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kanwal32
 one year ago
Two capacitors of 2uf and 3uf are charge 150 volt and 120 volt respectively. The plates of capacitor are connected as shown in the figure. A discharged capacitor of capacity 1.5uF falls to the free ends of the wire
kanwal32
 one year ago
Two capacitors of 2uf and 3uf are charge 150 volt and 120 volt respectively. The plates of capacitor are connected as shown in the figure. A discharged capacitor of capacity 1.5uF falls to the free ends of the wire

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kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434697500095:dw

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0@radar @IrishBoy123 @

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, when you complete the circuit, the current will flow until the voltages have balanced out. I'm assuming you want the steady state voltages? Essentially, if the voltages weren't equal in the steady state, there would be a potential difference. If there was a potential difference (especially in this resistance free circuit), current would flow and the voltage would balance back out.. So in this closed circuit there is no energy dissipation, and charge cannot be created or destroyed. So you know the total charge based on the given initial values, and you know that the final voltages are all equal. That should be enough to get you going :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0when we connect the two capacitors C_1 and C_2, a charge motion will occur between them, in order to establish a equilibrium potential V_e across each of them. So we can write: before connecting C_1 and C_2 , we have: \[\begin{gathered} {Q_1} = {C_1}{V_1} \hfill \\ {Q_2} = {C_2}{V_2} \hfill \\ \end{gathered} \] after connecting C_1 and C_2 together, we have: \[\begin{gathered} Q{'_1} = {C_1}{V_e} \hfill \\ Q{'_2} = {C_2}{V_e} \hfill \\ \end{gathered} \] now, since electric charge is conserved, we can write: \[{Q_1} + {Q_2} = Q{'_1} + Q{'_2}\] from which we get: \[{V_e} = \frac{{{C_1}{V_1} + {C_2}{V_2}}}{{{C_1} + {C_2}}}\]

radar
 one year ago
Best ResponseYou've already chosen the best response.0Need more info...... Is the problem what will be the voltage or charge on each capacitor when the uncharged 1.5 uF capacitor is connected to A and B ??

radar
 one year ago
Best ResponseYou've already chosen the best response.0Note that the voltage between A & B is 270 volts before any connection is made.

radar
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434718296697:dwIs the problem determine the voltages across each capacitor after the switches are closed connecting the 1.5uF capacitor in series with the other two series connected capacitors?

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0charge on 1.5uf capacitor charge on2uf and direction of flow of conductor

radar
 one year ago
Best ResponseYou've already chosen the best response.0First determine the present charge on the 2 and 3 uF capacitors. Q on 2Uf, Q=VC= 150 X 2UF=300 Micro Coulomb Q on 3uF, Q=VC = 120 X 3uF = 360 micro coulomg. I Have to run an errand will be back.

radar
 one year ago
Best ResponseYou've already chosen the best response.0The 1.5uF capacitor will have 0 charge, as it has 0 volts across it. Now the 1.5 uF capacitor is placed across points A and B. There will be a redistribution of both charge and voltage among the the three series capacitors. I would use the distribution of charge rather than voltage to solve this. Let us know how you make out.

radar
 one year ago
Best ResponseYou've already chosen the best response.0The total charge of 660 micro coulombs will redistribute in accordance with the rule for series connected capacitors, this means the charge Q for each will be 220 micro coulombs The voltage would distribute as follows: 1.5 uF =146 Volts the highest 2 uF = 110 volts (middle) 3 uF = 73 1/3 volts lowest Total approx 330 volts. Would like this to be reviewed by others and make any corrections needed.
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