- #1
fog37
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Hello Forum,
I am having issue understanding what is going on with the circuit below which is supposed to blink an LED by alternatively pulling the switch up and down which results in opening and closing the circuit the LED is part of (lower circuit portion).
This is what I am confused about: at time ##t=0##, there is a large initial current ##I(t_0)## passing through ##R2## resistor. At the top junction (positive terminal of the capacitor), the current ##I## splits and a portion of it goes to charge the capacitor ##C1## while the remaining current goes straight through the relay electromagnet which gets activated and pulls the switch upward turning the LED on.
But, simultaneously, this suddenly stops the current flow in the top part of the circuit where ##R2## and ##C2## are. That quick change in current through the electromagnet should causes an induced back voltage across the electromagnet, correct?
So is the purpose of the capacitor to prevent a spike due to that back emf? Or will the capacitor voltage, which will be less than ##9V## when the switch
pulled to the up position, keep the electromagnet energized for a little longer keeping the switch in the upper position and keeping the LED ON for a bit longer? Will the voltage across the capacitor be sufficiently large to activate the electromagnet?
Does the switch return to its lower position turning the LED off when the capacitor voltage has decreased enough and is too low to activate the electromagnet? I am little confused about the sequence of events...
thanks
I am having issue understanding what is going on with the circuit below which is supposed to blink an LED by alternatively pulling the switch up and down which results in opening and closing the circuit the LED is part of (lower circuit portion).
This is what I am confused about: at time ##t=0##, there is a large initial current ##I(t_0)## passing through ##R2## resistor. At the top junction (positive terminal of the capacitor), the current ##I## splits and a portion of it goes to charge the capacitor ##C1## while the remaining current goes straight through the relay electromagnet which gets activated and pulls the switch upward turning the LED on.
But, simultaneously, this suddenly stops the current flow in the top part of the circuit where ##R2## and ##C2## are. That quick change in current through the electromagnet should causes an induced back voltage across the electromagnet, correct?
So is the purpose of the capacitor to prevent a spike due to that back emf? Or will the capacitor voltage, which will be less than ##9V## when the switch
pulled to the up position, keep the electromagnet energized for a little longer keeping the switch in the upper position and keeping the LED ON for a bit longer? Will the voltage across the capacitor be sufficiently large to activate the electromagnet?
Does the switch return to its lower position turning the LED off when the capacitor voltage has decreased enough and is too low to activate the electromagnet? I am little confused about the sequence of events...
thanks
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