Troubleshooting a Complex Circuit Problem

[Gott_ist_tot]

I have a circuit problem. There is a 30 V power source conected in series with a switch and a ten ohm resistor. The circuit then spling to an inductor of 10 mH and a 20 ohm resistor. I am having difficulty finding the current across the 20 ohm resistor. I got the current immediately after the switch closes as 1 A. Then a long time after the switch closed the current across the 20 ohm resistor is 0 A. What is the current when the switch is reopened. The circuit is now an LR circuit L = 10mH and R = 20 ohms. by V=IR I thought that the current across the resistor would be 1.5 A. However, it is not. Is V of the inductor different than the power supply? If so, I con't find a rule governing that. Thanks for any help.

 

[lightgrav]

Just before the switch closes, the current through the inductor is 0 Amp.
The current after the switch closes is 1 Amp, as you calculated.
So the current through R20 = 1 Amp (at t=0, Current through L is zero).
What is the Voltage across R20 at t=0? _____
So, what is the Voltage across the inductor at t=0? _____
What is the rate of change in Current through L at t=0? _____

After a long time, the rate of change in Current through L becomes zero.
This does NOT imply that I through L is zero, only that it is constant.
So the Voltage across L is zero ... What Current goes through R10? ____
What current was going through the Inductor just before switch opens?

As the switch opens, there's no longer any power supply in the circuit.
(How would the Inductor Voltage be equal to some dis-connected thing?)
But the Inductor Current has not changed from the I just before opening.
What is the current through R20 now? ____
So, what is the Voltage across the inductor? _____
So, what is the rate of change in Current thru inductor? ___

 

[kyle8921]

I would first suggest double checking your calculations and ensuring that all components in the circuit are connected correctly. It is also important to consider any possible sources of error, such as faulty equipment or incorrect measurements.

In regards to your question about the current when the switch is reopened, it is important to understand the behavior of an LR circuit. When the switch is closed, the inductor initially acts as a short circuit, allowing the full 30 V to pass through the circuit. As the current begins to flow, the inductor builds up a magnetic field, causing the current to decrease. This is why you are seeing a decrease in current over time.

When the switch is reopened, the inductor will try to maintain the current flow, causing a brief spike in voltage across the inductor. This voltage spike can be calculated using the equation V = L di/dt, where L is the inductance and di/dt is the change in current over time. This spike in voltage can cause the current to briefly increase before eventually decreasing to 0 A.

In this case, the current across the 20 ohm resistor will not be 1.5 A when the switch is reopened. It will depend on the values of the inductance and the time since the switch was opened, as well as any other components in the circuit. It is important to consider the behavior of each component in the circuit and how they interact with each other.

In regards to your question about the voltage across the inductor, it is important to note that the voltage across any component in a series circuit is the same. However, the current through each component may be different. In this case, the voltage across the inductor may be different than the power supply voltage, but the current through both components should be the same.

I recommend further research and experimentation to fully understand the behavior of an LR circuit and how to calculate the current in this specific scenario. It may also be helpful to consult with other experts in the field or refer to reliable resources for additional guidance. Good luck with your troubleshooting!