In the figure a potential of $+$ $1200\, V$ is given to point $A$ and point $B$ is earthed, what is the potential at the point $P$....$V$
Medium
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(c) Given circuit can be reduced as follows
In series combination charge on each capacitor remain same. So using $Q = CV$
$==>$ ${C_1}{V_1} = {C_2}{V_2}$ $==>$ $3\,(1200 - {V_p}) = 6({V_P} - {V_B})$
$==>$ $1200 - {V_p} = 2{V_p}$
In series combination charge on each capacitor remain same. So using $Q = CV$
$==>$ ${C_1}{V_1} = {C_2}{V_2}$ $==>$ $3\,(1200 - {V_p}) = 6({V_P} - {V_B})$
$==>$ $1200 - {V_p} = 2{V_p}$
$==>$ ( ${V_B} = 0)$
$ 3V_p = 1200 ==> V_p = 400\, volt$
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