Inductive circuits

[Jk1996]

I’ve been trying to understand inductive loads and voltage drop. I though if voltage drops on an inductive circuit the current will also decrease. Is this correct? I understand if on phase of a 3 phase motor circuit one phase is high resistance this causes an imbalance causing the other two phases to draw excessive current due to the back EMF becoming smaller. Any clarification will be greatly appreciated.

Thanks again guys

 

[ETechnoG]

Yes, you are right, voltage drop in one phase can unbalance in other two phases.

 

[Jk1996]

But what happens if voltage drops in all phases equally?

 

[crshears]

Depends on what's causing the voltage drop . . .

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[bacon4life]

With respect to small voltage changes, loads are usually modeled as a combination of constant impedance, constant current, and constant power elements. Typical induction motor loads are modeled as constant power. For constant power loads, the current goes up when the voltage is reduced.

[tangent warning] Constant power loads provide destabilizing feedback because low voltage causes higher currents. Over the last decade there has been a huge shift from incandescent (i.e. constant impedance) lighting over to LED lighting that operates as constant power. This shift potentially leaves the grid more vulnerable to voltage collapse.

Although induction motors are typically modeled as constant power with respect to voltage, typical induction motors do provide damping with respect to frequency changes. Motor loads typically respond to declines in system frequency by slightly reducing power consumption. This feedback helps to achieve a stable balance between load and generation. When motors are put behind VFDs, the load become constant power with respect to both voltage and frequency. As more loads are put onto VFDs, controlling the frequency of the grid becomes harder.
[/tangent]

 

[davidbeach]

Not just LED lighting, but anything with some form of switchmode power supply. Motors with VFDs are a great example. Go back 10-15 years to when there was lots of electric resistive heating and compare that heat pumps with VFDs. Even older air conditioners had motors that were across the line and now they're on drives.

I’ll see your silver lining and raise you two black clouds. - Protection Operations

 

[Jk1996]

But say if a motors voltage dropped from 415v across lines to 300v dues to something like a contactor fault would this not increase the impedance of the circuit (high resistance at contactor) so the current would be reduced?

 

[bacon4life]

Keep in mind that the power needs of the load attached to the motor does not change just because the input voltage changed. As a first order approximation ignoring loss and changes in slip, dropping the voltage to 0.70 per unit increases the current to 1/0.7= 1.43 per unit. Resistive losses in the motor are proportional to current squared, so these losses would approximately double and likely motor burn out.

Dropping 115 volts across a motor contactor would lead to a very large amount of heat to dissipate within the contactor. Typically a "high resistant" in a contactor/joint/spice just means high in comparison to the expected value. For example, a joint might have 0.001 ohm of resistance instead of the expected 0.0001 ohm. This would increase the heat generated by the joint by 10x, but would have negligible impact on the rest of the circuit.

On the original question about a single phase with low voltage, the explanation of reduced back EMF is misleading. An alterative way to analyze imbalanced voltages is by splitting the voltages into positive and negative sequence components. Unbalanced voltages produce a negative sequence a torque that rotates opposite to the direction of the motor. The motor then draws an additional amount of positive sequence current to offset the negative sequence torque generated by the unbalanced voltage.

As a thought experiment, picture two identical motors attached to the same shaft. Apply normal voltage to the first motor. For the second motor, swap two of the phases so it tries to drive the shaft the opposite direct. For the second motor, apply 2% of rated voltage. The second motor will attempt to slow down the shaft, and this setup would lead to significant increase in total current feeding the two motors.

 

[Jk1996]

I’m really trying to understand this. How does the added resistance in the circuit effect the current?