Page 2 - Op. Sheet

Boucherot's theorem

Electrical Diagrams

Fig.3.1

Components List

R1 = 150 Ohm - 4W- body in ceramics
R2 = 100 Ohm - 4W - body in ceramics
R3 = 150 Ohm - 4W - body in ceramics
L1 = 4mH-1,8Ohm - toroidal
L2 = 4mH-1,8Ohm - toroidal
C1 = 15µF-63V - polyester

Calculation data

Total active power:	P_t = P₁+P₂+P₃
Total apparent power:	S_t = U · I₁
Total power factor:	cosj_t = P_t / S_t

Topographical diagram

Fig.3.2

Obtained Results

Tab. 3.2

I_read

I_1v_w

I₁

I₂

I₃

P_read

Ptvw

DIV

COST

DIV

COST

DIV

COST

Tab. 3.3
P₁=R₁·I₁²	P₂=R₂·I₂²	P₃=R₃·I₃²	P_t = P₁+P₂+P₃
W	W	W	W

EXPERIMENTATION

	insert the Module 8 in the console and set the main switch to ON;
	connect the ammeters, the voltmeter and the wattmeter as it is shown in Fig. 3.2;
	record the instrument indications and write them in Tab. 3.2;
	calculate the total actual, apparent power, the cosj_t and write the values in Tab. 3.2: the circuit power is practically entirely active;
	calculate the actual powers P₁, P₂, P₃ and write the values in Tab. 3.3;
	calculate the total actual power (P_t = P₁ + P₂ + P₃) and write the values in Tab. 3.3;
	compare the value of the total actual power with the one measured with the wattmeter: the values are equal.

QUESTIONS

The Boucherot’s theorem states that the active powers sum :
	Algebraically	Answer:
	Arithmetically
	Vectorially

An electrical line supplies three loads two of which ohmic-inductive and one ohmic-capacitive. The total reactive power is equal to :
	Q_t = Q₁+ Q₂- Q₃	Answer:
	Q_t = Q₁- Q₂+ Q₃
	Q_t = Q₁- Q₂- Q₃

TROUBLESHOOTING
Push the INSERT button to insert the fault in the circuit. Repeat the operations in the EXPERIMENTATION section to find the fault inserted in the circuit.

Which is the fault ?
	R1 faulty (in loss)	Answer:
	C1 short-circuited
	The circuit is not supplied
	L2 interrupted

Remove all the connections.