@24.29 should’ve 2c x Vx = C (1v ) + C (1v) since total charge is addition of charge across both caps because they are in parallel and charge across them is different so we add them ?
No, initially the switch is open so the entire voltage is across the first capacitor, so Qinitial = CV, then the switch was closed and total charge in the ckt = Q1 + Q2 = V * (C1 + C2) And by charge conservation, Qini = Qfin => CV = (2C)*Vx => Vx = CV/2 They're both in parallel so voltage is equal across both but charges depends on the capacitances.
at 27:52, why is voltage across the equivalent "2C" capacitor even changing? Assume u just replaced the 2, C capacitors with a single 2C capacitor, keeping the voltage source constant as 1V, then we shall have energy conservation of the system.
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Thankyou sir, being a student from tire 3 college I benefit a lot from your videoes
very intuitive video and has a wealth of information
Genuinely, thank you so much.
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@24.29 should’ve 2c x Vx = C (1v ) + C (1v) since total charge is addition of charge across both caps because they are in parallel and charge across them is different so we add them ?
No, initially the switch is open so the entire voltage is across the first capacitor, so Qinitial = CV,
then the switch was closed and total charge in the ckt = Q1 + Q2 = V * (C1 + C2)
And by charge conservation, Qini = Qfin => CV = (2C)*Vx => Vx = CV/2
They're both in parallel so voltage is equal across both but charges depends on the capacitances.
@@nabhay583 but cant the battery provide more charge for the second capacitor since its still connected?
@@nabhay583 but shouldnt the overall voltage = 1v and not 1/2v??
Thank u sir
at 27:52, why is voltage across the equivalent "2C" capacitor even changing? Assume u just replaced the 2, C capacitors with a single 2C capacitor, keeping the voltage source constant as 1V, then we shall have energy conservation of the system.
why did I find this video so late!!!!
❤
25:13