stevenal, agreed with your calculation that the PF of this circuit is .707
so. at the transformer's secondary side the power delivered to the circuit will be Irms.Vrmsx(.707)
Now what will be the exact tranformer's rating:
Is it the product of Irms.Vrms
Or the real power it delivers Irms.Vrmsx(.707)
JBinCA,
Thanks for the detailed explaination.
I still have one point is not clear.If the load is pure resistive,that means that the current and voltage are in phase and the power consumed by the load is real power,so if we calculate the power at the load should equal the power deliverd by the...
Actually, I did two calculations:
one at the load and the other at the transformer secondary.
The two calculations give different results.
No measurments are taken- Just mathematical calculations and neglect any losses.
This ciruit is very simple :
single phase transformer , its secondery is connected in series with a single diode and resistive load( simple half-wave rectifier)
The DC power Wdc = Idc x Vdc -- average values of the rectified voltage and current.
If you calculate the transformer's secondary VA...
We can use the following formulas for the instanious current and voltage:
(Esin x /R) and (Esin x) ,multiply them together and ,make the integral over one period cycle and devide over the period cycle and that would be the wattmeter reading( the mean value of the product of current and voltage)...
A single phase transformer is feeding a half-wave rctifier with a resistive load across it.
What is the real value of the power cosumed in the resistor load( dc + ripples ),is it to :
- Multiply transformer's secondary rms values of the voltage and current- assume 100% efficiency.
-Or to mutiply...
So,I understand that my equation should be corrected to:
Power=(Vdc+Vac_rms)'2/R
And the superposition theorem does not apply for power.
Is that what you mean ?
Thx for reply
Is that mean to measure the power by DVM we have to make 2 measurments across the resisor :
Vdc and Vac ,and then the power will be:
(Vdc'2 + Vac'2)/R
Regards
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