Five capacitors are connected as shown in the figure. Initially S is opened and all capacitors are uncharged. When S is closed, steady state is obtained. Then find out potential difference between M and N

Question

Khimraj · Answer

equivalent capacitance is ½ and EMF is 24 so charge through each capacitor = 12 C So volatge across 4 F is 3V and across 6 F is 2V So potential difference between M and N = 3 + 7+ 2 = 12V HOpe it clears.

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Five capacitors are connected as shown in the figure. Initially S is opened and all capacitors are uncharged. When S is closed, steady state is obtained. Then find out potential difference between M and N

Five capacitors are connected as shown in the figure. Initially S is opened and all capacitors are uncharged. When S is closed, steady state is obtained. Then find out potential difference between M and N

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Five capacitors are connected as shown in the figure. Initially S is opened and all capacitors are uncharged. When S is closed, steady state is obtained. Then find out potential difference between M and N

Five capacitors are connected as shown in the figure. Initially S is opened and all capacitors are uncharged. When S is closed, steady state is obtained. Then find out potential difference between M and N

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