OK you're looking for a graph that shows the capacitor discharging from the instant the switch is opened and then charging up to V when the switch is closed a short time later.
the discharge of the capacitor will take place through the series resistor shown in the diagram so you're looking for the characteristic shape of a RC circuit discharging on the left of the graph.
the question includes the phrase 'short time' which may be an indication that the capacitor doesn't have enough time to discharge fully.
the switch is then closed and the capacitor will charge up, the battery has negligable internal resistance so there is effectively no series resistance in the charging circuit... so the right of the graph will look like a capacitor charging up to full V in essentially no time at all.
i get the entire explanation
but doesn't charging up graphs look exponential in shape which is why i thought D should be the answer
but ofc the time is short so can not discharge fully but still why isnt the shape exponential
also during charging there is still a resistor in the circuit..i dont get the part where you say that there is no series resistance in the charging circuit
but doesn't charging up graphs look exponential in shape which is why i thought D should be the answer
but ofc the time is short so can not discharge fully but still why isnt the shape exponential
also during charging there is still a resistor in the circuit..i dont get the part where you say that there is no series resistance in the charging circuit
Original post by MARYAM1234567
i get the entire explanation
but doesn't charging up graphs look exponential in shape which is why i thought D should be the answer
but ofc the time is short so can not discharge fully but still why isnt the shape exponential
also during charging there is still a resistor in the circuit..i dont get the part where you say that there is no series resistance in the charging circuit
but doesn't charging up graphs look exponential in shape which is why i thought D should be the answer
but ofc the time is short so can not discharge fully but still why isnt the shape exponential
also during charging there is still a resistor in the circuit..i dont get the part where you say that there is no series resistance in the charging circuit
When the capacitor is discharging, because the switch is open, the resistor is effectively in series. So the shape is exponential on the left hand side of the graph.
And when charging, you have to consider the separate 'loops' of the circuit - there's one loop that contains the resistor, and one that doesn't. So some of the charge goes straight to the capacitor, bypassing the resistor.
Original post by Joinedup
OK you're looking for a graph that shows the capacitor discharging from the instant the switch is opened and then charging up to V when the switch is closed a short time later.
the discharge of the capacitor will take place through the series resistor shown in the diagram so you're looking for the characteristic shape of a RC circuit discharging on the left of the graph.
the question includes the phrase 'short time' which may be an indication that the capacitor doesn't have enough time to discharge fully.
the switch is then closed and the capacitor will charge up, the battery has negligable internal resistance so there is effectively no series resistance in the charging circuit... so the right of the graph will look like a capacitor charging up to full V in essentially no time at all.
the discharge of the capacitor will take place through the series resistor shown in the diagram so you're looking for the characteristic shape of a RC circuit discharging on the left of the graph.
the question includes the phrase 'short time' which may be an indication that the capacitor doesn't have enough time to discharge fully.
the switch is then closed and the capacitor will charge up, the battery has negligable internal resistance so there is effectively no series resistance in the charging circuit... so the right of the graph will look like a capacitor charging up to full V in essentially no time at all.
hello??
Original post by MARYAM1234567
hello??
The resistor isn't in series with the capacitor during charging so it's not having any limiting effect on the current going into the capacitor...
if you did a real world experiment with this circuit you'd actually get a *very rapid* exponential charging of the capacitor (that would probably look instantaneous on an oscilloscope) and this is because real world batteries always have some small internal resistance and real world copper wires always have some small resistance... and that'd be the resistance in your RC circuit diring charging.
in the real world version you'd be looking for an exponential discharge curve with a fairly high time constant, probably not reaching zero, followed by an exponential charging curve where the time constant is so small it'd probably look vertical.
Original post by Joinedup
The resistor isn't in series with the capacitor during charging so it's not having any limiting effect on the current going into the capacitor...
if you did a real world experiment with this circuit you'd actually get a *very rapid* exponential charging of the capacitor (that would probably look instantaneous on an oscilloscope) and this is because real world batteries always have some small internal resistance and real world copper wires always have some small resistance... and that'd be the resistance in your RC circuit diring charging.
in the real world version you'd be looking for an exponential discharge curve with a fairly high time constant, probably not reaching zero, followed by an exponential charging curve where the time constant is so small it'd probably look vertical.
if you did a real world experiment with this circuit you'd actually get a *very rapid* exponential charging of the capacitor (that would probably look instantaneous on an oscilloscope) and this is because real world batteries always have some small internal resistance and real world copper wires always have some small resistance... and that'd be the resistance in your RC circuit diring charging.
in the real world version you'd be looking for an exponential discharge curve with a fairly high time constant, probably not reaching zero, followed by an exponential charging curve where the time constant is so small it'd probably look vertical.
oh right! thanks so much
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