Hi Meg. When the switch is closed the inductor produces a voltage (back emf) that opposes the increase of current. This inductor voltage is proportional to current change. Current change is at a maximum initially, therefore so too is the inductor voltage (specifically 12V). The current change reduces to zero over time, and therefore so too does the inductor voltage. If we consider Kirchoff's voltage law, we know that the inductor and lamp voltage must add to our source voltage of 12V. If the inductor voltage is initially 12V, the lamp voltage must initially be 0V. Hope this helps!
Hi Yaya, the difference is whether you pull the negative out of VC or leave it contained in the variable. For example 5-3 vs 5+(-3). Both ways are perfectly fine. At higher levels we tend more towards the latter way, as it's valid for both VL
Using part of my response to an earlier question: The voltage after 1 time constant is 12V x 0.63 = 7.56V. After 2 time constants the voltage will increase by 63% of the difference between 7.56V and max voltage 12V, which is calculated as 0.63(12V - 7.56V). Therefore the voltage after 2 time constants will be 7.56V + 0.63(12V - 7.56V).
just a question about 3d, how come we subtract the 3.5 ohm resistor but add the 2.9 ohm resistor in the second loop? I understand that we must subtract 2.9 ohm resistor as it the current is moving in the opposite direction, but why do we not do the same for the IR?
@@vincentball-lahood1885 Aha. As a general rule we add for voltage sources and subtract for resistors, unless the voltage source opposes the current or the current flows in the opposite direction, then we do the opposite. Let's go through the components in order: The 6V voltage source is working against the current so instead of adding, we do the opposite and subtract. Consider how an incoming positive charge would be repelled by the positive terminal when the voltage source is in this orientation, subtracting energy from the flow. We subtract the 3.5ohm resistor voltage as per usual, no funny business here. Rather than subtract the 2.9ohm resistor voltage, we do the opposite and add it. We do this because the current is traveling in the opposite direction on this branch of the circuit. Hope this helps!
Good question, the voltage after 1 time constant is 7.56V. After 2 time constants the voltage will increase by 63% of the difference between 7.56V and max voltage 12V, which is calculated as 0.63(12V - 7.56V). Therefore the voltage after 2 time constants will be 7.56V + 0.63(12V - 7.56V). Hope that helps!
Excellent question! Either way works. If you choose to do it the other way the terms you added will be subtracted and the terms you subtracted will be added. With a bit of rearranging you'll wind up with the same equation regardless.
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Babe wake up Mr Whitley posted
Your videos have really come in handy, my school doesn't teach electricity, along with most my papers, Thank you very much
You're welcome Twistsy, these papers are tough even with a teacher! I'm glad these videos have been useful, let me know if you have any questions.
Thank you for the help with the videos over the years Sir, these videos have been immensely useful and much needed :)
Thanks Nina, that's great to hear.
for 2bi, how do we know the voltage starts at 0 and how do we know that its not decreasing exponentially?
Hi Meg.
When the switch is closed the inductor produces a voltage (back emf) that opposes the increase of current.
This inductor voltage is proportional to current change. Current change is at a maximum initially, therefore so too is the inductor voltage (specifically 12V). The current change reduces to zero over time, and therefore so too does the inductor voltage.
If we consider Kirchoff's voltage law, we know that the inductor and lamp voltage must add to our source voltage of 12V. If the inductor voltage is initially 12V, the lamp voltage must initially be 0V.
Hope this helps!
For 1d, shouldn't the opposite side of the triangle be VL -Vc? I'm wondering why you did it as Vc + VL and how you justify it.
Hi Yaya, the difference is whether you pull the negative out of VC or leave it contained in the variable. For example 5-3 vs 5+(-3). Both ways are perfectly fine.
At higher levels we tend more towards the latter way, as it's valid for both VL
@@MrWhibleyPhysics Ok I see so it’s a matter of convenience. Cheers sir, wish me luck 🙏
You're welcome Yaya, good luck!
For question 2bi, why did you choose to times 0.63 by 12 and not other numbers?
Using part of my response to an earlier question: The voltage after 1 time constant is 12V x 0.63 = 7.56V. After 2 time constants the voltage will increase by 63% of the difference between 7.56V and max voltage 12V, which is calculated as 0.63(12V - 7.56V). Therefore the voltage after 2 time constants will be 7.56V + 0.63(12V - 7.56V).
Hi sir just wondering, is back emf the same as induced voltage, many thanks.
Hi Jonathan, yes it is.
@@MrWhibleyPhysics 👍👍Cheers, sir.
just a question about 3d, how come we subtract the 3.5 ohm resistor but add the 2.9 ohm resistor in the second loop? I understand that we must subtract 2.9 ohm resistor as it the current is moving in the opposite direction, but why do we not do the same for the IR?
Hi Vincent, what do you mean by IR?
@@MrWhibleyPhysics the internal resistance of the 6v battery
@@vincentball-lahood1885 Aha. As a general rule we add for voltage sources and subtract for resistors, unless the voltage source opposes the current or the current flows in the opposite direction, then we do the opposite.
Let's go through the components in order:
The 6V voltage source is working against the current so instead of adding, we do the opposite and subtract. Consider how an incoming positive charge would be repelled by the positive terminal when the voltage source is in this orientation, subtracting energy from the flow.
We subtract the 3.5ohm resistor voltage as per usual, no funny business here.
Rather than subtract the 2.9ohm resistor voltage, we do the opposite and add it. We do this because the current is traveling in the opposite direction on this branch of the circuit. Hope this helps!
@@MrWhibleyPhysics thank you, very helpful 😊
For question 2 bi , how do you find the 2nd time constant, Iam confused about the method you used?
Good question, the voltage after 1 time constant is 7.56V. After 2 time constants the voltage will increase by 63% of the difference between 7.56V and max voltage 12V, which is calculated as 0.63(12V - 7.56V). Therefore the voltage after 2 time constants will be 7.56V + 0.63(12V - 7.56V). Hope that helps!
@@MrWhibleyPhysics Thank you that helped alot
For 3d how do we know which way the loop goes or does it not matter.
Excellent question! Either way works. If you choose to do it the other way the terms you added will be subtracted and the terms you subtracted will be added. With a bit of rearranging you'll wind up with the same equation regardless.
you're a king sir very helpful video
Cheers JJ Duck.
incredibly useful video. Thank you!!!
Cheers GalactikNZ, glad you found it helpful!
excellent video as always!
Thanks Stefan! Fond memories I'm sure.
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