color\red{Z}\;\color\green{=}\;\color\red{\sqrt{R^2\;+\;(X_L\;-\;X_C)^2}}\] \[\color\green{in\; parallel\;circuit\;}\color\red{ Z}\;\color\green{ =}\;\color\red{ \dfrac{V}{I_Z}}\]
\[\color\green{=}\;\color\red{ \dfrac{1}{\sqrt{{\left\{\dfrac{1}{R}\right\}^2}\;+\;{\left\{\dfrac{1}{X_L}\;-\;\dfrac{1}{X_C}\right\}^2}}}}\]

where Z is the impedance in ohm, XL is the inductive reactance in ohm and XC is capacitive reactance in ohm. The relation is shown in impedance triangle as shown in fig.: 17.5.

The phase angle of VZ is given by

\[ \color\red{\phi}\color\green{ =}\;\color\red{ tan^{-1}\;\left\{\dfrac{V_L-V_C}{V_R}\right\}}\] \[\color\green{=}\;\color\red{ tan^{-1}\left\{\dfrac{IX_L-IX_C}{IR}\right\}}\] \[\color\green{=}\; \color\red{tan^{-1}\left\{\dfrac{X_L-X_C}{R}\right\}}\] \[\color\green{ =} \;\color\red{tan^{-1}\left\{\dfrac{X}{R}\right\}}\]

where X is the net reactance in ohm.

from impedance triangle

\[\color\red{Cos\phi}\;\color\green{=}\;\color\red{\dfrac{R}{Z}}\] \[\color\green{=}\;\color\red{\dfrac{R}{\sqrt{R^2+(X_L-X_C)^2}}}\]

Procedures: –

  1. Connect all the instruments as per circuit diagram given above.
  2. Before switch on the main power supply make sure that single-phase auto transformer knob is at zero position.
  3. Now slowly increase the supply voltage to the circuit after giving supply to the single-phase auto-transformer.
  4. Take all the corresponding readings of the connected instruments in the circuit as per observation table.
  5. Now calculate VZ, power factor Cosɸ and % error as per formula given in observation table.

Observation Table: –

For R, L, C Series circuit-

Sl.  No. V 
(in volts.) 		I
(in Amp.) 		VR
(in volts.) 		VL
(in volts.) 		VC
(in volts.) 		VZ
(in volts.) 		Cosɸ= (VR/VZ) % Error
[{(V-VZ )/V}*100]
1
2
3
4
5
6
7
8
Table 17.2

For R, L, C parallel circuit-

Sl. No.V (in volts.)I (in Amp.)IR (in Amp.)IL (in Amp.)IC (in Amp.)IZ (in Amp.)Cosɸ= (IR/IZ)% Error
[{(I-IZ )/I}*100]
1
2
3
4
5
6
7
8
Table No. 17.3

Precaution: –

  1. Don’t switch on power supply without concerning teacher.
  2. Single phase autotransformer must be kept at minimum potential point before switch on the experiment.

Conclusion:

To be written by student.

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