Use voltmeter, ammeter, wattmeter to determine active, reactive and apparent power
consumed in given R-L-C series circuit. Draw phasor diagram

Experiment No.: 4

Experiment Name:

Use of Voltmeter, Ammeter, Wattmeter to Determine Active, Reactive and Apparent Power
Consumed in Given R-L-C Series Circuit and Draw Phasor Diagram.


  1. To use voltmeter, ammeter, wattmeter to determine active, reactive and apparent power
    consumed in given R-L-C series circuit.
  2. To draw the phasor diagram of that circuit.


Let V volts (rms) be applied to an R-L-C series circuit and the current flowing through the circuit be I amps. The impedance Z=V/I. If voltage across resistor is VR then R=VR/I

If voltage across inductor is V r,L, the impedance of the inductor coil is given by


Again the voltage across the resistance R and coil VR, L, then

Vr,L/I=√((R+r)2+XL2 )

From the above two equation the value of R and XL can be found out. If the voltage across the capacitor is VC then the reactance of the capacitor is given by XC=VC/I. hence the value of capacitance C=1/ωVC where ω=2πf is the angular frequency. the power factor is given by (R+r)/Z. If VC>VL, The pf is leading otherwise lagging.

The power consumed P=I2(R+r) or VI cos Ø

Circuit Diagram:


  1. The circuit is connected as shown in the circuit diagram with resistor, the inductor coil and capacitor in series.
  2. The variac is adjusted to zero output position and the circuit is switched on.
  3. A suitable voltage is applied from the variac so that a reasonable current flows through the circuit. The output voltage of the variac and voltage across the resistor, inductor and the capacitor are noted along with the current.
  4. Different readings are taken by varying voltage from the variac.
  5. Readings are noted from the data sheet.

Observation Table:

Sl noVoltage (V)Current (mA)Frequency (kHz)R (Ω)L (mH)C (nF)VR (V)VR (V)VC (V)PF = cos ØActive power
(mW) [VIcosØ]
Reactive power
(mVAR) [VIsinØ]
Apparent power
(mVA) [VI] 5467.7522.08  4.674.22 4.20.9844.719.034 45.62


XL = 2πfL = 2π×13.05×103×7.75×10-3= 635.46 Ω

XC = 1/2πfC = 1/(2π×13.05×103×22.08×10-9)= 552.34 Ω

cos Ø = R/Z = R/√(R2+(XL – XC)2) = 546/√(5462+(635.46 – 552.34)2) = 546/552.29 = 0.98 lag

Active power = VIcosØ = 5.02×9.09×10-3×0.98 = 44.71 mW

Rective power = VIsinØ = 5.02×9.09×10-3×0.198 = 9.034 mVAR

Apparent power = VI = 5.02×9.09×10-3 = 45.62 mVA

Phasor Diagram:

Apparatus Used:

Sl. No.Name of the apparatusQuantitySpecificationMakes name
1.Single phase variac1230/0-270 V ac
2.LCR Trainer Kit Sushama Electronics 
3.Digital Multimeter30-750 V, 10 A Akademika 
5.LCR MeterDigital Metravi 
6. Connecting Probes10  RGP-2 


  1. The phasor diagram of different voltage and current for at least two observation are drawn in the graph paper.
  2. The value of power factor and magnitude of the supplied voltage are found from phasor diagram.

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