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What is the voltage across the inductor immediately after the switch is closed?
Before the switch is closed, there is no voltage or current across either the resistor or the inductor. When the switch is first closed, the current through the inductor is zero, because it cannot change instantaneously. This means that the inductor acts like an open circuit, so all the voltage is across the inductor.
Why is the switch closed for a short period of time?
Closing the switch completes the conductive path in this flashlight, allowing electrons to flow. Sometimes open circuits are created by accident. A short circuit is a direct connection between two points in a circuit that aren’t supposed to be directly connected, such as the two terminals of a power supply.
Why does an open switch have voltage?
The switch is open. This means, the charges could not flow through the circuit and complete the circuit. This is due to the gap in between the switch. So the applied energy is not converted to current even though the resistance is still there.
When is the voltage across the inductor zero?
No because at t = 0, there is no current flowing through the inductor but there is a current flowing through both resistors. Extra credit: if the switch is closed for a long time, the voltage accross the inductor is 0.
What happens to the voltage when the switch is closed?
Extra credit: if the switch is closed for a long time, the voltage accross the inductor is 0. Hence no current is flowing through the 3kOhm resistor, and the current flowing through the inductor is 5/2k = 2.5mA. The switch is opened: the current flowing through the inductor is still 2.5mA.
Why does an inductor not like a change in current?
Inductors do not like changes in current, since a change in current means the magnetic field linking the inductor is changing and this generates a back emf that opposes the change.
What is the current flowing through the 3kohm resistor?
Hence no current is flowing through the 3kOhm resistor, and the current flowing through the inductor is 5/2k = 2.5mA. The switch is opened: the current flowing through the inductor is still 2.5mA. Because the switch is open, this is also the current flowing through the 3kOhm resistor. Hence the voltage accross the inductor is – (3k*2.5mA) = -7.5V.