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Because the power of a resistive component is directly proportional to the square of the voltage across that component.
The power factor of a purely resistive circuit is 1.0.
No. If the load is truly resistive, just measure the voltage across the load (in volts)and the current flowing in it (in amps) and multiply them. eg: 115 volts at 1.5 amps = 172.5 watts.
With a pure resistive load the Power Factor should be 1.
Incandescent lamps are nearly pure resistive loads with a power factor of 1
Because the power of a resistive component is directly proportional to the square of the voltage across that component.
A 40 Watt bulb emits 40 Joules each second. ( the operating voltage is irrelevant )
You must know the current or resistance to convert voltage to power.
The power factor of a purely resistive circuit is 1.0.
Power = E I = (110) x (8) = 880 watts
The PF will increase
Power factor is:the ratio of true power to apparent powerthe ratio of resistance to impedancethe ratio of the voltage across a circuit's resistive component to the supply voltagethe cosine of the phase angleetc.
No. If the load is truly resistive, just measure the voltage across the load (in volts)and the current flowing in it (in amps) and multiply them. eg: 115 volts at 1.5 amps = 172.5 watts.
ratio between true power and apparent power is called the power factor for a circuit Power factor =true power/apparent power also we conclude PF=power dissipated / actual power in pure resistive circuit if total resistance is made zero power factor will be zero
With a pure resistive load the Power Factor should be 1.
real P= V * I *cos(phase angle between V and I)for purely resistive loads or DC voltages this equals real power P=V*I = 120*5= 600Wattsfor not pure resistive loads you'd have to measure the phase angle between Voltage and Current to get real power.However, at home, the utility company charges for Complex power = V*I.So you'd still pay for V*I.
When it supplies a resistive load.