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Electronics Help

 Posts: 12,788 Add as FriendChallenge to a DebateSend a Message 10/11/2013 10:03:52 AMPosted: 4 years agoLet's assume that you have two equivalent capacitors connected in parallel to a 9v battery supply.Question 1:If you let them charge, will each capacitor have 9v potential difference?Question 2:Does the current released from both of these two capacitors act as if they were two parallel batteries? For example, if the voltage on both is 9v and there are X amp hours, the amp hours would be doubled, instead of the voltage.I know that capacitance increases in parallel, but I fail to see how that has anything to do with this."Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 10/11/2013 10:27:45 AMPosted: 4 years ago1. Yes2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.
 Posts: 12,788 Add as FriendChallenge to a DebateSend a Message 10/11/2013 10:49:27 AMPosted: 4 years agoAt 10/11/2013 10:27:45 AM, Enji wrote:1. Yes2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
 Posts: 12,788 Add as FriendChallenge to a DebateSend a Message 10/11/2013 10:50:38 AMPosted: 4 years agoOr can the capacitors only output at the input current?"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:00:38 AMPosted: 4 years agoAt 10/11/2013 10:49:27 AM, Lordknukle wrote:At 10/11/2013 10:27:45 AM, Enji wrote:1. Yes2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.
 Posts: 12,788 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:10:16 AMPosted: 4 years agoAt 10/11/2013 11:00:38 AM, Enji wrote:At 10/11/2013 10:49:27 AM, Lordknukle wrote:At 10/11/2013 10:27:45 AM, Enji wrote:1. Yes2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.So if I want my capacitors to produce 18v, I'd connect them in series; if I want them to produce 2x amp/hours I'd connect them in parallel?"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
 Posts: 12,788 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:14:26 AMPosted: 4 years agoAlso, just one more question, what exactly is the point of capacitance ratings? It seems as if you can get all the information you need by knowing the output voltage and the nature of the connection."Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
 Posts: 18,870 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:15:31 AMPosted: 4 years agoGo here:http://www.falstad.com...When the app loads (new window) go to: File > ImportPaste:\$ 1 5.0E-6 10.20027730826997 50 5.0 43v 464 256 464 320 0 0 40.0 9.0 0.0 0.0 0.5w 464 208 512 208 0w 464 160 512 160 0c 512 160 576 160 0 9.999999999999999E-6 -8.999999999998936c 512 208 576 208 0 1.0E-5 -8.999999999999291w 576 160 624 160 0w 576 208 624 208 0w 624 160 624 208 0w 624 208 624 368 0w 624 368 464 368 0w 464 368 464 320 0r 464 256 464 208 0 100.0r 464 208 464 160 0 100.0o 0 64 0 35 10.0 9.765625E-5 0 -1o 4 64 0 35 10.0 9.765625E-5 1 -1o 3 64 0 35 10.0 9.765625E-5 2 -1Click "Import"
 Posts: 18,870 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:16:14 AMPosted: 4 years agoHad to add some resistors but the answer to the first question seems to be yes. Not sure how to test the second.
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:25:07 AMPosted: 4 years agoAt 10/11/2013 11:10:16 AM, Lordknukle wrote:At 10/11/2013 11:00:38 AM, Enji wrote:At 10/11/2013 10:49:27 AM, Lordknukle wrote:At 10/11/2013 10:27:45 AM, Enji wrote:1. Yes2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.So if I want my capacitors to produce 18v, I'd connect them in series; if I want them to produce 2x amp/hours I'd connect them in parallel?No, if you had a 9V battery and equivalent capacitors in series, then the total voltage across both capacitors would still be 9V; the sum of the voltages across the capacitors equals the total voltage in series [http://en.wikipedia.org...].
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:27:57 AMPosted: 4 years agoAt 10/11/2013 11:14:26 AM, Lordknukle wrote:Also, just one more question, what exactly is the point of capacitance ratings? It seems as if you can get all the information you need by knowing the output voltage and the nature of the connection.The discharge voltage varies with time, and the rate of decay depends upon the capacitance.
 Posts: 12,788 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:33:37 AMPosted: 4 years agoAt 10/11/2013 11:25:07 AM, Enji wrote:At 10/11/2013 11:10:16 AM, Lordknukle wrote:At 10/11/2013 11:00:38 AM, Enji wrote:At 10/11/2013 10:49:27 AM, Lordknukle wrote:At 10/11/2013 10:27:45 AM, Enji wrote:1. Yes2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.So if I want my capacitors to produce 18v, I'd connect them in series; if I want them to produce 2x amp/hours I'd connect them in parallel?No, if you had a 9V battery and equivalent capacitors in series, then the total voltage across both capacitors would still be 9V; the sum of the voltages across the capacitors equals the total voltage in series [http://en.wikipedia.org...].Ok. So I would get the 18v output with a parallel circuit?"Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
 Posts: 12,788 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:35:25 AMPosted: 4 years agoAt 10/11/2013 11:15:31 AM, drafterman wrote:Go here:http://www.falstad.com...When the app loads (new window) go to: File > ImportPaste:\$ 1 5.0E-6 10.20027730826997 50 5.0 43v 464 256 464 320 0 0 40.0 9.0 0.0 0.0 0.5w 464 208 512 208 0w 464 160 512 160 0c 512 160 576 160 0 9.999999999999999E-6 -8.999999999998936c 512 208 576 208 0 1.0E-5 -8.999999999999291w 576 160 624 160 0w 576 208 624 208 0w 624 160 624 208 0w 624 208 624 368 0w 624 368 464 368 0w 464 368 464 320 0r 464 256 464 208 0 100.0r 464 208 464 160 0 100.0o 0 64 0 35 10.0 9.765625E-5 0 -1o 4 64 0 35 10.0 9.765625E-5 1 -1o 3 64 0 35 10.0 9.765625E-5 2 -1Click "Import"Won't let me c/p."Easy is the descent to Avernus, for the door to the Underworld lies upon both day and night. But to retrace your steps and return to the breezes above- that's the task, that's the toil."
 Posts: 1,023 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:36:14 AMPosted: 4 years agoAt 10/11/2013 11:33:37 AM, Lordknukle wrote:At 10/11/2013 11:25:07 AM, Enji wrote:At 10/11/2013 11:10:16 AM, Lordknukle wrote:At 10/11/2013 11:00:38 AM, Enji wrote:At 10/11/2013 10:49:27 AM, Lordknukle wrote:At 10/11/2013 10:27:45 AM, Enji wrote:1. Yes2. Remember Kirchhoff's Rules - at any branch point the current in equals the current out. So if the current in the input wire is I_t, then the sum of the currents across each capacitor must equal I_t. Since the two capacitors are equivalent, the current across each is the same, which means that each has half of the total current so the total current remains the same.Right, but when you disconnect the power source, do the capacitors send electricity at 18v and x amps, or 9v and 2x amps? Would this be different if they were connected in series?The current of the output wire would equal the sum of the currents of the two capacitors, so you'd get 9v and 2x amps where x is the current across one capacitor. This would be different if you connected the capacitors in series; the sum of the voltages across each capacitor would equal the voltage of the battery and the current would be the same across each component.So if I want my capacitors to produce 18v, I'd connect them in series; if I want them to produce 2x amp/hours I'd connect them in parallel?No, if you had a 9V battery and equivalent capacitors in series, then the total voltage across both capacitors would still be 9V; the sum of the voltages across the capacitors equals the total voltage in series [http://en.wikipedia.org...].Ok. So I would get the 18v output with a parallel circuit?In parallel the voltage across each capacitor would be the same as the voltage of the output wire.
 Posts: 18,870 Add as FriendChallenge to a DebateSend a Message 10/11/2013 11:38:30 AMPosted: 4 years agoAt 10/11/2013 11:35:25 AM, Lordknukle wrote:At 10/11/2013 11:15:31 AM, drafterman wrote:Go here:http://www.falstad.com...When the app loads (new window) go to: File > ImportPaste:\$ 1 5.0E-6 10.20027730826997 50 5.0 43v 464 256 464 320 0 0 40.0 9.0 0.0 0.0 0.5w 464 208 512 208 0w 464 160 512 160 0c 512 160 576 160 0 9.999999999999999E-6 -8.999999999998936c 512 208 576 208 0 1.0E-5 -8.999999999999291w 576 160 624 160 0w 576 208 624 208 0w 624 160 624 208 0w 624 208 624 368 0w 624 368 464 368 0w 464 368 464 320 0r 464 256 464 208 0 100.0r 464 208 464 160 0 100.0o 0 64 0 35 10.0 9.765625E-5 0 -1o 4 64 0 35 10.0 9.765625E-5 1 -1o 3 64 0 35 10.0 9.765625E-5 2 -1Click "Import"Won't let me c/p.You can build it manually. Right click on the workspace to add components. RIght click on a component to add it to the scope for monitoring.